E Buforn_Etal-Solved Problems in Geophysics-Cambridge University Press (2012)

Solved Problems in Geophysics

Solving problems is an indispensable exercise for mastering the theory underlying the

various branches of geophysics. Without this practice, students often find it hard to

understand and relate theoretical concepts to their application in real-world situations.

This book is a collection of nearly 200 problems in geophysics, which are solved in

detail showing each step of their solution, the equations used and the assumptions made.

Simple figures are also included to help students understand how to reduce a problem to its

key elements. The book begins with an introduction to the equations most commonly used

in solving geophysical problems. The subsequent four chapters then present a series of

exercises for each of the main, classical areas of geophysics – gravity, geomagnetism,

seismology and heat flow and geochronology. For each topic there are problems with

different degrees of difficulty, from simple exercises that can be used in the most elemen-

tary courses, to more complex problems suitable for graduate-level students.

This handy book is the ideal adjunct to core course textbooks on geophysical theory. It is

a convenient source of additional homework and exam questions for instructors, and

provides students with step-by-step examples that can be used as a practice or revision aid.

Elisa Buforn is a Professor of geophysics at the Universidad Complutense de Madrid (UCM)

where she teaches courses on geophysics, seismology, physics, and numerical methods.

Professor Buforn’s research focuses on source fracture processes, seismicity, and seismo-

tectonics, and she is Editor in Chief of Física de la Tierra and on the Editorial Board of the

Journal of Seismology.

Carmen Pro is an Associate Professor at the University of Extremadura, Spain, where she has

taught geophysics and astronomy for over 20 years. She has participated in several

geophysical research projects and is involved in college management.

Agustín Udías is an Emeritus Professor at UCM and is the author of a large number of papers

about seismicity, seismotectonics, and the physics of seismic sources, as well as the

textbook Principles of Seismology (Cambridge University Press, 1999). He has held

positions as Editor in Chief of Física de la Tierra and the Journal of Seismology and as

Vice President of the European Seismological Commission.

Solved Problems in Geophysics

ELISA BUFORNUniversidad Complutense, Madrid

CARMEN PROUniversidad de Extremadura, Spain

AGUSTÍN UDÍASUniversidad Complutense, Madrid

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# Elisa Buforn, Carmen Pro and Agustín Udías 2012

This publication is in copyright. Subject to statutory exception

and to the provisions of relevant collective licensing agreements,

no reproduction of any part may take place without

the written permission of Cambridge University Press.

First published 2012

Printed in the United Kingdom at the University Press, Cambridge

A catalogue record for this publication is available from the British Library

Library of Congress Cataloging-in-Publication Data

Buforn, E.

Solved problems in geophysics / Elisa Buforn, Carmen Pro, Agustín Udías.

p. cm.

Includes bibliographical references.

ISBN 978-1-107-60271-7 (Paperback)

1. Geophysics–Problems, exercises, etc. I. Pro, Carmen. II. Udías Vallina, Agustín. III. Title.

QC807.52.B84 2012

550.78–dc23

2011046101

ISBN 978-1-107-60271-7 Paperback

Cambridge University Press has no responsibility for the persistence or

accuracy of URLs for external or third-party internet websites referred to

in this publication, and does not guarantee that any content on such

websites is, or will remain, accurate or appropriate.

Contents

Preface page vii

1 Introduction 1

Gravity 1

Geomagnetism 4

Seismology 6

Heat flow 10

Geochronology 11

2 Gravity 13

Terrestrial geoid and ellipsoid 13

Earth’s gravity field and potential 25

Gravity anomalies. Isostasy 53

Tides 95

Gravity observations 116

3 Geomagnetism 121

Main field 121

Magnetic anomalies 142

External magnetic field 156

Main (internal), external, and anomalous magnetic fields 174

Paleomagnetism 201

4 Seismology 208

Elasticity 208

Wave propagation. Potentials and displacements 211

Reflection and refraction 224

Ray theory. Constant and variable velocity 243

Ray theory. Spherical media 277

Surface waves 307

Focal parameters 324

5 Heat flow and geochronology 335

Heat flow 335

Geochronology 345

Bibliography 352

v

Preface

This book presents a collection of 197 solved problems in geophysics. Our teaching

experience has shown us that there was a need for a work of this kind. Solving problems

is an indispensable exercise for understanding the theory contained in the various branches

of geophysics. Without this exercise, the student often finds it hard to understand and relate

the theoretical concepts with their application to practical cases. Although most teachers

present exercises and problems for their students during the course, the hours allotted to the

subject significantly limit how many exercises can be worked through in class. Although

the students may try to solve other problems outside of class time, if there are no solutions

available this significantly reduces the effectiveness of this type of study. It helps, there-

fore, both for the student and for the teacher who is explaining the subject if they have

problems whose solutions are given and whose steps can be followed in detail. Some

geophysics textbooks, for example, F.D. Stacey, Physics of the Earth; G.D. Garland,

Introduction to Geophysics; C.M. Fowler, The Solid Earth: An Introduction to Global

Geophysics; and W. Lowrie, Fundamentals of Geophysics, contain example problems, and,

in the case of Stacey’s, Fowler’s, and Lowrie’s textbooks, their solutions are provided on

the website of Cambridge University Press. The main difference in the present text is the

type of problems and the detail with which the solutions are given, and in the much greater

number.

All the problems proposed in the book are solved in detail, showing each step of their

solution, the equations used, and the assumptions made, so that their solution can be

followed without consulting any other book. When necessary, and indeed quite often, we

also include figures that allow the problems to be more clearly understood. For a given

topic, there are problems with different degrees of difficulty, from simple exercises that can

be used in the most elementary courses, to more complex problems with greater difficulty

and more suitable for teaching at a more advanced level.

The problems cover all parts of geophysics. The book begins with an Introduction

(Chapter 1) that includes the equations most used in solving the problems. The idea of

this chapter is not to develop the theory, but rather to simply give a list of the equations

most commonly used in solving the problems, at the same time as introducing the reader to

the nomenclature. The next four chapters correspond to the division of the problems into

the four thematic blocks that are classic in geophysics: gravity, geomagnetism, seismology,

and heat flow and geochronology. We have not included problems in geodynamics, since

this would depart too much from the approach we have taken, which is to facilitate

comprehension of the theory through its application to specific cases, sometimes cases

which are far from the real situation on Earth. Indeed, some of the problems may seem a bit

artificial, but their function is to help the student practise with what has been seen in the

vii

theory. Neither did we want to include specific problems of geophysical prospecting as this

would have considerably increased the length of the text, and moreover some of the topics

that would be covered in prospecting, such as gravimetric and geomagnetic anomalies, are

already included in other sections of this work.

Chapter 2 contains 68 problems in gravity divided into five sections. The first section is

dedicated to the terrestrial geoid and ellipsoid, proposing calculations of the parameters

that define them in order to help better understand these reference surfaces. The second

corresponds to calculating the gravitational field and potential for various models of the

Earth, including the existence of internal structures. Gravity anomalies are dealt with in the

third section, with a variety of problems to allow students to familiarize themselves with

the corrections to the observed gravity, with the concept of isostasy, and with the Airy and

Pratt hypotheses. The fourth section studies the phenomenon of the Earth’s tides and their

influence on the gravitational field. The last section is devoted to the observations of

gravity from measurements made with different types of gravimeters and the corrections

necessary in each case. We also include the application of these observations to the

accurate determination of different types of height.

Chapter 3 contains 42 problems in geomagnetism divided into five sections. The first is

devoted to the main (internal) field generated by a tilted dipole at the centre of the Earth. It

includes straightforward problems that correspond to the calculation of the geomagnetic

coordinates of a point and the theoretical components of the magnetic field. This section

also introduces the student to the use of the principal units used in geomagnetism. The

second considers the magnetic anomalies generated by different magnetized bodies and

their influence on the internal field. The third section is devoted to the external field and its

variation with time. In the fourth section, we propose problems of greater complexity

involving the internal field, the external field, and anomalous magnetized bodies at the

same time. The last section is devoted to problems in paleomagnetism.

Chapter 4 contains 69 problems in seismology divided into seven sections. The first

presents some simple exercises on the theory of elasticity. The second addresses the problem

of the propagation of seismic energy in the form of elastic waves, resolving the problems on

the basis of potentials, and calculating the components of their displacements. We study the

reflection and refraction of seismic waves in the third section. The fourth is devoted to the

problem of wave propagation using the theory of ray paths in a planemedium of constant and

variable velocity of propagation. The fifth studies the problem of the propagation of rays in a

spherical medium of either constant or variable propagation velocity, with the calculation of

the travel-time curves for both plane and spherical media. The sixth section contains

problems in the propagation of surface waves in layered media. The seventh section is

devoted to problems of calculating the focal parameters and the mechanism of earthquakes.

Chapter 5 includes 11 problems in heat flow with the propagation of heat in plane and

spherical media, and seven problems in geochronology involving the use of radioactive

elements for dating rocks.

Finally, we provide a bibliography of general textbooks on geophysics and of specific

textbooks for the topics of gravity, geomagnetism, and seismology. We have tried to

include only those most recent and commonly used textbooks which are likely to be found

in university libraries.

viii Preface

In sum, the book is a university text for students of physics, geology, geophysics,

planetary sciences, and engineering at the undergraduate or Master’s degree levels. It is

intended to be an aid to teaching the subjects of general geophysics, as well as the specific

topics of gravity, geomagnetism, seismology, and heat flow and geochronology contained

in university curricula.

The teaching experience of the authors in the universities of Barcelona, Extremadura,

and the Complutense of Madrid highlighted the need for a work of this kind. This text is

the result of the teaching work of its authors for over 20 years. Thanks are due to

the generations of students over those years who, with their comments, questions, and

suggestions, have really allowed this work to see the light. We are also especially grateful

to Prof. Greg McIntosh who provided us with some problems on paleomagnetism, to

Prof. Ana Negredo for her comments on heat flow and geochronology problems, and to

Dr R.A. Chatwin who worked on translating our text into English.

The text is an extension of the Spanish edition published by Pearson (Madrid, 2010).

E. BUFORN, C. PRO AND A. UDÍAS

ix Preface

1 Introduction

Gravity

As a first approximation the Earth’s gravity is given by that of a rotating sphere. The

gravitational potential of a sphere of mass M is:

V ¼GM

r

where r is the position vector (Fig. A) and G the universal gravitational constant.

If the sphere is rotating with angular velocity o the centrifugal potential at a point on the

surface is given by

F ¼1

2o2

r2sin2y

where y is the angle that r forms with the axis of rotation.

The gravity potential is their sum U ¼ V þ F.

The value of the acceleration due to gravity (the gravity ‘force’) is given by the gradient

of the potential:

g ¼ rU

The radial component of the gravity force is given by

gr ¼ GM

r2

þ ro2sin2y

The potential of the Earth to a first-order approximation corresponds to that of a rotating

ellipsoid, and is given by

U ¼GM

a

a

rJ2

2

a

r

3

3sin2’ 1

þm

2

r

a

2

cos2’

where ’ ¼ 90º y is the geocentric latitude and a the equatorial radius.

The coefficient m is the ratio between the centrifugal and gravitational forces on the

sphere of radius a at the equator:

m ¼a3o2

GM

1

The dynamic form factor J2 is defined as

J2 ¼C A

a2M

where C and A are the moments of inertia about the axis of rotation and an equatorial axis.

The flattening of the ellipsoid (the shape of the Earth to a first-order approximation) of

equatorial and polar radius a and c is:

a ¼a c

aIn terms of J2 and m,

a ¼3

2J2 þ

m

2

The dynamic ellipticity is

H ¼C A

C

The gravity flattening is

b ¼gp ge

ge

where gp and ge are the normal values of gravity at the pole and the equator, respectively.

w

q

North Pole

Equator

P

a

r

l

Fig. A

2 Introduction

The gravity at a point of geocentric latitude ’ = 90º y is

g ¼ ge 1þ bsin2’

The geocentric latitude of a point is the angle between the equator and the radius vector of

the point. The geodetic latitude is defined as the angle between the equatorial plane and the

normal to the ellipsoid surface at a point. Astronomical latitude is the angle between the

equatorial plane and the observed vertical at a point.

The normal or theoretical gravity at a point of geocentric latitude ’ referred to the

GRS1980 reference ellipsoid is

g ¼ 9:780327 ð1þ 0:0053024 sin2’ 0:0000059 sin22’Þms2

The effect of the Sun and Moon on the Earth is to produce the phenomenon of the tides.

If one considers more generally the tidal effect due to an astronomical body of mass M at a

distance R from the centre of the Earth, one must add the corresponding potential, which, in

the first-order approximation, is given by

c ¼GMr

2

2R33cos2# 1

where r is the geocentric radius vector of the point, and # is the angle the position vector r

forms with the distance vector R.

Gravity anomalies, defined as Dg ¼ g –g, are the effects of the existence of anomalous

masses inside the Earth. The gravity anomaly along the Z (vertical) axis at a point distance

x along the horizontal axis produced by a sphere of radius R, density contrast Dr, and

buried at a depth d, is given by

g x; zð Þ ¼@Va

@z¼

GMðzþ dÞ

x2 þ ðzþ dÞ2h i3=2

where Va is the potential produced by the anomalous spherical mass DM ¼ 4/3pR3 Dr.

For problems in two dimensions, one uses the anomaly produced by an infinite horizon-

tal cylinder at depth d, perpendicular to the plane under consideration. The anomalous

potential is given by

Va ¼ 2pGra2ln1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ ðzþ dÞ2q

0

B

@

1

C

A

and the anomaly by

gðx; zÞ ¼ @Va

@z¼

2pGra2ðzþ dÞ

x2 þ ðzþ dÞ2

To correct for the height above sea level at which measurements are made, one uses the

concepts of the free-air and Bouguer anomalies. The free-air anomaly is

gFA ¼ g gþ 3:086h

3 Gravity

where g is the observed gravity, h the height in metres, and the anomaly is obtained in

gu (gravity units) mms2.

The Bouguer anomaly is

gB ¼ g gþ ð3:086 0:419rÞh

with r being the density of the plate of thickness h.

To account for isostatic compensation at height in mountainous areas, one adds an

isostatic correction which can be calculated assuming either the Airy or Pratt hypotheses.

With the Airy hypothesis, the root t of a mountain is given by

t ¼rc

rM rch

where rc and rM are the densities of the crust and mantle, and h is the height of the

mountain. For an ocean zone, with water density ra, the anti-root is

t0 ¼

rc rarM ra

h0

With the Pratt hypothesis, the density contrast in a mountainous area is

r ¼ r r0 ¼h

Dþ hr0

where D is the level of compensation, h the height of the mountain, and r0 the density at

sea level. For an oceanic zone of depth h0:

r0 ¼r0D rah

0

D h0

r ¼ r0 r0

The isostatic correction can be calculated using a cylinder of radius a and height b, whose

base is located at a distance c beneath the point, and with density contrast Dr:

CI ¼ 2pGr bþ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ c bð Þ2q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ c2p

For mountainous zones, with the Airy hypothesis: b ¼ t, c ¼ h þ H þ t (H ¼ crustal

thickness, h ¼ height of the point); and with the Pratt hypothesis: b ¼ D, c ¼ D þ h.

Geomagnetism

To a first approximation, the internal magnetic field of the Earth can be approximated by a

centred dipole inclined at 11.5 to the axis of rotation. The potential created by a magnetic

dipole at a point distant r from its centre and forming an angle ywith the axis of the dipole is

F ¼Cm cos y

r2

4 Introduction

where C ¼ m0/4p with m0 ¼ 4p 107 Hm1, and m is the dipole moment in units

of Am². The product Cm is given in Tm3.

The components of the magnetic dipole field B are:

Br ¼ @F

@r¼

2Cm cos y

r3

By ¼ 1

r

@F

@y¼

2Cm sin y

r3

In the centred dipole approximation for the Earth’s magnetic field, the geomagnetic

coordinates (f*, l*) of a point (y ¼ 90º f*) in terms of its geographic coordinates

(f, l) and those of the Geomagnetic North Pole (GMNP) (fB, lB) can be calculated using

the expressions of spherical trigonometry (Fig. B):

sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ

sin l ¼sinðl lBÞ cosf

cosf

The vertical and horizontal components of the field, the geomagnetic constant B0, and the

total field are given by:

Z ¼ 2B0 sinf

H ¼ B0 cosf

B0 ¼Cm

a3

F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

H2 þ Z2p

¼ B0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ 3sin2fq

The units used for the components of the magnetic field are the tesla T and the nanotesla

nT ¼ 109 T. The NS (X*) and EW (Y*) components are

X ¼ H cosD

Y ¼ H sinD

and the declination and inclination are given by

90º – fB

90º – f

GMNP

GNP

180º – l∗

q = 90º – f∗ D∗

l – lB

P

Fig. B

5 Geomagnetism

sinD ¼ cosfB sinðl lBÞ

cosf

tan I ¼ 2 tanf

The radius vector at each point of the line of force is:

r ¼ r0cos2f ¼ r0sin

2y

where r0 is the radius vector of the point of the line of force located at the geomagnetic

equator.

Magnetic anomalies are produced by magnetic materials within the Earth. The anomal-

ous potential due to a vertical dipole buried at depth d is

FA ¼Cmcosy

r2

¼Cmðzþ dÞ


The vertical (z) and the horizontal (x) components of the magnetic anomaly at the surface

(z ¼ 0) produced by a vertical magnetic dipole at depth d are:

Z ¼Cmð2d2 x2Þ

ðx2 þ d2Þ5=2

X ¼3Cmxd

ðx2 þ d2Þ5=2

The Earth is affected by an external magnetic field produced mainly by the activity of the

Sun. This field is variable in time, with distinct periods of variation. The most noticeable is

the diurnal variation (Sq) with a maximum at 12 noon local time. The most important non-

periodic variations are the so-called magnetic storms.

Seismology

Earthquakes produce elastic waves which propagate through the interior and along the

surface of the Earth. Using the plane-wave approximation, the displacements of the

internal P- and S-waves (uiP and ui

S) can be obtained from a scalar potential and a vector

potential:

ui ¼ uiP þ ui

S ¼ r’ð Þi þ r cj

i

’ ¼ A exp ika gjxj at

cj ¼ Bj exp ikb gjxj bt

where A and Bj are the amplitudes, xj the coordinates of the observation point, ka and kb the

wavenumbers, gj are the direction cosines defined from the azimuth az and angle of

incidence i of the ray as:

6 Introduction

g1 ¼ sin i cos az

g2 ¼ sin i sin az

g3 ¼ cos i

and a and b are the P- and S-wave velocities of propagation, respectively, defined from

the Lamé coefficients (l and shear modulus m) and the density r:

vP ¼ a ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

lþ 2m

r

s

vS ¼ b ¼

ffiffiffi

m

r

r

Units used are: displacement amplitudes (u) in µm; potential amplitudes (A, Bi) in 103 m2;

wavenumber (k) in km1; and wave velocity (a, b) in km s1.

Poisson’s ratio is defined in terms of the Lamé coefficients as

s ¼l

2 lþ mð Þ

The angle of polarization of S-wave e is defined as

e ¼ tan1 uSH

uSV

where uSH is the amplitude of the SH component, and uSV that of the SV component. SH

and SV are the horizontal and vertical components of the S-wave on the wavefront plane.

The coefficients of reflection V and transmission W are given by the respective

ratios between the amplitudes of the reflected or transmitted potentials and the incident

potential:

V ¼A

A0

W ¼A0

A0

where A0 is the amplitude of the incident wave potential, A that of the reflected potential,

and A0 of the transmitted potential.

Snell’s law for plane media is expressed as

p ¼sin i

v

and for spherical media

p ¼r sin i

v

where p is the ray parameter, i the angle of incidence, v the propagation velocity of the

medium, and r the position vector along the ray.

7 Seismology

In the case of plane media with propagation velocity varying with depth v(z), the

epicentral distance and the travel time of a ray for a surface focus are given by

x ¼ 2

ðh

0

pdzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 p2p

t ¼ 2

ðh

0

2dzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 p2p

where ¼ v1 and h is the depth of maximum penetration of the ray. The variation of the

epicentral distance x with the ray parameter p is given by

dx

dp¼

2

B0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

20 p2p þ 2

ðB

0

dB

dzdz

B2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 p2p

where

B ¼1

v

dv

dz

In spherical media with velocity varying with depth v(r), the epicentral distance,

trajectory along the ray, and travel time are given by

¼ 2

ð

r0

rp

p

r

drffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 r2p

s ¼ 2

ð

r0

rp


2 r2

p

t ¼ 2

ð

r0

rp

dr

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 r2

p

where ¼ rv1, r0 is the radius at the surface of the Earth, and rp is the radius at the point

of maximum penetration of the ray.

The variation of the distance from the epicentre D with the ray parameter p in a spherical

medium is

d

dp¼

2

ð1 B0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

20 p2p þ 2

ðB

0

dB

drdr

ð1 B2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 p2p

where

B ¼r

v

dv

dr

The radial and vertical components (u1 and u3) of surface waves can be obtained from

the potentials ’ and c. The transverse component (u2) is kept apart

u1 ¼@’

@x1

@c

@x3¼ ’;1 c;3

u2 ¼ C exp iksx3 þ ikðx1 ctÞ½

u3 ¼@’

@x3þ

@c

@x1¼ ’;3 þ c;1

8 Introduction

where c is the wave propagation velocity and

’ ¼ A exp ikrx3 þ ikðx1 ctÞ½

c ¼ B exp iksx3 þ ikðx1 ctÞ½

r ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffi

c2

a2 1

r

s ¼


c2

b2 1

s

For surface waves, c < b < a, and hence r and s are imaginary.

For dispersive waves, the relationship between the phase velocity c and the group

velocity U is

U ¼ cþ kdc

dk

where k is the wavenumber.

The position of the seismic focus is given by the coordinates of the epicentre (’0, l0)

and the depth h. The time is that of the origin of the earthquake t0. The size is given by

the magnitude which is proportional to the logarithm of the amplitude of the recorded

waves. For surface waves this is:

Ms ¼ logA

Tþ 1:66 logþ 3:3

where A is the amplitude of ground motion in microns, T is the period in seconds, and ∆ the

epicentral distance in degrees.

The magnitude of the moment is given by

Mw ¼2

3logM0 6:1

whereM0 is the seismic moment in Nm (newtonmetres). The seismic moment is related to

the displacement of the fault ∆u and its area S:

M0 ¼ muS

The mechanism of earthquakes is given by the orientation of the fracture plane (fault)

defined by the angles ’ (azimuth), d (dip), and l (slip angle or rake), or by the vectors n

(the normal to the fault plane) and l (the direction of slip).

The elastic displacement of the waves produced by a point shear fault is

uk xs; tð Þ ¼ mu tð ÞS linj þ ljni @Gki

@xj

where Gki is the medium’s Green’s function which, for an isotropic, homogeneous, infinite

medium, and P-waves in the far-field regime, is given by

GPki ¼

1

4pra2rgigkd t

r

a

9 Seismology

The P-wave displacements are given by:

uPk xs; tð Þ ¼_u tð ÞS

4pra3rm linj þ ljni

gigjgk

This equation can be expressed also in terms of the moment tensor Mij

uPk xs; tð Þ ¼_M ij tð Þ

4pra3rgigjgk

Mij is a more general representation of a point source.

Heat flow

The Fourier law of heat transfer by diffusion states that the heat flux _q is proportional to the

gradient of the temperature T:

_q ¼ KrT

where K is the thermal conductivity coefficient. The units of heat flow are Wm2.

The heat diffusion equation, assuming that K is constant, is given by

kr2T þe

rCv

¼@T

@t

where Cv is the specific heat, r the density, e the heat generated per unit volume and unit

time (heat sources), and k the thermal diffusivity:

k ¼K

rCv

If there are no heat sources, the diffusion equation is

kr2T ¼@T

@t

In the case of one-dimensional flow with periodic variation of temperature over time,

one has:

T z; tð Þ ¼ T0 exp

ffiffiffiffiffiffi

o

2k

r

zþ i

ffiffiffiffiffiffiffiffi

o

2kz

r

þ ot

where z is the vertical direction (positive towards the nadir) and o the angular frequency.

In the case of stationary one-dimensional solutions (T constant in time) one obtains from

the diffusion equation:

T ¼ e

2Kz2 þ

_q0Kzþ T0

10 Introduction

where T0 and _q0 are the temperature and flow at the surface (z ¼ 0).

For a spherical Earth, assuming that the thermal conductivity is constant, and that

the amount of heat per unit volume depends only on time, the diffusion equation takes the

form:

K@2T

@r2þ2

r

@T

@r

þ eðtÞ ¼ rCv

@T

@t

where r is the radial direction.

For the stationary case, the above equation reduces to

1

r2

d

drr2 dT

dr

¼ e

K

Integrating twice, one has

T ¼ T0 þe

6KR2 r

2

where T0 is the temperature at the surface (r ¼ R).

Geochronology

Geochronology is based on determining the age of a rock by measuring the decay of its

radioactive elements. In a sample of radioactive material, the number of atoms that have

yet to disintegrate after time t is given by

nt ¼ n0elt

where n0 is the initial number of atoms, and l the decay constant. The rate of decay dn/dt is

the activity R, so that

R ¼ R0elt

where R0 is the initial activity (at t ¼ 0).

The half-life (or period) of the sample is the time it takes for the activity R to fall to half

its initial value. It is given by:

T1=2¼

0:693

l

The mean life-time t of one of the atoms that existed at the start is given by:

t ¼1

l

If a sample consists of NR radioactive nuclei and NE stable nuclei, the time to arrive at the

propotion NE/NR is given by

11 Geochronology

t ¼1

l1þ

NE

NR

If the rubidium–strontium (Rb-Sr) method is used to date a sample, a correction must be

made for the contamination of the stable 86Sr isotope relative to the radioisotope 87Sr:

87Sr86Sr

total ¼87Sr86Sr

initial þ87Rb86Sr

elt 1

This expression corresponds to a straight line (isochrone) of slope elt 1

and intercept

corresponding to the initial content87Sr86Sr

initial.

12 Introduction

2 Gravity

Terrestrial geoid and ellipsoid

1. Calculate the geodetic and geocentric latitudes of a point P on the ellipsoid

whose radius vector is 6370.031 km, given that m ¼ 3.4425 103, GM ¼ 39.86005

1013 m3 s2, and 6356.742 km is the polar radius. Determine J2 and b.

The major semi-axis (equatorial radius) is a and the minor (polar radius) is c, the geocentric

latitude is ’, and the geodetic latitude is ’d (Fig. 1).

The coefficient m is given by the equation

m ¼o2a3

GM

where G ¼ 6.671011 m3 kg1 s2 is the gravitational constant, M the Earth’s mass, and

the angular velocity is o = 2p/T, where T is the rotation period (T ¼ 24 h). We obtain for

the semi-axis a the value

a ¼mGMT 2

4p2

1=3

¼ 6378:127 km

The Earth’s flattening a can be obtained directly since we already know a and c so

a ¼a c

c¼

6378:127 6356:742

6378:127¼ 3:3529 103

The radius vector to the point P is given by the equation r ¼ a(1 a sin2 ’ )

From this equation we can calculate the geocentric latitude ’:

6370:031 ¼ 6378:127 1 3:3539 103 sin2 ’

’ ¼ 37 580 2200

The relation between the geocentric ’ and geodetic ’d latitudes is given by

tan’d ¼1

1 að Þ2tan’

Substituting the already obtained values for a and ’

’d ¼ 38 090 3500

13

The dynamic form factor J2 can be obtained from the equation

a ¼3

2J2 þ

m

2

Then

J2 ¼2

3a

m

2

¼ 1:0878 103

From this value we can determine the gravity flattening b using the equation

b ¼5

2m a ¼ 5:2533 103

2. Taking the first-order approximation, let two points of the ellipsoid at 45 N and 30 S

be situated at distances of 6367.444 km and 6372.790 km from the centre, respectively.

If the normal gravity values are 9.806193 m s2 for the first and 9.793242 m s2 for

the second, calculate: the flattening, gravity flattening, coefficient m, equatorial radius,

polar radius, dynamic form factor, and the Earth’s mass.

Data

r1 ¼ 6367.444 km w1 = 45 N g1 = 9.806193 m s2

r2 ¼ 6372.790 km w2 = 30 S g2 = 9.793242 m s2

The normal or theoretical gravity at a point can be expressed in terms of the normal gravity

at the equator ge, the gravity flattening b, and the latitude of the point ’:

g1 ¼ ge 1þ b sin2 ’1

g2 ¼ ge 1þ b sin2 ’2

If we divide both expressions we obtain:

g1g2

¼1þ b sin2 ’1

1þ b sin2 ’2

P

a

j jd

rC

Fig. 1

14 Gravity

From this expression we can obtain the gravity flattening, since we already know g1, g2,’1,’2:

b ¼g1 g2

g2 sin2 ’1 g1 sin

2 ’2

¼ 5:297 103

The distance r from the centre of the ellipsoid to points on its surface can be given as a

function of the flattening a, the equatorial radius a, and the latitude ’:

r1 ¼ að1 a sin2 ’1Þ

r2 ¼ að1 a sin2 ’2Þ

If we divide both expressions

r1

r2

¼1 asin2’1

1 asin2’2

Thus we obtain the value of the flattening,

a ¼ 3:353 103

From this value we find the equatorial radius,

a ¼r1

1 asin2’1

¼ 6378:137 km

The polar radius c can be found from this value and the flattening:

a ¼a c

aand c ¼ að1 aÞ ¼ 6356:751 km

The coefficient m is obtained from a and b:

aþ b ¼5

2m m ¼

2

5ðaþ bÞ ¼ 3:460 103

From this value we can obtain the value of the Earth’s mass M from

m ¼o2a3

GM

with o ¼2p

Tp where T ¼ 24 hours.

Therefore

M ¼4p2a3

T2Gm¼ 5:946 1024kg

3. Obtain the value of the terrestrial flattening in the first-order approximation, given

that the normal gravity values for two points of the ellipsoid are:

Point 1: w1 ¼ 42º 200 g1 ¼ 980.389 063 Gal

Point 2: w2 ¼ 47º 300 g2 ¼ 980.854 830 Gal

Take the equatorial radius to be 6378.388 km.

15 Terrestrial geoid and ellipsoid

For this problem we use the equations of Problems 1 and 2:

g1 ¼ geð1þ bsin2’1Þ

g2 ¼ geð1þ bsin2’2Þ

If we divide these expressions:

g1g2

¼1þ bsin2’1

1þ bsin2’2

and we can solve for the gravity flattening, b:

b ¼ 5:288 2675 103

Using the gravity flattening b we can determine the value of gravity at the equator, ge:

ge ¼g1

1þ bsin2’1

¼ 978:043 614Gal

Using the following equations

aþ b ¼5

2m

a ¼3

2J2 þ

m

2

m ¼o2a3

GM

we derive the expression

ge ¼GM

a2ð1 bÞ þ o2a

and substituting the values we obtain GM ¼ 3.986 5415 1014 m3 s2

From this value, taking T ¼ 24 hours, we obtain

m ¼o2a3

GM¼

4p2a3

T2GM¼ 3:442 5698 103

And finally the Earth’s flattening is

a ¼5

2m b ¼ 3:318 1575 103

4. P is a point of the terrestrial ellipsoid at latitude 60 ºS and distance to the centre of

6362.121 km. The Earth’s mass is 5.9761 1024 kg and the ratio between the polar

and equatorial semi-axes is 0.9966. Taking the first-order approximation, calculate:

(a) The flattening and the coefficient J2.

(b) The value of normal gravity in mGal at P.

As in the previous problems we use the equations given in Problems 1 and 2.

16 Gravity

(a) From the equation for the flattening (Problem 2)

a ¼ 1c

a¼ 3:4 103

The value of the equatorial radius a is obtained from the equation

a ¼r

1 asin2’¼ 6 378 386m

The coefficient m is given by

m ¼o2a3

GM¼

4p2a3

T2GM¼ 3:4429 103

The gravity flattening b is given by

b ¼5

2m a ¼ 5:2072 103

The dynamic form factor J2 is found from the relation

J2 ¼2a m

3¼ 1:1190 103

(b) The normal gravity at that point is given by

g ¼ geð1þ bsin2’Þ ¼ 981 856:3mGal

5. At a point P on the ellipsoid at latitude 50 ºS, the value of normal gravity is 9.810

752 m s2 and the distance to the centre of the Earth is 6365.587 km. Given that

the mass of the Earth is 5.976 1024 kg and the ratio between the minor and major

semi-axes is c/a ¼ 0.996 6509, calculate:

(a) The flattening, equatorial radius, gravity flattening, dynamic form factor, and

coefficient m.

(b) The normal gravity at the equator.

(c) The centrifugal force at P.

Data

’ ¼ 50 S r ¼ 6365.587 km g ¼ 9.810 752 m s2.

(a) According to Problem 1, the flattening is given by

a ¼ 1c

a¼ 3:349 103

and the equatorial radius a is

a ¼r

1 asin2’¼ 6378:122 km


Taking T ¼ 24 hours, the coefficient m is then given by

m ¼4p2a3

T2GM¼ 3:4425 103

b and J2 can be obtained from the equations

b ¼5

2m a ¼ 5:2571 103 J2 ¼

2

3a

m

2

¼ 1:0852 103

(b) The normal gravity ge at the equator is given by

ge ¼g

1þ bsin2’¼ 9:780 579m s2

(c) The centrifugal force at point P is given by its radial and transverse components

f ¼ frer þ fyey

where, since 90 y = ’,

fr ¼ o2r sin2 y ¼ o2

r cos2 ’ ¼ 0:013 909m s2

fy ¼ o2r sin y cos y ¼ o2

r cos’ sin’ ¼ 0:016 576m s2

6. Taking the first-order approximation, calculate the Earth’s flattening a, gravity

flattening b, dynamic form factor J2, and polar radius c, given that:

ge ¼ 978.032 Gal (normal gravity at the equator)

a ¼ 6378.136 km (equatorial radius)

GM ¼ 39:8603 1013 m3 s2

and that for a point on the ellipsoid at latitude 60 ºN the normal gravity value is

981 921 mGal.

Calculate also the radius vector of this point and the gravitational potential.

Assuming a first-order approximation, the expression for the normal gravity is

g ¼ geð1þ b sin2 ’Þ

The value of b is given by

b ¼g ge

gesin2’

¼ 5:302 103

The coefficient m, taking T ¼ 24 hours, is

m ¼4p2a3

T2GM¼ 3:442 103

a and J2 are determined from the equations

a ¼5

2m b ¼ 3:303 103 J2 ¼

2

3a

m

2

¼ 1:055 103

18 Gravity

The polar radius c is determined from the flattening

a ¼a c

a) c ¼ að1 aÞ ¼ 6357:069 km

The radius vector at the point of latitude 60 ºN is given by

r ¼ að1 asin2’Þ ¼ 6362:335 km

The gravity potential at the same point, in the first-order approximation, is found using

Mac Cullagh’s formula,

U ¼GM

r1

J2

2

a

r

2

3 sin2 ’ 1

þr

a

3m

2cos2 ’

¼ 6:263 57 107 m2 s2

7. Assuming that the Moon is an ellipsoid of equatorial radius 1738 km and polar

radius 1737 km, with J2 = 3.8195 104 and a mass of 7.3483 1022 kg, calculate its

period of rotation.

First, we calculate the lunar flattening a,

a ¼a c

a¼ 5:7537 104

The coefficient m is found from the values of a and J2:

a ¼3

2J2 þ

m

2) m ¼ 2a 3J2 ¼ 7:5900 106

From the value of m we find the period of rotation T:

m ¼4p2a3

T2GM) T ¼

4p2a3

mGM

12

¼ 27:32 days

8. Calculate, in the first-order approximation, the latitude and radius vector of a point

P of the terrestrial ellipsoid for which the value of normal gravity is 979.992 Gal, given

that the Earth’s mass is 5.976 1024 kg and that normal gravity for another point Q at

50º S latitude is 981.067 Gal, for the equator is 978.032 Gal, and that J2 is 1.083 103.

The gravity flattening b is found from the normal gravity at Q with latitude ’1 = 50:

b ¼

g1ge

1

sin2 ’Q

¼ 5:288 103

Using the same expression and the normal gravity at P we calculate its latitude

sin2’P ¼

gAge

1

b; ’P ¼ 37 590 46:5700

The Earth’s flattening is given by

a ¼15

8J2 þ

b

4¼ 3:353 103


and the value of the coefficient m by

m ¼ 2a 3J2 ¼ 3:457 103

From the value of m we obtain the equatorial radius a:

a ¼mGM

o2

13

¼ 6 387 062:758m

where we have substituted o = 2p/T, taking T ¼ 24 hours.

Finally, we find the radius vector of point P

r ¼ að1 asin2’Þ ¼ 6 378 946:678m

9. At a point P on the terrestrial ellipsoid of latitude 70 ºS and radius vector 6359.253

km, the value of normal gravity is 982.609 Gal. If the mass of the Earth is 5.9769 1024

kg and the equatorial radius is 6378.136 km, calculate the value of normal gravity at

the Pole, the dynamic form factor, and the centrifugal force at the Pole and the equator.

The flattening a is given by

a ¼1

sin2’1

r

a

¼ 3:3528 103

Putting T ¼ 24 hours, the coefficient m is obtained from the equation

m ¼o2a3

GM¼ 3:4425 103

From a and m we find the gravity flattening b:

b ¼5

2m a ¼ 5:2535 103

Normal gravity at the equator is found from the value of the gravity at point P:

ge ¼g

1þ b sin2 ’¼ 9:780 72m s2

From this value we find the normal gravity at the Pole:

gp ¼ ge 1þ bð Þ ¼ 9:832 10m s2

The dynamic form factor is found from the values of a and m:

J2 ¼2

3a

m

2

¼ 1:0877 103

The centrifugal force is given by the expression:

f ¼ o2r sin2 yer þ o2

r sin y cos yey

At the pole, y = 90 ’ = 0 ! f = 0.

At the equator, y ¼ 90º ! f = o2 a er = 0.033 73 er m s2

20 Gravity

10. Let two points of the ellipsoid be of latitudes w1 and w2, with radius vectors

6372.819 km and 6362.121 km, respectively. The ratio of the normal gravities is

0.997 37, the flattening 3.3529 103, and the gravity flattening 5.2884 103.

Calculate:

(a) The Earth’s mass.

(b) The latitude of each point and the dynamic form factor.

(a) The equatorial radius a can be obtained from the ratio of the two normal gravities

g1g2

¼1þ bsin2’1

1þ bsin2’2

¼1þ b

a r1

aa

1þ ba r2

aa

¼ l

where l ¼ 0.997 37. We solve for a and obtain

a ¼b lr2 r1ð Þ

aþ bð Þ l 1ð Þ¼ 6382:94 km

We calculate the mass of the Earth from the flattening and gravity flattening as in Problem 2:

m ¼2

5aþ bð Þ ¼ 3:4565 103 M ¼

4p2a3

T2Gm¼ 5:9653 1024 kg

(b) The latitudes at each point are calculated from the radius vectors

sin2’1 ¼a r1

aa! ’1 ¼ 43 270 5800

sin2’2 ¼a r2

aa! ’2 ¼ 80 330 4600

The dynamic form factor is obtained from the values of a, b, and m:

b ¼ 2aþ m9

2J2

J2 ¼ 2aþ m bð Þ2

9¼ 1:0831 103

11. Let a point A have a value of gravity of 9793 626.8 gu and a geopotential number

of 32.614 gpu. Calculate the gravity at a point B, knowing that the increments in

dynamic and Helmert height over point A are 271.116 m and 271.456 m, respectively.

Take g45 ¼ 9.806 2940 m s2. Give the units for each parameter.

The dynamic heights at points A and B are given by:

HAD ¼

CA

g45

HBD ¼

CB

g45

where C is the value of the geopotential at each pointP

N

j¼1

gjdhj

!

and g45 the normal

gravity for a point on the ellipsoid at 45º latitude.


Subtracting both equations,

HBD HA

D ¼CB CAð Þ

g45

Solving for CB,

CB ¼ CA þ g45 HBD HA

D

¼ 298:478 gpu

If heights are given in kmand normal gravity inGal, geopotentials are in gpu (geopotential units)

1 gpu ¼ 1 kGal m ¼ 1 Gal km

The Helmert orthometric height H is given by

H ¼C

g þ 0:0424Hð11:1Þ

where C is in gpu, g in Gal, and H in km.

Solving for H:

H ¼g

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

g2 þ 4 0:0424Cp

2 0:0424

Since the point A is above the geoid (CA > 0), we take the positive solution,

HA ¼ 33:301m

Then, the Helmert height at point B is

HB ¼ HA þHBA ¼ 304:757m

The gravity at point B is calculated using Equation (11.1)

gB ¼CB

HB

0:0424HB ¼ 979:382 75Gal

12. Calculate the value of gravity in gravimetric units and mGal of a point on the

Earth’s surface whose orthometric (Helmert) and dynamic heights are 678.612 m and

679.919 m, respectively, taking g45 ¼ 9.806 294 m s2.

The geopotential is calculated from the dynamic height,

HD ¼C

g45) C ¼ HDg45 ¼ 666:748 gpu

where HD is given in km and g45 in Gal.

Knowing the geopotential, we calculate the gravity from the orthometric (Helmert)

height H, using its definition,

H ¼C

g þ 0:0424H) g ¼

C

H 0:0424H ¼ 982:489 34Gal ¼ 9 824 893:4 gu

13. If at a point on the surface of the Earth of Helmert height 1000 m one observes

a value of gravity of 9.796 235 m s2, calculate the average value of gravity

22 Gravity

between that point and the geoid along the direction of the plumb-line, and

the point’s geopotential number.

The mean value of gravity between a height H and the surface of the geoid is given by

g ¼1

H

ðH

0

gðzÞdz

where g(z) is the value of gravity at a distance z from the geoid along the vertical path to a

point of height H. This value can be obtained using the Poincaré and Prey reduction from

the value of g observed at the Earth’s surface at a point of height H,

gðzÞ ¼ g þ 0:0848ðH zÞ

Then

g ¼1

H

ðH

0

gðzÞdz ¼1

H

ðH

0

g þ 0:0848 H zð Þ½ dz

¼1

Hgzþ 0:0848 Hz 0:0424z2 H

0

g ¼ g þ 0:0424 H ¼ 979:6659Gal

where g is given in Gal and H in km.

The geopotential C can be obtained from the formula for the Helmert height,

H ¼C

g þ 0:0424H) C ¼ ðg þ 0:0424 HÞ H ¼ 979:666 gpu

14. For two points A and B belonging to a gravity measurement levelling line, one

obtained:

gA = 9.801 137 6 m s2

CA = 933.316 gpu

Gross increment elevation: DhBA ¼ 20:340

Increment in dynamic height: HAD HB

D ¼ 20:340m.

Given that the normal gravity at 45 º latitude is 9806 294 gu, calculate the Helmert

heigth of point B.

As in Problem 11, the dynamic heights at A and B are given by

HAD ¼

CA

g45

HBD ¼

CB

g45Subtracting both equations:

HBD HA

D ¼CB CAð Þ

g45

Solving for CB:

CB ¼ CA þ g45 HBD HA

D

¼ 913:371 gpu


The geopotential at B can be obtained from the gross increment in elevation betweenA and B,

CB ¼ CA þgA þ gB

2

hBA

and, solving for gB,

gB ¼2ðCB CAÞ

hBA gA ¼ 980:103 08Gal

Finally we calculate the orthometric Helmert height at point B,

H ¼C

g þ 0:0424H

Substituting the values for point B, and solving for H we obtain

H ¼gB

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

gB2 þ 4 0:0424CB

p

2 0:0424¼ 931:875m

15. A, B, and C are points connected by a geometric levelling line. Given that the

normal gravity at a latitude of 45º is 980.6294 Gal, complete the following table:

Station A

Dynamic height:

HAD ¼

CA

g45¼ 678:118m

Helmert height:

H ¼C

g þ 0:0424H

H ¼g

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

g2 þ 4 0:0424 Cp

2 0:0424¼ 678:611m

ð15:1Þ

Station B

The geopotential number is found from the dynamic height:

CB ¼ g45HBD ¼ 664:452 gpu

Station

Gravity

(Gal)

Height

Increment (m)

Geopotential

Number (gpu)

Dynamic

Height (m)

Helmert

Height (m)

A 979.88696 – 664.982 ? ?

B ? 0.541 ? 677.577 ?

C 979.88665 ? ? ? 657.134

24 Gravity

From this value and the difference in height with respect to station Awe find the gravity at B:

CB ¼ CA þgA þ gB

2

hAB

from which we get gB = 979.877 84 Gal

The Helmert height of B is found as in station A:

HB ¼ 678:077m

Station C

From the known values of gravity and Helmert height we find the geopotential number

(Equation 15.1)

CC ¼ gHC þ 0:0424H2C ¼ 661:574 gpu

To calculate the difference in height of C with respect to B we begin with the expression

CC ¼ CB þgB þ gC

2

hCB

from which

hCB ¼2 CC CBð Þ

gB þ gC¼ 2:937m

The dynamic height is found directly from the geopotential number:

HCD ¼

CC

g45¼ 674:642m

The complete table is:

Earth’s gravity field and potential

16. Suppose an Earth is formed by a sphere of radius a and density r, and within it

there are two spheres of radius a/2 with centres located on the axis of rotation. The

density of that of the northern hemisphere is 5r and of that of the southern hemi-

sphere is r /5. The value of the rotation is such that m ¼ 0.1. Determine:

(a) The potential U in the r3 approximation.

(b) The values of gr and gu for a point on the equator in the r2

approximation.

Station

Gravity

(Gal)

Height

increment (m)

Geopotential

number (gpu)

Dynamic

height (m)

Helmert

height (m)

A 979.88696 – 664.982 678.118 678.611

B 979.87784 0.541 664.452 677.577 678.077

C 979.88665 2.937 661.574 674.642 657.134

25 Earth’s gravity field and potential

(c) The error made in (b) with respect to the exact solution.

(d) The deviation of the vertical from the radial at the equator.

(a) The gravitational potential is the sum of the potentials of the three spheres

V ¼ V1 þ V2 þ V3 ¼GM

rþGM1

qþGM2

q0ð16:1Þ

where r is the distance from a point P to the centre of the sphere of radius a and mass M,

where M is given by

M ¼4

3pra3

q and q0 are the distances to the centres of the two spheres in its interior in the northern and

southern hemispheres which have differential masses M1 and M2, respectively (Fig. 16).

The differential masses are those corresponding to the difference in density in each case

with respect to the large sphere:

M1 ¼4

3pð5r rÞ

a3

8¼

M

2

differential mass of the sphere in the northern hemisphere

M2 ¼4

3p

r

5 r

a3

8¼

M

10

differential mass of the sphere in the southern hemisphere

The distance q can be calculated using the cosine law

q ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

r2 þa

2

2

2a

2r cos y

r

¼ r

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þa

2r

2

2a

2r

cos y

r

Considering this expression, 1/q corresponds to one of the generating functions of the

Legendre polynomials. Then 1/q, in the first-order approximation, is given by

1

q¼

1

r1þ

a

2rcos yþ

1

2

a

2r

2

3cos2y 1

Since cos y0 = cos y, 1/q0 is given by

1

q0¼

1

r1

a

2rcos yþ

1

2

a

2r

2

3cos2y 1

P

θ

θ'a

r

r 2

r 3

a/2

α

q

q '

ρ

ρ/5

5ρ

Fig. 16

26 Gravity

If we substitute in Equation (16.1), the potentials for each sphere are given by

V1 ¼GM

rV2 ¼

GM

2

1

rþ

a

2r2cos yþ

a2

8r33cos2y 1

V3 ¼ GM

10

1

r

a

2r2cos yþ

a2

8r33cos2y 1

Then, the total gravity potential is the sum of the three gravitational potentials plus the

potential of the centrifugal force due to the rotation:

U ¼ GM 1þ1

2

1

10

1

rþ

1

4þ

1

20

a

r2cos y

þ1

16

1

80

a2

r3

3 cos2 y 1

þ1

2r2o2 sin2 y

In terms of the coefficient m, given here by m ¼o2a3

GM,

U ¼GM

a

7

5

a

rþ

3

10

a2

r2cos yþ

1

20

a3

r3

3cos2y 1

þ1

2m

r

a

2

sin2y

(b) Using this first-order approximation of the potential, the radial and tangential

components of gravity at the equator, r = a and y = 90, putting m ¼ 0.1, are

gr ¼@U

@r¼ 1:3

GM

a2

gy ¼1

r

@U

@y¼ 0:3

GM

a2

(c) To calculate exactly the value of gr at the equator we have to calculate

the exact contribution of each of the three spheres plus the centrifugal force (m¼ 0.1):

gr ¼ g1rþ g2

rþ g3

r m

GM

a2

g1r¼

GM

a2

g2r¼

GM

2r22cos a

g3r¼

GM

10r23cos a

where

r2 ¼


a2 þa2

4

r

¼ r3

and a is the angle which forms r2 and r3 with the equator (Fig. 16)


sin a ¼a=2

r2

¼1ffiffiffi

5p

Then

gr ¼ GM

a21þ

4

5ffiffiffi

5p

4

25ffiffiffi

5p m

¼ 1:19GM

a2

gy ¼ g1y þ g2y ¼ GM

25

4a2

1ffiffiffi

5p þ

GM

105

4a2

1ffiffiffi

5p

0

B

@

1

C

A

¼GM

a2

2

5ffiffiffi

5p þ

2

25ffiffiffi

5p

¼ 0:14GM

a2

The error made in the first-order approximation with respect to the exact solution is

gr ¼ 1:19þ 1:3ð ÞGM

a2¼ 0:11

GM

a2

gy ¼ ð0:3þ 0:14ÞGM

a2¼ 0:16

GM

a2

(d) The deviation of the vertical with respect to the radial direction is given by the angle i

which is determined from the gravity components gr and gy. At the equator this angle is:

• Using the first order approximation

tan i ¼gy

gr¼

0:3

1:3) i ¼ 13:0

• Using the exact values

tan i ¼0:16

1:19) i ¼ 7:6

17. A spherical planet is formed by a sphere of radius a and density r, and inside it a

sphere of radius a/2 and density 5r centred at the midpoint of the radius of the

northern hemisphere. There is no rotation.

(a) Determine J0, J1, and J2.

(b) What is the deviation of the vertical from the radial at the equator?

(a) The total gravitational potential is the sum of the potentials of the two spheres

(Fig. 17) where g, is the attraction due to the potential V1 and g2 that due to the

potential V2:

V ¼ V1 þ V2 ¼GM

rþGM 0

q

where r and q are the distances from a point P to the centres of the large and small spheres,

respectively.

28 Gravity

As we did in Problem 16, for the small sphere of radius a/2 we take the differential

mass M 0

M 0 ¼4

3pa3

85r rð Þ ¼

16

24pra3 ¼

M

2

where M is the mass of the sphere of radius a and density r.

For 1/q we take the first-order approximation of the Legendre polynomial, as we did in

Problem 16:

1

q¼

1

r1þ

a

2rcos yþ

1

2

a

2r

2

3cos2y 1

Then, the expression for the gravitational potential V is:

V ¼GM

rþGM

2

1

rþ

a

2r2cos yþ

1

4

a2

4r33cos2y 1

¼ GM3

2rþ

a

4r2cos yþ

a2

32r33cos2y 1

(b) We know that the potential can be expressed by an expansion in zonal

spherical harmonics (Legendre polynomials) given in the first-order approxima-

tion by

V ¼GM

aJ0

a

r

þa

r

2

J1 cos yþa

r

3

J21

23cos2y 1

Comparing the two expressions we obtain,

5r qP

r

a /2

g1

g2

a

r

q

i

Fig. 17


J0 ¼3

2

J1 ¼1

4

J2 ¼1

16

The components of gravity at the surface of the large sphere (r ¼ a) are:

gr ¼@V

@r¼

GM

a23

21

2cos y

3

323cos2y 1

gy ¼1

r

@V

@y¼

GM

a21

4sin y

3

16cos y sin y

and at the equator, y ¼ 90:

gr ¼ 45GM

32a2

gy ¼ GM

4a2

At the equator the deviation of the vertical with respect to the radial direction is

tan i ¼gy

gr¼

8

45

i ¼ 10:08


there are two spheres of radius a/2 and density 2r with centres located on the axis of

rotation in each hemisphere. If M is the mass of the sphere of radius a, calculate:

(a) ThepotentialU(r,u) and the formof the equipotential surface passing through thePoles.

(b) The component gr of gravity in the first-order approximation for points on the surface.

(c) Calculate gr directly at the Pole and the equator, and compare with the first-order

approximation.

(a) This problem is similar to Problem 16, but now the density of the two spheres is the

same. The total gravity potential is the sum of the gravitational potentials of the

three spheres (V, V1 and V2) plus the potential due to the rotation F:

U ¼ V þ V1 þ V2 þ F ð18:1Þwhere

V ¼GM

r

V1 ¼GM 0

q1

V2 ¼GM 0

q2

F ¼1

2o2

r2sin2y

30 Gravity

and whereM is the mass of the large sphere of radius a andM 0 the differential mass of each

of the small spheres of radius a/2, r is the distance from a point P to the centre of the large

sphere, and q1 and q2 the distances from P to the centres of the small spheres (Fig. 18). As

in Problem 16 the differential mass is given by the difference in density between the large

and the small spheres:

M 0 ¼4

3p 2r rð Þ

a

2

3

¼M

8

The inverse of the distance 1/q can be approximated by

1

q1¼

1

r1þ

a

2rcos yþ

a

2r

2 1

23cos2y 1

and since cos y0 = cos y

1

q2¼

1

r1

a

2rcos yþ

a

2r

2 1

23cos2y 1

The potential of the rotation can be written in terms of the coefficient m = a3o2/GM,

F ¼GM

r

1

2

1

a3a3o2

GMr3

sin2y ¼GM

r

m

2

r

a

3

sin2y

Substituting in Equation (18.1)

U ¼GM

r

10

8þ1

8

a

2r

2

3cos2y 1

þr

a

3 m

2sin2y

ð18:2Þ

r

a

a/2

2r

2r

q

q’

q2

q1

a

r

P

q

Fig. 18


At the Poles, r ¼a and y = 0, and the potential is

Upoles ¼GM

a

10

8þ

2

32

¼21

16

GM

a

The form of the equipotential surface which passes through the Poles (r ¼ a) is obtained

from Equation (18.2)

r ¼GM

Upoles

10

8þ1

8

a

2r

2

3cos2y 1

þr

a

3 m

2sin2y

Making the approximation r ¼ a in the right-hand side:

r ¼32

42a

10

8þm

2sin2yþ

1

323cos2y 1

and substituting cos2 y = 1 sin2 y, we obtain

r ¼ a 132

42

3

32m

2

sin2y

This is the equation of an ellipse with flattening a ¼ (32/42)(3/32m/2). Since there is

symmetry with respect to the axis of rotation, the equipotential surface is an ellipsoid of

revolution.

At the poles: y ¼ 0 ) rp ¼ a.

At the equator: y ¼ 90 ) re ¼ a 1þ32

42

m

2

3

32

Depending on the value of m, we have the following cases,

m

2¼

3

32) re ¼ a ) sphere

m

2<

3

32) re < a ) prolate ellipsoid

m

2>

3

32) re > a ) oblate ellipsoid

m ¼ 0 ) r ¼ a 13

42

< a ) prolate ellipsoid

(b) For the gravity at the Pole, in the first-order approximation, we take the derivative

of the potential (18.2) and substitute y ¼ 0 ) rp ¼ a:

gr ¼@U

@r¼

GM

a210

8

6

32

¼ 1:4375GM

a2

(c) The exact solution for the gravity at the pole is the sum of the attractions of the

three spheres:

gr ¼ GM

a2GM

2a2

GM

18a2¼ 1:5555

GM

a2

32 Gravity

At the equator we take the derivative of the potential and substitute r = a and

y = 90:

gr ¼@U

@r¼

GM

a210

8þ

3

32þ m

¼ GM

a237

32 m

¼ GM

a21:1562 m½

For the exact solution we write

gr ¼ GM

a22GM

8q2cos aþ o2a

From Fig. 18 the distance q is given by

q2 ¼a2

4þ a2 ¼

5

4a2

cos a ¼

ffiffiffi

4

5

r

Therefore

gr ¼ GM

a21þ

8

40

ffiffiffi

4

5

r

m

" #

¼ GM

a21:1789 m½

The approximated values are smaller than the exact solutions.

19. For the case of Problem 18, if GM = 4 103 m3 s2, a = 6 103 km, and v = 7

105 s1, calculate the values of J2, a, m, H, and b.

From the definition of m we obtain

m ¼a3o2

GM¼

216 1018 49 1010

4 1014¼ 2:6 103

The value of J2 is obtained by comparing the two expressions for the potential U

(Problem 18):

U ¼GM

r1þ

a

r

2

J21

23cos2y 1

þm

2

r

a

3

sin2y

U ¼5

4

GM

r1þ

1

40

a

r

2

3cos2y 1

þm

2

r

a

3

sin2y

Then, J2 = 0.05.

The flattening is obtained from the relation

a ¼3

2J2 þ

m

2

a ¼3

2 50 103 þ

2:3 103

2¼ 0:0765


The gravity flattening is given by

b ¼gp ge

ge¼

1:555 1:175

1:175¼ 0:323

where we have used the values of gravity at the Pole and equator obtained in Problem 18,

and in the latter we have substituted the value obtained for the coefficient m.

The dynamic ellipticity H is defined as the ratio of the moments of inertia with respect to

the polar and equatorial radius (Fig. 19a):

H ¼C A

C

where A and C are the moments of inertia of a sphere respect to the polar and equatorial

radi: (axes x1 and x3). The moment of inertia of a sphere of radius R is

Isph ¼2

5MR2

We have to add to the moment of inertia of the sphere of radius a the moments of inertia of

the two internal spheres of radius a/2. For the C-axis (x3) we have

IC ¼ Isph a þ 2Isph a2¼

2

5Ma2 þ 2

2

5

M

8

a2

4¼ 0:425Ma2

For the A-axis (x1) the moment of inertia of each of the small spheres is given by

(Fig. 19b)

X3

X1

2r

2r

r

a/2

a

Fig. 19a

34 Gravity

I ¼ ICM þMh2

since in this case the A axis does not coincide with the centre of mass, where R ¼ a/2

and h ¼ a/2:

Isph a=2 ¼2

5

M

8

a2

4þM

8

a2

4¼ 0:044Ma2

IA ¼ Isph a þ 2Isph a=2 ¼2

5Ma2 þ 2 0:044Ma2 ¼ 0:488Ma2

Finally

H ¼C A

C¼

IC IA

IC¼

0:425 0:488

0:425¼ 0:147


there is a sphere of radius a/2 and density 5r centred at the midpoint of the northern-

hemisphere polar radius. If m ¼ 1/8 and M is the mass of the sphere of radius a,

determine:

(a) The form of the equipotential surface passing through the North Pole.

(b) For latitude 45º, the astronomical latitude and the deviation of the vertical from

the radial.

(a) The gravitational potential is the sum of the potentials for the sphere of radius a

and that of the sphere of radius a/2 (Fig. 20):

V ¼ V1 þ V2 ¼GM

rþGM1

q

As in the previous problems the potential of the small sphere is given in terms of

differential mass M1:

a

R

h

b

Fig. 19b


M1 ¼4

3p 5r rð Þ

a

2

3

¼M

2

and for the inverse of the distance 1/q we use the approximation

1

q¼

1

r1þ

a

2rcos yþ

a

2r

2 1

23cos2y 1

Then, the total gravitational potential is

V ¼GM

r

3

2þ

a

4rcos yþ

1

16

a

r

2

3cos2y 1

The total potential U is the sum of the gravitational potential V plus the potential of rotation

F, where

F ¼1

2r2o2sin2y

and using the coefficient m = o2a3/GM = 1/8, we have

U ¼GM

r

3

2þ

a

4rcos yþ

1

16

a

r

2

3cos2y 1

þr

a

3 m

2sin2y

At the North Pole, y = 0 and r ¼ a, and the value of the potential is

Up ¼GM

a

3

2þ1

4þ1

8

¼15GM

8a

a/2

a

5r

r

q j

qg

gr

gq

fa

i

P

Fig. 20

36 Gravity

The form of the equipotential surface is found by putting U ¼ Up:

15GM

8a¼

GM

r

3

2þ

a

4rcos yþ

1

16

a

r

2

3cos2y 1

þr

a

3 m

2sin2y

Putting r ¼ a inside the square brackets and solving for r we find

r ¼ a4

51þ

1

6cos yþ

1

12cos2y

(b) The deviation of the vertical with respect to the radial direction is given by the angle i:

tan i ¼gy

gr

To find this value we have to calculate the two components of gravity

gr ¼@U

@r¼ GM

3

2

1

r22a

4

1

r3cos y

3a2

16

1

r43 cos2 y 1

þ2r

16a3sin2 y

gy ¼1

r

@U

@y¼

GM

r

a

4r2sin y

a2

16r36 cos y sin yþ

r2

8a3sin y cos y

For a point on the surface we put r ¼a:

gr ¼ GM

a219

16þ1

2cos yþ

11

16cos2y

gy ¼ GM

a21

4sin yþ

1

4sin y cos y

and for latitude 45º

gr ¼ GM

a21:88

gy ¼ GM

a20:30

Then the angle i is given by tan i ¼0:30

1:88) i ¼ 9:0.

The astronomical latitude is

fa ¼ 90 y i ¼ 36:0

21. If the internal sphere of Problem 20 is located on the equatorial radius at

longitude zero, find expressions for the components of gravity: gr , gu , gl.

As in the previous problem the differential mass of the small sphere M1 is (Fig. 21a):

M1 ¼M

2

The total potential U is the sum of the gravitational potentials V and V1, and the potential

due to rotation F. According to Fig. 21b, using the relations of spherical triangles, if ’ and

l are the coordinates of the point where the potential is evaluated, then


cosc ¼ cos 90º cos(90º’)þ sin 90º sin(90º’) cosl

cosc ¼ cos’ cosl

Using the expression for 1/q as in Problem 16,

1

q¼

1

r1þ

a

2rcoscþ

a

2r

2 1

23cos2c 1

The gravitational potential of the small sphere is given by

V1 ¼GM

2r1þ

a

2rcos’ cos lþ

a2

8r23cos2’cos2l 1

The total potential U is given by:

U ¼GM

r

3

2þ

a

4rcos’ cos lþ

a2

16r23cos2’cos2l 1

þ1

2r2o2cos2’

l = 0° A

l

a/2

y

P

B

rq

j

Fig. 21a

90° 90°– j

y

l

B

P

A

Fig. 21b

38 Gravity

Using the coefficient m and sin y = cos ’, we obtain

U ¼ GM3

2rþ

a2

4r2sin y cos lþ

a2

16r33sin2ycos2l 1

þr2

a3m

2sin2y

The three components of gravity are found by differentiating U with respect to r, y, and l:

gr ¼@U

@r¼ GM

3

2r2

a2

2r3sin y cos l

3a2

16r43sin2ycos2l 1

þr

a3msin2y

gy ¼1

r

@U

@y¼ GM

a2

4r3cos y cos lþ

a2

16r46 cos l sin y cos yþ

r

a3m sin y cos y

gl ¼1

r sin y

@U

@l¼ GM

a2

4r3sin l

3a2

8r4sin y cos l sin l

22. A planet is formed by a sphere of radius a and density r, with a spherical core of

density 5r and radius a/2 centred on the axis of rotation in the northern hemisphere

and tangential to the equator. The planet rotates with m ¼ 1/4. For the point at

coordinates (45º N, 45º E), calculate:

(a) The astronomical latitude.

(b) The deviation of the vertical from the radial.

(c) The angular velocity of rotation that would be required for this deviation to be zero.

(a) The gravitational potential is the sum of the potentials of the two spheres (Fig. 22):

V ¼ V1 þ V2 ¼GM

rþGM 0

qð22:1Þ

5r

a

r

a/2

q

ggr

gq

r

j

P

i

Fig. 22


As in Problem 16, the inverse of the distance from a point P to the centre of the small

sphere, 1/q, can be approximated by

1

q¼

1

r1þ

a

2rcos yþ

1

2

a

2r

2

3cos2y 1

As in previous problems we use the differential mass of the small sphere,

M 0 ¼4

3pa3

8ð5r rÞ ¼

2

3pra3

and since M ¼4

3pra3, then M 0 = M/2.

Substituting in Equation (22.1) we obtain

V ¼GM

rþGM

2

1

rþ

a

2r2cos yþ

1

2

a2

4r33cos2y 1

¼ GM3

2rþ

a

4r2cos yþ

a2

16r33cos2y 1

The total potential U is the sum of V plus the potential due to rotation F:

U ¼ V þ F ¼GM

r

3

2þ

a

4rcos yþ

a2

16r23cos2y 1

þ1

2r2o2sin2y

The components of gravity gr and gy are

gr ¼@U

@r¼ GM

3

2r2

a

2r3cos y

3a2

16r43cos2y 1

þ ro2sin2y ð22:2Þ

gy ¼1

r

@U

@y¼ GM

a

4r3sin y

a2

16r46 cos y sin y

þ ro2 sin y cos y ð22:3Þ

For a point on the surface of the large sphere and coordinates 45º N, 45º E, we have that

y = 45, r ¼ a, and

m ¼a3o2

GM¼

1

4

Putting these values in (22.2) and (22.3), we obtain

gr ¼ 1:82GM

a2and gy ¼ 0:24

GM

a2

To find the astronomical latitude we first have to find the deviation of the vertical with

respect to the radial:

tan i ¼gy

gr¼ 0:13 ; i ¼ 7:5

The astronomical latitude is, then, f = ’ i = 45 7.5 = 37.5.

(b) Thedeviation of the verticalwith respect to the radial direction, as already found, is i=7.5.

(c) If we want the deviation of the vertical to be null, i ¼ 0, this implies gy = 0.

40 Gravity

writing Equation (22.3) in terms of the coefficient m, where

m ¼a3o2

GMð22:4Þ

we have

gy ¼ 0 ¼GM

a22ffiffiffi

2p

þ 3

16m

2

and solving for m gives m ¼ 0.73.


o ¼


GM

a3

r

0:85

23. A planet consists of a very thin spherical shell of mass M and radius a, within

which is a solid sphere of radius a/2 and mass M 0 centred at the midpoint of the

equatorial radius of the zero meridian. The planet rotates with angular velocity v

about an axis normal to the equatorial plane. Calculate:

(a) The potential at points on the surface as a function of latitude and longitude.

(b) The components of the gravity vector.

(c) If M 0 ¼ 10 M, what is the ratio between the tangential and radial components of

gravity at the North Pole?

(a) The potentialU is the sumof the gravitational potentials due to the spherical shellV1, and

to the interior sphere V2, plus the potential due to the rotation of the planet F (Fig. 23):

U ¼ V1 þ V2 þ F

F ¼1

2o2

r2cos2’

V1 ¼GM

r

V2 ¼GM 0

q

ð23:1Þ

where r is the distance from a point P on the surface of the planet to its centre, q is the

distance from point P to the centre of the interior sphere, and ’ the latitude of point P.

Using the cosine law,

q ¼


r2 þ

a2

4 ar cosc

r

where c is the angle between r and the equatorial radius, and its inverse can be approxi-

mated by (Problem 16)

1

q¼

1

r1þ

a

2rcoscþ

a2

8r23cos2c 1

ð23:2Þ

Using the relation for spherical triangles

cos a ¼ cos b cos cþ sin b sin c cosA


putting b ¼ 90º ’, c ¼90º, A ¼ l, and a ¼ c, where l is the longitude of P, then

cos c = cos ’ cos l

Substituting in (23.1), the potential due to the small sphere is

V2 ¼ GM 0

r1þ

a

2rcos’ cos lþ

a2

8r23cos2’cos2l 1

The total potential U is

U ¼GM

rþGM 0

r1þ

a

2rcos’ cos lþ

a2

8r23cos2’cos2l 1

þ1

2o2

r2cos2’

ð23:3Þ

(b) The components of the gravity vector are obtained from Equation (23.3):

gr ¼@U

@r

¼GM

r2

þ GM 0 1

r2

a

r3cos’ cos l

3a2

8r43cos2’cos2l 1

þ o2rcos2’

gy ¼1

r

@U

@y¼

1

r

@U

@’

¼GM 0

r

a

2r2sin’ cos lþ

a2

8r36 cos’cos2l sin’

þ o2r cos’ sin’

a

l

l

a/2

y

y

r q

j

P

90°–

j

x

Fig. 23

42 Gravity

gl ¼1

r cos’

@U

@l¼

GM 0

r cos’

a

2r2cos’ sin l

a2

8r36cos2’ cos l sin l

(c) At the North Pole, ’ = 90 and r ¼ a. Putting M 0 ¼10M and substituting in the

previous equations we obtain

gr ¼ GM

a2GM 0

a2þ3GM 0a2

8a4¼ 7:25

GM

a2

gy ¼GM 0

2a2¼

5GM

a2

The ratio between the radial and the tangential components of gravity at the North

Pole is

gr

gy¼ 1:45

24. An Earth consists of a sphere of radius a and density r, within which there are two

spheres of radius a/2 centred on the axis of rotation and tangent to each other. The

density of that of the northern hemisphere is 4r and that of the southern hemisphere

is r/4.

(a) Express the gravitational potential in terms ofM (the mass of the large sphere) up

to terms of 1/r3.

(b) What astronomical latitude corresponds to points on the equator (without

rotation)?

(c) What error is made by using the 1/r3 approximation in calculating the value of gr

at the equator?

(a) The total gravitational potential V is the sum of the potentials of the sphere of

radius a (V0) and of the two spheres of radius a/2 situated in the northern (V1) and

southern (V2) hemispheres (Fig. 24):

V ¼ V0 þ V1 þ V2

As in previous problems the large sphere is considered to have uniform density

r and the effect of the two interior spheres is calculated using their differential

masses

M ¼4

3pra3

M1 ¼4

3p 4r rð Þ

a3

8¼

3M

8

M2 ¼4

3p

r

4 1

a3

8¼

3M

32

The potentials are


V0 ¼GM

r

V1 ¼GM1

r1

¼3GM

8

1

rþ

a

2r2cos yþ

a2

8r33cos2y 1

V2 ¼GM2

r2

¼ 3GM

32

1

r

a

2r2cos yþ

a2

8r33cos2y 1

where r1 and r2 have been calculated as in Problem 16. Then, the total gravitational

potential in the 1/r3 approximation is

V ¼ GM41

32rþ15

64

a

r2cos yþ

9

256

a2

r3

3cos2y 1

(b) The components of the gravity vector, taking into account that there is no rotation, are

gr ¼@V

@r¼ GM

41

32r2

15a

32r3cos y

27a2

256r43cos2y 1

gy ¼1

r

@V

@y¼ GM

15a

64r3sin y

27a2

128r3cos y sin y

ð24:1Þ

At the equator, r ¼ a and y = 90 and we obtain

r

r/4

4r

a/2

a

r2

r1

a

x

x

Fig. 24

44 Gravity

gr ¼ 1:175GM

a2

gy ¼ 0:243GM

a2

The astronomical latitude (’a) is the angle between the vertical and the equatorial plane. In

our case at the equator this is given by the deviation of the vertical from the radial

direction:

tan’a ¼gy

gr¼ 0:207

Then

’a ¼ 11:68 N

(c) If we want to calculate the exact value of gr at the equator, we calculate the exact

attractions of each sphere and add them:

g0r¼

GM

a2g1r¼

3GM

8r21cos a g2

r¼

3GM

32r22cos a ð24:2Þ

where r1 and r2 are the distances from the centre of each of the two interior

spheres (Fig. 24):

r1 ¼ r2 ¼


a2 þa2

4

r

¼affiffiffi

5p

2

and a is the angle which r1 and r2 form with the equatorial plane:

sin a ¼a=2

r1

¼1ffiffiffi

5p

The radial component of gravity is given by

gTr¼ g0

rþ g1

rþ g2

r¼ 1:335

GM

a2

The error we make using the approximation is

gapprox gexact ¼ 0:160GM

a2; that is; 16%:

25. An Earth consists of a sphere of radius a and density r within which there are

two spheres of radius a/2 centred on the axis of rotation and tangent to each other.

The density of that of the northern hemisphere is 2r and that of the southern

hemisphere is r/2.

(a) Express the potential V in terms ofM (the mass of the large sphere), G, and r up to

terms in 1/r3.


(b) According to the value of this potential V, which astronomical latitudes corres-

pond to the geocentric latitudes 45º N and 45º S?

(c) What must the rotation period be for the astronomical and geocentric latitudes to

coincide?

(d) What error is made by the 1/r3 approximation in calculating the value of gr at the

equator? And at the North Pole?

(a) As in previous problems the effect of the interior spheres is given in terms of their

differential masses (Fig. 25):

M1 ¼4

3p 2r rð Þ

a

2

3

¼M

8

M2 ¼4

3p

r

2 r

a

2

3

¼ M

16

The distances q1 and q2 from the centre of each sphere to an arbitrary point P are found

using the cosine law:

q1 ¼ r2 þ

a2

4 2

ar

2cos y

q2 ¼ r2 þ

a2

4þ 2

ar

2cos y

r

r

2r

a/2

a

ja

qj

b

gr

gq gr

r/2

q2q1

x

x

P

Fig. 25

46 Gravity

Using the approximation for 1/q (Problem 16), the total gravitational potential V is the sum

of the potentials of the three spheres:

V ¼GM

rþGM1

q1þGM2

q2

¼GM

r

17

16þ

3

32

a

rcos yþ

1

128

a

r

2

3cos2y 1

(b) The components of the gravity vector are given by

gr ¼@V

@r¼ GM

1

r2

17

16þ

6

32

a

r3cos yþ

3

128

a2

r43cos2y 1

gy ¼1

r

@V

@y¼

GM

r

3

32

a

r2sin yþ

1

128

a2

r36 cos y sin y

ð25:1Þ

If the point P is at the surface, r ¼ a, then

gr ¼ GM

a217

16þ

3

16cos yþ

3

1283cos2y 1

gy ¼ GM

a23

32sin yþ

6

128cos y sin y

At geocentric latitude 45º N, y ¼ 45º,

gr ¼ 1:21GM

a2

gy ¼ 0:09GM

a2

The deviation of the vertical with respect to the radial direction i is given by

tan i ¼gy

gr¼ 0:074 ) i45 ¼ 4:2

According to Fig. 25, the astronomical latitude ’a can be determined from the deviation of

the vertical i,

’a þ iþ 180 ’ð Þ ¼ 180 ) ’a ¼ ’ i ¼ 45 4:2 ¼ 40:8 N

In the same way, for geocentric latitude 45º S (y = 135)

gr ¼ 0:94GM

a2

gy ¼ 0:04GM

a2

Then tan i135 ¼gy

gr¼ 0:043 ) i135 ¼ 2:5

Then the astronomical latitude is

’a ¼ 45 2:5 ¼ 47:5 ¼ 47:5 S


(c) If we want the astronomical and geocentric latitudes to coincide, then the

deviation of the vertical must be null, i ¼ 0º. This implies that gtotaly must be

zero. To do this by means of the rotation, we have to make the tangential

component of the centrifugal force gRy be equal and of opposite sign to that of

the gravitational potential gVy :

gytotal ¼ gy

V þ gyR ¼ 0 ) gy

V ¼ gyR

The tangential component due to rotation is

gRy ¼1

r

@F

@y

where F ¼1

2o2

r2sin2y. Then

gRy ¼ o2r cos y sin y

For a point on the surface at latitude 45º N, r = a and y = 45, so

gyV ¼ gy

R ) 0:09GM=a2 ¼ o2 a=2

From here we can calculate the period of rotation

T ¼2p

o¼

2pffiffiffiffiffiffiffiffiffi

0:18p


a3

GM

r

For a point at latitude 45º S, r = a and y = 135, so

gRy ¼ gVy ) 1

2o2a ¼ 0:04

GM

a2) T ¼

2pffiffiffiffiffiffiffiffiffi

0:08p


a3

GM

r

(d) The value of the radial component of gravity at the equator, r ¼ a, y = 90, by

substitution in (25.1), is

gr ¼ 1:04GM

a2

If we calculate the exact value by adding the contributions of the three spheres

(Fig. 25)

gexactr

¼ gMrþ g1

rþ g2

r

gMr

¼ GM

a2

g1r¼ g1 cos b

g2r¼ g2 cos b

where

48 Gravity

cos b ¼affiffiffi

5

4

r

a

¼2ffiffiffi

5p

g1 ¼ GM

8

4

5a2

g2 ¼GM

16

4

5a2

gexactr

¼ GM

a20:96

The error in the approximation is:

gerrorr

¼ 0:96GM

a2 1:04

GM

a2

¼ 0:08GM

a2

In a similar way, for a point at the North Pole, r ¼ a, y ¼ 0º:

gr ¼ 1:30GM=a2

gexactr

¼ gMrþ g1

rþ g2

r

gMr

¼ GM

a2

g1r¼

GM

8a

2

2¼

GM

2a2

g2r¼

GM

16 aþa

2

2¼

GM

36a2

gexactr

¼ GM

a21:47

The error in the approximation is

gerrorr

¼ 1:47GM

a2 1:30

GM

a2

¼ 0:17GM

a2

26. A sphericalEarth of radius ahas a core of radius a/2whose centre is displaced a/2 along

the axis of rotation towards the North Pole. The core density is twice that of the mantle.

(a) What should the period of rotation of the Earth be for the direction of the plumb-

line to coincide with the radius at a latitude of 45º S?

(b) What are the values of J0, J1, J2, and m?

(a) As in previous problems we calculate the gravitational potential by the sum of the

potentials of the two spheres, using for the core the differential mass (Fig. 26):

V ¼ V1 þ V2 ¼GM

rþGM 0

q

M 0 ¼4

3p 2r rð Þ

a2

8¼

M

8


As we saw in Problem 16, we use for 1/q the first-order approximation

V ¼GM

rþGM

8

1

rþ

a

2r2cos yþ

1

2

a2

4r33cos2y 1

ð26:1Þ

The total potential U is the sum of the gravitational potential V and the potential due to

rotation

F ¼1

2r2o2sin2y

U ¼ GM9

8

1

rþ

a

16r2cos yþ

a2

64r33cos2y 1

þ1

2r2o2sin2y

In order that the direction of the plumb-line coincides with the radial direction, the

tangential component of gravity, gy, must be null:

gy ¼1

r

@U

@y¼GM

r

a

16

1

r2sin y

a2

64

1

r36 cos y sin y

þ ro2 sin y cos y

For a point on the surface at latitude 45º S, the tangential component of gravity is, with

r = a, y = 135,

gy ¼ 0:003GM

a2ao2

2

a

r

q j

r

gq

gr

i

g

q

x

45°

2r

a/2 P

Fig. 26

50 Gravity

Putting this component equal to zero, we find the value of the period of rotation T:

gy ¼ 0:003GM

a2ao2

2¼ 0 ) o2 ¼ 0:006

GM

a3) T ¼

2p

0:077


a3

GM

r

(b) The gravitational potential V of Equation (26.1) can be written as

V ¼GM

r

9

81þ

a

18rcos yþ

a2

72r23cos2y 1

We obtain the values of J1 and J2 by comparison with the equation

V ¼GM

rJ0 þ J1

a

rP1 þ J2

a

r

2

P2

Since the total mass is (9/8)M, we obtain

J0 ¼ 1

J1 ¼1

18

J2 ¼1

72

27. Within a spherical planet of radius a and density r there are two spherical cores of

radius a/2 and density r 0 with centres located on the axis of rotation at a/2 from the

planet’s centre, one in the northern hemisphere and the other in the southern hemisphere.

(a) Neglecting rotation of the planet, calculate what the ratio r0/r should be for the

gravity flattening to be 1/8.

(b) If the planet rotates so that m ¼ 1/16, and the ratio of the densities is that found in

part (a), calculate the astronomical latitude which corresponds to the geocentric

latitude 45º N.

(a) Since there is no rotation the total potential U is the sum of the gravitational

potentials of the three spheres (Fig. 27). As in previous problems we use the mass

M of the planet with uniform density r and for the two cores the differential

masses M 0. For 1/q we use the approximation as in Problem 16:

M 0 ¼4

3pa3

8r0 rð Þ ¼

4

3pa3

8r0 rð Þ

r

r¼

M

8

r0

r 1

ð27:1Þ

The potential U is

U ¼GM

rþGM 0

r2þ

a

2r

2

3cos2y 1

The radial components of gravity at the equator and the Pole are found by taking the

derivative of the potential U:

gr ¼@U

@r¼

GM

r2

þ GM 0 2

r23a2

4r43cos2y 1


On the surface r ¼ a, and at the equator y = 90 and at the Pole y = 0, so

ger¼

GM

a2GM 0

a25

4

gpr¼

GM

a2GM 0

a214

4

The gravity flattening is given by

b ¼gp ge

ge¼

1

8

By substituting the values of gravity we find the relation between M and M 0:

1

8¼

M 7

2M 0 þM þ

5

4M 0

M 5

4M 0

) M ¼67

4M 0

Putting M 0 in terms of M from Equation (27.1) we find the ratio of the densities:

M 0 ¼4

67M ¼

M

8

r0

r 1

)r0

r¼ 1:48

(b) For a rotating planet we add to the potential U the rotational potential, F:

U ¼GM

rþGM 0

r2þ

a

2r

2

3cos2y 1

þGM

r

r

a

3 m

2sin2y

The radial and tangential components of gravity are now

gr ¼@U

@r¼

GM

r2

þ GM 0 2

r23a2

4r43cos2y 1

þ GMrm

a3sin2y

¼ 1:11GM

r2

gy ¼1

r

@U

@y¼GM 0

r2

a

2r

2

6 cos y sin y

þGM

r2

r

a

3

m sin y cos y

¼ 0:013GM

r2

From Fig. 27 we see that the relation between the geocentric and astronomical latitudes is

’a ¼ ’ i

where i is the deviation of the vertical with respect to the radial direction, which is given by

tan i ¼gy

gr¼ 0:012 ) i ¼ 0:7

52 Gravity

Then the astronomical latitude for geocentric latitude 45º is

’a ¼ 45 0:7 ¼ 44:3

Gravity anomalies. Isostasy

28. For two-dimensional problems, the gravitational potential of an infinite horizontal

cylinder of radius a is

V ¼ 2pGra2 ln1

r

where r is the distance measured perpendicular to the axis. Assume that a horizontal

cylinder is buried at depth d as measured from the surface to the cylinder’s axis.

(a) Calculate the anomaly along a line of zero elevation on the surface perpendicular

to the axis of the cylinder.

(b) At what point on this line is the anomaly greatest?

(c) What is the relationship between the distance at which the anomaly is half the

maximum and the depth at which the cylinder is buried?

(d) For a sphere of equivalent mass to produce the same anomaly, would it be at a

greater or lesser depth?

r

r

r’

a

x

x

a/2

ja

ggr

gq

j = 45°

i

Fig. 27

53 Gravity anomalies. Isostasy

(a) The gravity anomaly produced by an infinite horizontal cylinder buried at depth d, with

centre at x ¼ 0 (Fig. 28), is given by the derivative in the vertical direction (z-axis) of

the gravitational potential V:

V ¼ 2pGra2 ln1


x2 þ ðzþ dÞ2q

0

B

@

1

C

A

g ¼ gz ¼ @V

@z¼

2prGa2d

x2 þ ðzþ dÞ2ð28:1Þ

For points on the surface (z ¼ 0):

g ¼2prGa2d

x2 þ d2

(b) To find the point at which the anomaly has its maximum value, we take the

derivative with respect to x and put it equal to zero:

@g

@x¼ 0 ) 2prGa2d2x ¼ 0 ) x ¼ 0

Substituting x ¼ 0 in (28.1):

gmax ¼2pGra2

d

(c) The distance at which the anomaly has a value equal to half its maximum value

gives us the depth d at which the cylinder is buried:

x

x

d

a

r

gr

gz

P

r

∆g

x

∆gmax

x1/2

(1/2)∆gmax

Fig. 28

54 Gravity

gmax

2¼ g )

2pGra2

2d¼

2pGra2d

x2 þ d2) x1=2 ¼ d

(d) The gravitational potential produced by a sphere of differential mass DM buried at

depth d under x ¼ 0 is given by

V ¼GM

r¼

GM

x2 þ ðzþ dÞ2 1=2

The gravity anomaly is

gz ¼@V

@z¼

GMðzþ dÞ


and for a point on the surface z ¼ 0,

gz ¼GMd

x2 þ d2ð Þ3=2

The maximum value for x ¼ 0 is

gmax ¼GM

d2

The distance at which the anomaly has half its maximum value is

GMd

x21=2 þ d2

3=2¼

GM

2d2

x1=2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

22=3 1p

d ¼ 0:766d

Therefore, the sphere is at a greater depth than the cylinder.

29. At a point at latitude 42º 290 19 00 and height 378.7 m the value of gravity is

observed to be 980 252.25 mGal. Calculate in gravimetric units (gu):

(a) The free-air anomaly.

(b) The Bouguer anomaly if the density of the crust is 2.65 g cm3.

(a) We first calculate the normal or theoretical value of gravity given by the expression

g ¼ 9:7803268 1þ 0:00530244sin2’ 0:0000058sin22’

ms2

We substitute for ’ its value 42º 290 1900 and obtain

g ¼ 9:803 9299m s2

The free-air anomaly, using the free-air correction, is

gFA ¼ g þ 3:086h g ¼ 238:7 gu


(b) The Bouguer anomaly is calculated from the free-air anomaly using the Bouguer

correction with a crust density of 2.65 g cm3:

gB ¼ g þ 3:086h 2pGrh g ¼ gFA 2pGrh ¼ 659:3 gu

30. An anomalous mass is formed by two equal tangent spheres of radius R, with

centres at the same depth d ( d R ) and density contrast Dr.

(a) Calculate the Bouguer anomaly at the surface (z ¼ 0) produced by the mass

anomaly along a profile passing through the centres of the two spheres.

(b) Represent it graphically for x ¼ 0 (above the tangent point), 500, 1000, and 2000 m

taking R ¼ 1 km, d ¼ 3 km, and Dr ¼ 1 g cm3.

(a) For one sphere the anomaly for points on the surface (z ¼ 0) is (Problem 28)

g ¼GMd

x2 þ d2ð Þ3=2

For two spheres the anomaly is the sum of the attractions of the two spheres (Fig. 30a):

g ¼GMd

r31

þGMd

r32

where

r1 ¼


x Rð Þ2 þ d2q

r2 ¼


xþ Rð Þ2 þ d2q

Then

g ¼GMd

x Rð Þ2þd2h i3=2

þGMd

xþ Rð Þ2þd2h i3=2

R

d r2r1

xP(x,0)

x x

Fig. 30a

56 Gravity

(b) To represent graphically the curve of the anomaly (Fig. 30b), we first find the point

at which it is a maximum:

@g

@x¼

@

@x

GMd

x Rð Þ2þd2h i3=2

þGMd

xþ Rð Þ2þd2h i3=2

0

B

@

1

C

A¼ 0

xþ Rð Þ2 x Rð Þ2 þ d2h i5

¼ x Rð Þ2 xþ Rð Þ2 þ d2h i5

) x ¼ 0maximum

Using the data given in the problem, we find the values of the anomaly for the five points,

with Dr = 1 g cm3, R = 1 km, d = 3 km

M ¼4

3pR3

r ¼4

3p 109 103 ¼ 4:19 1012 kg

x (m) Dg (gu)

0 53.0

500 52.0

1000 48.9

1500 43.9

2000 37.5

55

50

45

Anom

aly

(g

µ)

40

–2000 –1000 0

Distance (m)

1000 2000

Fig. 30b


31. At a point at geocentric latitude 45º N and height 2000 m the observed value of

gravity is g ¼ 6690 000 gu. Taking the approximation that the Earth is an ellipsoid of

equatorial radius a ¼ 6000 km, density ¼ 4 g cm3, J2 ¼ 103, and m ¼ 103,

calculate for that point:

(a) The free-air and the Bouguer anomalies.

(b) The distance from the free surface to that of the sphere of radius a (precision 1 gu).

(a) The volume of an ellipsoid is:

V ¼4

3pa3 1þ 2að Þ

The flattening is

a ¼3J2

2þm

2¼ 2 103

and the mass is

Me ¼ Vr ¼4

3pa3 1þ 2að Þr ¼ 3:624 1024 kg

Using G = 6.67 1011 m3 kg1 s2

GM

a2¼

6:67 1011 3:624 1024

36 1012¼ 6:732 994m s2

The value of gravity at the equator in the first-order approximation is given by

ge ¼GM

a21þ

3

2J2 m

¼ 6:736 361m s2

For a point at latitude 45ºN the radial component of gravity is

gr ¼ ge 1þ b sin2 ’

¼ 6:738 045m s2

where we have used the value of the gravity flattening b given by

b ¼5

2m a ¼ 0:5 103

The free-air correction is

CFA ¼ 2GM

a3h ¼ 2:24 106h m s2 ¼ 2:244h gu

Then, the free-air anomaly at that point is

gFA ¼ g gþ CFA

¼ 6 690 000 6 738 045þ 2:244 2000 ¼ 43 557 gu

In order to calculate the Bouguer anomaly, we first calculate the Bouguer correction

CB ¼ 2pGrh ¼ 1:676 106 hms2 ¼ 1:676h gu

58 Gravity

Then, the Bouguer anomaly is:

gB ¼ g gþ CAL 2pGrh ¼ gFA CB

¼ 45 557 1:676 2000 ¼ 46 909 gu

(b) If we call N the distance at the given point between the free surface and the surface

of the sphere of radius a (Fig. 31), this is given by:

N ¼ a r h

where r is the radius of the ellipsoid at latitude 45º N which to a first approximation is

r ¼ a 1 asin2’

¼ 6000 1 2 103 1

2

¼ 5994 km

Then,

N ¼ 6000 5994 2 ¼ 4 km

32. Beneath a point A at height 400 m there exists an anomalous spherical mass of

radius 200 m, density 3.5 g cm3, whose centre is 200 m below the reference level.

A point B is located at a height of 200 m and a horizontal distance of 400 m from A,

and a third point C is at a height of 0 m and at a horizontal distance of 800 m from

A. The density of the medium above the reference level is 2.6 g cm3, and below the

reference level it is 2.5 g cm3. The theoretical value of gravity is 980 000 mGal.

Calculate:

(a) The values of gravity at A, B, and C.

(b) The Bouguer anomalies at these points.

a

a

h

N

r

Fig. 31


Precision 1 gu.

(a) The gravity at each point is given by

g ¼ g CFA þ CB þ Cam

where

Normal gravity: g ¼ 9800 000 gu

Free-air correction: CFA ¼ 3.086 h

Bouguer correction: CB ¼ 0.419 r1 h

r1 ¼ 2.6 g cm3 is the density of the material above the reference level

C am is the anomaly produced by the buried sphere at a point at height h and a horizontal

distance x from its centre:

Cam ¼GMðhþ dÞ

x2 þ ðhþ dÞ2 3=2

¼G4

3pR3ðrsph r2Þðhþ dÞ


ð32:1Þ

where d is the depth to the centre from the reference level; and rsph and r2 are the densities

of the sphere and of the medium where it is located, respectively. In our case: d ¼ 200 m,

rsph ¼ 3.5 g cm3, and r2 ¼ 2.5 g cm3.

For point A, x ¼ 0, we obtain

CFA ¼ 1234 gu CB ¼ 436 gu

Cam ¼GM

hþ dð Þ2¼

G4

3pR3ðrsph r2Þ

hþ dð Þ2¼ 6 gu

The value of gravity is gA ¼ 9 799 208 gu.

At point B:

CFA ¼ 617 gu

CB ¼ 218 gu

The anomaly produced by the sphere is calculated by Equation (32.1), substituting

x ¼ xB ¼ 400 m and h ¼ hB ¼ 200 m

C am ¼ 5 gu

We obtain gB ¼ 9 799 606 gu.

At point C:

The free-air and Bouguer corrections are null, because the point is at the reference level.

The anomaly due to the sphere, by substitution in Equation (32.1), x ¼ xC ¼ 800 m, and

h ¼ hC ¼ 0, is

C am ¼ 1 gu

The value of gravity is: gC ¼ 9800 001 gu.

(b) The Bouguer anomaly is given by

DgB ¼ g þ CFA– CB

– g

60 Gravity

By substitution of the values for each point we obtain that the anomalies correspond to

those produced by the sphere:

gBA ¼ 6 gu

gBB ¼ 5 gu

gBC ¼ 1 gu

33. For a series of points in a line and at zero height which are affected by the

gravitational attraction exerted by a buried sphere of density contrast 1.5 g cm3, the

anomaly versus horizontal distance curve has a maximum of 4.526 mGal and a point

of inflexion at 250 m from the maximum. Calculate:

(a) The depth, anomalous mass, and radius of the sphere.

(b) The horizontal distance to the centre of the sphere of the point at which the

anomaly is half the maximum.

(a) We know that the inflection point of the curve of the anomaly produced by a

sphere buried at depth d corresponds to the horizontal distance d/2. Then

xinf ¼d

2) d ¼ 2xinf ¼ 2 250 ¼ 500m

The maximum value of the anomaly at x ¼ 0 is

gmax ¼GM

d2¼ 4:526mGal ¼ 45:26 gu

and solving for DM

DM ¼45:26 106 m s2 5002 m2

6:67 1011 m3 s2 kg1¼ 1:6964 1011 kg

From this value we calculate the radius of the sphere:

M ¼4

3pR3

r ) R ¼3M

4pr

1=3

¼3 1:6964 1011 kg

4 3:14 1:5 103 kgm3

1=3

¼ 300m

(b) In order that g ¼ 12gmax with z ¼ 0 we write

GMd

x21=2 þ d2

3=2¼

1

2

GM

d2) x1=2 ¼ 383m

34. At a point at height 2000 m, the measured value of gravity is 9.794 815 m s2.

The reference value at sea level is 9.8 m s2. The crust is 10 km thick and of density

2 g cm3, and the mantle density is 3 g cm3. Calculate:

(a) The free-air, Bouguer, and isostatic anomalies. Use the Pratt hypothesis with a

cylinder of radius 10 km and a 40 km depth of compensation.

(b) If beneath this point there is a spherical anomalous mass of GDM ¼ 160 m3 s2 at

a 2000 m depth, what should the compensatory cylinder’s density be for the

compensation to be total?


(a) The free-air anomaly is

gFA ¼ g gþ CFA ¼ 9 794 815 9 800 000þ 3:086 2000 ¼ 987 gu


gB ¼ g gþ CFA CB ¼ gFA 0:4191rh

¼ 987 0:4191 2 2000 ¼ 689 gu

To calculate the isostatic anomaly (Fig. 34a) we begin with the calculation of the isostatic

correction assuming Pratt’s hypothesis and using only a vertical cylinder of radius 10 km

under the point and the compensation level at 40 km. In this way, the correction consists

of the gravitational attraction of a cylinder of radius a and height b at a point at distance c

from the base of the cylinder, which is given by

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð34:1Þ

where Dr is the contrast of densities, which according to Pratt’s hypothesis is given by

r ¼hr0Dþ h

ð34:2Þ

where r0 is the density for a block at sea level, which in our case is formed by a crust

of density 2 g cm3 and thickness 10 km over a mantle of density 3 g cm3 and thickness

30 km. For the whole 40 km we use a mean value of density

r0 ¼2 10þ 3 30

40¼ 2:75 g cm3

P

C a

b

Fig. 34a

62 Gravity

By substitution in (34.2) we obtain

r ¼2 2:75

42¼ 0:13 g cm3

The isostatic correction (34.1) is

CI ¼ 2 3:1416 6:67 1011 0:13 103b40þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

100þ 4p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

100þ 1764p

c 103

¼ 382 gu

Finally the isostatic anomaly is given by

gI ¼ g gþ CFA þ CB þ CI ¼ gB þ CI ¼ 689þ 382 ¼ 307 gu

(b) If under the point considered there is an anomalous spherical mass (Fig. 34b) at

depth d ¼ 2 km, the anomaly it produces is

gam ¼GM

d2¼

160

2000ð Þ2¼ 40 gu

The total anomaly now is the Bouguer anomaly plus the anomaly due to the sphere:

g ¼ 689 40 ¼ 729 gu

If the isostatic compensation is total (isostatic anomaly equal to zero), this anomaly must

be compensated by the cylinder. Thus, the necessary contrast of densities Dr to do this can

be calculated using expression (34.1):

729 106 ¼ 2 3:1416 6:67 1011

40þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

100þ 4p


100þ 1764p

103 r

10 km

D = 40 km

P

h

drc= 2 g cm–3

rM= 3 g cm–3

Fig. 34b


so

r ¼ 0:25 g cm3

As the mean value (crust–mantle) of the density is 2.75 g cm3, the density of the cylinder

must now be

r0 r ¼ r ¼ 0:25 ¼ 2:75 r ) r ¼ 2:50 g cm3

35. At a point on the Earth at height 1000 m, the observed value of gravity is 979 700

mGal. The value at sea level is 980 000 mGal.

(a) Calculate the free-air and Bouguer anomalies.

(b) According to the Airy hypothesis, which is the state of compensation of that height?

(c) What should the depth of the root be for the compensation to be total?

To calculate the compensation, use cylinders of radius 40 km, crustal thickness

H ¼ 30 km, crust density 2.7 g cm3, and mantle density 3.3 g cm

3.

(a) The free-air anomaly is

gFA ¼ g gþ CFA ¼ g gþ 3:086h ¼ 86 gu


gB ¼ g gþ CFA CB ¼ gFA 2p G r h ¼ 1046 gu

(b) To calculate the isostatic anomaly according to the Airy hypothesis we first need to

obtain the value of the root given by the equation

t ¼rc

rM rch ¼ 4500m

The isostatic correction is given by

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

Substituting the values

Dr ¼ rM – rC ¼ 600 kg m3

b ¼ t ¼ 4500 m

c ¼ h þ H þ t ¼ 35 500 m

a ¼ 40 km

we obtain:

CI ¼ 409 gu

the isostatic anomaly is:

gI ¼ g gþ CFA þ CB þ CI ¼ gB þ CI ¼ 637 gu

The negative value of the anomaly indicates that the zone is overcompensated.

64 Gravity

(c) If we want the compensation to be total, the value of the isostatic correction

must be

DgI ¼ DgB þ CI ¼ 0 ¼> CI ¼ – DgB ¼ 1046 gu

Since the isostatic correction under the Airy hypothesis is

CI ¼ 2pGr t þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a2 þ hþ Hð Þ2q


a2 þ ðhþ H þ tÞ2q

substituting and solving for t, we obtain

t ¼ 13 068m

For a total isostatic compensation the value of the root (13 068 m) must be much larger

than that corresponding to the 1000 m height, which is only 4500 m.

36. Gravity measurements are made at two points A and B of altitude 1000 m and

1000 m above the reference level, respectively, 2 km apart along a W-E profile at

latitude 38.80º N. Below a point C located in the direction AB and 1 km from A

is buried a sphere of radius 1 km and centre 3 km below the reference level, of density

r ¼ 1.76 g cm3. Calculate:

(a) The value of gravity at A and B.

(b) Using the Airy assumption and neglecting the sphere, calculate the root at A and B.

Crustal density rC ¼ 2.76 g cm3, mantle density rM ¼ 3.72 g cm3, a ¼ 10 km, and

H ¼ 30 km.

(a) The gravity observed at points A and B is given by

gA ¼ g CFA þ CB þ Cam

gB ¼ gþ CFA CB þ Cam

where g is the theoretical gravity, CFA the free-air correction, CB the Bouguer correction,

and C am the attraction due to the anomalous mass.

The theoretical gravity at the observation point at latitude 38.80º N is

g ¼ 9:780 32 1þ 0:005 3025sin2’ 0:000 0058sin22’

¼ 9:800714m s2

The free-air and Bouguer corrections are:

CFA ¼ 3:806h ¼ 3:806 1000 ¼ 3806 gu

CB ¼ 0:419rCh ¼ 0:419 2:76 1000 ¼ 1156 gu

The attraction due to the spherical anomalous mass (Fig. 36) is given by

Cam ¼GMðzþ dÞ



For points A and B, by substitution of the values

M ¼4

3pR3

r

R ¼ 1000m; r ¼ 1000 kgm3

zA ¼ h ¼ 1000m

zB ¼ h ¼ 1000m

xA ¼ xB ¼ 1000m

d ¼ 3000m

we find

CamA ¼ 16 gu

CamB ¼ 50 gu

Then the values of gravity at both points are

gA ¼ 9800 627:9 3806þ 1156:4 16 ¼ 9:798 048m s2

gB ¼ 9800 627:9þ 3806 1156:4 50 ¼ 9:803 314m s2

(b) To calculate the value of the root under A and B according to the Airy hypothesis we

use the equation

t ¼rC

rM rCh

A

C

h

d

R

–h

B

XB

XA

ρ

r c = 2.76 g / cm3

x

Fig. 36

66 Gravity

where rC and rM are the crust and mantle densities. Then we find

tA ¼2:76

3:72 2:76 1000 ¼ 2875m

tB ¼2:76

3:72 2:76 ð1000Þ ¼ 2875m

37. At a point at latitude 43º N, the observed value of gravity is 9800 317 gu, and the

free-air anomaly is 1000 gu.

(a) Calculate the Bouguer anomaly. Take rC ¼ 2.67 g cm3.

(b) If the isostatic compensation is due to a cylinder of radius 10 km which is beneath

the point of measurement, what percentage of the Bouguer anomaly is compen-

sated by the classical models of Airy and Pratt?

(c) According to the Pratt hypothesis, what density should the cylinder have for the

compensation to be total?

(a) First we calculate the normal gravity at latitude 43ºN:

g ¼ 978:0320 1þ 0:005 3025sin2’ 0:000 0058sin22’

Gal

g ¼ 9804 385 gu

The height of the point is determined from the free-air anomaly,

gFA ¼ g gþ 3:086h ¼ 9 800 317 9 804 385þ 3:086h ¼ 1000 gu

and solving for h,

h ¼ 1642m

From this value we calculate the Bouguer anomaly

gB ¼ g gþ 1:967h ¼ 838 gu

(b) To apply the isostatic compensation using the Airy hypothesis we first calculate the

root corresponding to the height h ¼ 1642 m:

t ¼ 4:45h ¼ 7307m

The isostatic correction is determined using Equation (34.1) of Problem 34:

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð37:1Þ

where

a ¼ 10 km; b ¼ t ¼ 7307m; c ¼ t þ 30 000þ h ¼ 38 949m

r ¼ rM rC ¼ 3:27 2:67 ¼ 0:6 g cm3

which results in CI ¼ 70 gu. This represents 8% of the observed Bouguer anomaly.


If we use the Pratt hypothesis, the contrast of densities corresponding to h ¼ 1642 m is

given by

r ¼hr0Dþ h

¼ 0:043 g cm3

where we have used r0 ¼ rC ¼ 2.67 g cm3 as the density of the crust. We now substitute

in Equation (37.1), b = D = 100 km, c = D þ h = 101.642 km and the obtained value of

Dr ¼ 0.043 g cm3, and obtain

CI ¼ 148 gu

We have to determine again the Bouguer anomaly using the density according to the Pratt

hypothesis

r ¼Dr0Dþ h

¼ 2:63 g cm3

gB ¼ gAL 2pGrh ¼ 810 gu

The isostatic correction corresponds now to 18% of the Bouguer anomaly.

(c) If the compensation is total the isostatic correction must be equal to the Bouguer

anomaly with changed sign:

CI ¼ gB

Using the Pratt hypothesis in order to calculate the density r of the cylinder under the

point, we have to take into account that this density must also be the density used in the

determination of the Bouguer anomaly. Then we write

CI ¼ gB ¼ gFA þ 2p G r h

2pGðr0 rÞ bþ


a2 þ c bð Þ2q


a2 þ c2p

¼ gFA þ 2pGrh

and putting

N ¼ bþ


a2 þ c bð Þ2q


a2 þ c2p

and solving for r, we obtain

r ¼gFA þ 2pGr0N

2pGðhþ NÞ¼ 2:46 g cm3

where we have used the values r0 ¼ 2.67 g cm3 and DgFA ¼ 1000 gu.

38. At a point on the Earth’s surface, a measurement of gravity gave a value of

9795 462 gu. The point is 2000 m above sea level. At sea level the crust is 20 km thick

and of density rC ¼ 2 g cm3. The density of the mantle is rM ¼ 4 g cm

3.


68 Gravity

(b) Calculate the isostatic anomaly according to the Airy and Pratt assumptions.

Use cylinders of 10 km radius and compensation depth of 60 km.

(c) Beneath the point, there is an anomalous spherical mass of GDM ¼ 1200 m3 s2.

How deep is it?

Take g ¼ 9.8 m s2.

(a) The free-air anomaly is given by

gFA ¼ g gþ 3:086h ¼ 9 795 462 9 800 000þ 3:086h ¼ 1634 gu

For the Bouguer anomaly we first calculate the Bouguer correction

CB ¼ 0:419rh ¼ 0:419 2 2000 ¼ 1676 gu

Then we obtain

gB ¼ g gþ CFA CB ¼ gFA CB ¼ 42 gu

(b) To calculate the isostatic anomaly according to the Airy hypothesis we determine

first the value of the root corresponding to the height 2000 m:

t ¼rc

rM rch ¼

2

4 22000 ¼ 2000m

The isostatic correction, using a single cylinder under the point, is given by

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð38:1Þ

where (Fig. 38a)

P

C a

b

Fig. 38a


a ¼ 10 km; b ¼ t ¼ 2 km; c ¼ H þ t þ h ¼ 20þ 2þ 2 ¼ 24 km

Calling A the term inside the brackets in Equation (38.1)

A ¼ 2þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

100þ 484p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

100þ 576p

¼ 0:166 km

The isostatic correction is, then, given by

CI ¼ 2 3:1416 6:67 108 cm3=gs2 2 g=cm3 166 102 cm ¼ 139 gu

Finally, the isostatic anomaly using the Airy hypothesis is

gI ¼ g gþ CFA CB þ CI ¼ 97 gu

According to the Pratt hypothesis, the regional density is given by

r ¼D

Dþ hr0 ¼

60

60þ 23:33 ¼ 3:22 g cm3

where D is the compensation depth (in this problem 60 km) and for r0 (Fig. 38b) we have

used the mean value of the density of the crust (2 g cm3) and of the mantle (4 g cm3)

along the compensation depth

r0 ¼1

32þ

2

34 ¼ 3:33 g cm3

The contrast of densities is

r ¼ 3:33 3:22 ¼ 0:11 g cm3

For the isostatic correction, using the Pratt hypothesis, the term A is now

A ¼ 60þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

102 þ 62 60ð Þ2q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

102 þ 622p

¼ 7:4 km

and the correction

20 km

40 km

h

P

rM= 4 g cm–3

rC= 2 g cm–3

x

Fig. 38b

70 Gravity

CI ¼ 2 3:1416 6:67 1011 m3=kg1s2 0:11 103 kg=m3 7:4 103 m

¼ 341 gu

Since according to the Pratt hypothesis, the density of the compensating cylinder extends

to the surface of the height 2000 m, we have to calculate again the Bouguer anomaly using

this density (3.33 g cm3). We find for the Bouguer and isostatic anomalies the values

CB¼ 2 p G r h ¼ 2699 gu

DgB ¼ DgFA – CB ¼ 1065 gu

DgI ¼ DgB þ CI ¼ 724 gu

(c) If we assume that the isostatic anomaly is produced by a spherical anomalous mass

buried under the point at a depth d under sea level its gravitational effect is given by

gmax ¼GM

ðhþ dÞ2¼ gI ¼ 724 gu

Solving for d we obtain

hþ d ¼ 3517m ) d ¼ 1517m

39. At a point P at height 2000 m above sea level, a measurement is made of gravity.

The crust at sea level, where gravity is 9.8 m s2, is 20 km thick and of density

3 g cm3, and the density of the mantle is 4 g cm3. Below the point P, at 2000 m depth

under sea level, is an anomalous spherical mass of GDM ¼ 1200 m3 s2.

(a) Neglecting the isostatic compensation, what would be the value of gravity at the

point P?

(b) With isostatic compensation, what now is the value of gravity at that point?

Use the Airy and Pratt assumptions for the isostatic compensation (Pratt

depth of compensation, 100 km) with single cylinders of 20 km radius under the

point.

(a) Without isostatic compensation, the gravity observed at point P is equal to the sum of

the normal gravity plus the free-air and Bouguer corrections and the effect of the

anomalous mass. Remember that the free-air correction has negative sign:

gP ¼ g CFA þ CB þ Cam ð39:1Þ

where,

g ¼ 9800 000 gu

CFA ¼ 3:086 h ¼ 3:086 2000 ¼ 6172 gu

CB ¼ 0:419rh ¼ 0:419 3 2000 ¼ 2514 gu

The anomaly due to the anomalous mass is given by

Cam ¼ gmax ¼GM

hþ dð Þ2ð39:2Þ


By substitution of the values,

Cam ¼GM

ðhþ dÞ2¼ 75 gu

The gravity observed a P is, then, given by

gP ¼ 9 800 000 6172þ 2514þ 75 ¼ 9 796 417 gu

(b) If there is isostatic compensation, according to the Airy hypothesis, we determine first

the depth of the root, using the density of the crust rC and of the mantle rM:

t ¼rC

rM rCh ¼

3

4 3 2000 ¼ 6000m

The isostatic correction is

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð39:3Þ

where a ¼ 20 km, b ¼ 6 km, c ¼ 2 þ 20 þ 6 ¼ 28 km, Dr ¼ 1 g cm3, so

CI¼ 23:146:67 1011 1103 6þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

202 þ 28 6ð Þ2q


202 þ 282p

103

¼ 553 gu

Then the observed gravity at point P is

gP ¼ 9796 417 553 ¼ 9795 864 gu

According to the Pratt hypothesis we first determine the contrast of densities

r ¼hr0Dþ h

where D is the level of compensation (100 km) and r0 is the mean density for the crust and

mantle down to depth 100 km:

r0 ¼20 3þ 80 4

100¼ 3:8 g cm3

Substituting we find

r ¼2 3:8

100þ 2¼ 0:074 g cm3

The isostatic correction is

CI ¼ 2 3:14 6:67 1011 0:074 103

100þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

202 þ 102 100ð Þ2q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

202 þ 1022p

103

¼ 501 gu

72 Gravity

where we have used in Equation (39.3) the values

a ¼ 20 km; b ¼ 100 km; c ¼ 100þ 2 ¼ 102 km

The value of gravity at point P is now

gP ¼ 9796 417 501 ¼ 9795 916 gu

which is larger by 52 gu than using the Airy hypothesis

40. A point P is at altitude 1000 m above sea level. Beneath this point is a sphere

of 1 km radius and GDM ¼ 650 m3 s2, with its centre 4 km vertically below the

point P. Given that the density of the sphere is twice that of the crust and 3/2 that of

the mantle, calculate:

(a) The density of the sphere, crust, and mantle.

(b) The value of gravity that would be observed at P for the isostatic compensation to

be total including the sphere.

(c) The radius of the sphere for the root to be null. Comment on the result.

Use the Airy hypothesis for the isostatic compensation with H ¼ 30 km, 20 km radius

of the cylinder, and theoretical gravity g ¼ 980 Gal.

(a) If we know GDM we can calculate the contrast of densities between the anomalous

mass and the crust

GM ¼4

3prR3G ) r ¼

3GM

4pR3G¼

3 650

4 3:1416 109 6:67 1011

¼ 2:326 g cm3

Since the density of the sphere rsph is double that of the crust rc, the densities of the crust

and mantle are

rsph ¼ 2rC ) r ¼ rsph rC ¼ 2rC rC ¼ rC ¼ 2:326 g cm3

rsph ¼ 4:652 g cm3

rM ¼2

3rsph ¼ 3:101 g cm3

(b) If isostatic compensation is total we have

gI ¼ 0 ¼ gþ gP þ CFA þ CB þ CI þ Cam ð40:1Þ

where g is the normal gravity, gP the observed gravity at point P, CFA the free-air correction,

CB the Bouguer correction, CI the isostatic correction and C am the gravitational effect of

the anomalous mass.

The free-air and Bouguer corrections are given by

CFA ¼ 3:806h ¼ 3:806 1000 ¼ 3086 gu

CB ¼ 0:419r h ¼ 0:419 2:326 1000 ¼ 975 gu


The effect of the spherical mass is

Cam ¼GM

hþ dð Þ2¼

650

4ð Þ2 106¼ 41 gu

We calculate the isostatic correction using the Airy hypothesis and taking into account the

presence of the spherical anomalous mass. Thus, according to Fig. 40, the equilibrium

between the gravity at P and at sea level far from P is given by

pa2rCH þ pa2rMt ¼ pa2rC hþ H þ tð Þ þ4

3pR3 rsph rC

and solving for t:

t ¼a2hrC þ

4

3R3 rsph rC

a2 rM rCð Þ¼ 3011m

As in previous problems we calculate the isostatic correction using a cylinder under point P

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

P

h

H

t

R

rM

rCrsph

a

d

x

Fig. 40

74 Gravity

where a ¼ 20 km, b ¼ t ¼ 3011 m, c ¼h þH þ t ¼ 34 011 m and Dr ¼ rM – rC ¼

0.775 g cm3, resulting in

CI ¼ 145 gu

If compensation is total, the isostatic anomaly must be null. This implies that the Bouguer

anomaly is equal, with opposite sign, to the isostatic correction

gI ¼ 0 ¼ gB CI ) gB ¼ 145gu

But the Bouguer correction can be obtained from Equation (40.1):

gI ¼ 0 ¼ gþ gP þ CFA þ CB þ CI þ Cam

Solving for CB we obtain

CB ¼ 975 gu

From the definition of the Bouguer anomaly we can find the value gP of gravity at P:

gB ¼ gP gþ CFA CB þ Cam

145 ¼ gP 9800 000þ 3086 975þ 41 ) gP ¼ 9797 703 gu

(c) Since the density of the sphere is greater than the density of the crust, there is an excess

of gravity at P with respect to other points at sea level far from P, which must be

compensated by a root of crustal material inside the mantle with negative gravitational

influence. In this situation the root can never be null.

41. At 10 km beneath sea level vertically under a point P of height 2000 m there exists

an anomalous spherical mass GDM ¼ 104 m3 s2. At sea level, gravity is 9800 000 gu

and the crustal thickness 20 km. The density of the crust is 2 g cm3, and of the

mantle 4 g cm3. Using the Airy assumption for the isostatic compensation with a

cylinder of 10 km radius, calculate for that point:

(a) The observed gravity.

(b) The free-air, Bouguer, and isostatic anomalies.

(a) For point P the Bouguer correction is

CB ¼ 0:419rh ¼ 0:419 2 h ¼ 0:838h ¼ 1676 gu

The gravity at point P, if there is no isostatic compensation and other effects, can be

obtained from the normal gravity and the free-air and Bouguer corrections

gP ¼ g CFA þ CB ¼ 9800 000 3:086 2000þ 0:838 2000

¼ 9795504 guð41:1Þ

Since there is an anomalous mass under point P we have to add its gravitational contribu-

tion to the gravity at P. For a spherical mass at depth h þ d under P the gravitational

attraction is


Cam ¼GM

hþ dð Þ2¼

104 m3 s2

108 m2¼ 104 ms2 ¼ 100 gu

We calculate the root corresponding to the isostatic compensation, assuming the Airy

hypothesis, and taking into account the presence of the anomalous mass in the same way as

in Problem 40:

pa2rCH þ pa2rMt ¼ pa2rC hþ H þ tð Þ þ4

3pR3 rsph rC

¼ pa2rC hþ H þ tð Þ þMa

t ¼pa2rChþMa

pa2ðrM rCÞ¼ 2239m

The isostatic correction using a cylinder is given by

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð41:2Þ

where a¼ 10 km, b¼ t¼ 2.239 km, c¼ 20þ 2þ 2¼ 24.239 km, Dr¼ 4 2¼ 2 g cm3,

resulting in

CI ¼ 154 gu

The gravity at point P is the value obtained in (41.1) plus the contribution of the anomalous

mass and minus the isostatic correction:

gP ¼ 9795 504þ 100 154 ¼ 9795 450 gu

(b) The free-air anomaly is equal to this observed value minus the normal gravity plus

the free-air correction:

gFA ¼ gP gþ CAL

Substituting the values we obtain

gFA ¼ 9795 450 9800 000þ 3:086 2000 ¼ 1622 gu

The Bouguer anomaly is given by

gB ¼ gP gþ CAL CB ¼ 54 gu

Finally the isostatic anomaly is the Bouguer anomaly plus the isostatic correction:

gI ¼ 54þ 154 ¼ 100 gu

This value corresponds to the gravitational contribution of the anomalous mass.

42. At a point P of height 2000 m above sea level the measured value of gravity is

979.5717Gal. Beneath P is a sphere centred at a depth of 12 kmbelow sea level, 1 g cm3

density, and radius 5 km. Assuming the Airy hypothesis (H ¼ 30 km, rC ¼ 2.5 g cm3),

rM ¼ 3.0 g cm3), calculate the isostatic anomaly at the point in gu and mGal.

76 Gravity

For the compensation, assume cylinders of the same radius as the sphere. Normal

gravity g ¼ 9.8 m s2.

We first calculate the root t, assuming the Airy hypothesis, corresponding to the height

2000 m of point P. If the situation is of total isostatic equilibrium, we have to introduce the

effect produced by the sphere in the determination of the root (Fig. 40):

4

3p rsph rC

a3 þ pa2rChþ pa2rCH þ pa2rCt ¼ pa2HrC þ pa2rMt

so

t ¼rChþ

4

3a rsph rC

rM rC¼ 10 000m ð42:1Þ

The negative value of t (anti-root) is due to the deficit of mass produced by the presence of

the sphere (rsph < rC) under point P.

The isostatic correction, as in previous problems, is calculated taking a cylinder under the point:

CI ¼ 2pG rM rCð Þ bþ


a2 þ c bð Þ2q


a2 þ c2p

where b ¼ t ¼ 10 000 m, a ¼ 5000, and c ¼ H þ h ¼ 32 000 m.

We obtain CI ¼ 36 gu.

The isostatic anomaly at P is equal to the observed gravity minus the normal gravity and

the free-air, Bouguer, and isostatic corrections, and the attraction of the spherical mass:

gI ¼ gP gþ CFA CB CI Cam ð42:2Þ

The effect of the anomalous mass is given by

Cam ¼G4

3pa3 rsph rC

hþ dð Þ2¼ 267 gu

By substitution in (42.2)

gI ¼ 9 795 717 9 800 000þ 3:086 2000 0:419 2:5 2000þ 267 36

¼ 25 gu

43. A point A on the Earth’s surface is at an altitude of 2100 m above sea level.

Calculate:

(a) The value of gravity at A if the isostatic anomaly is 2.5 mGal. Assume the Airy

hypothesis (rC ¼ 2.6 g cm3, rM ¼ 3.3 g cm3, H ¼ 30 km).

(b) If the previous value had been measured with a Worden gravimeter of constant

0.301 82 mGal/division giving a reading of 630.6, calculate the value of gravity at

another point B at which the device reads 510.1 (both readings corrected for drift).

(c) At what depth is the centre of a sphere of density 4 g cm3 and radius 5 km which

is buried in the crust, given that the anomaly created at a point A, 12 km from the


centre of the sphere, not in the same vertical, is 321 gu. Also calculate the

horizontal distance from the centre to point A.

Take, for compensation, cylinders of 10 km radius. g ¼ 9.8 m s2

(a) The isostatic anomaly is given by

gI ¼ gA gþ 3:086h 0:419rChþ CI ð43:1Þ

To calculate the isostatic correction CI, assuming the Airy hypothesis, we must first

calculate the root t that corresponds to the height h

t ¼rCh

rM rC¼ 7800m

As in other problems the isostatic correction is calculated using



a2 þ c bð Þ2q


a2 þ c2p

and substituting the values

b ¼ t ¼ 7800m; c ¼ hþ H þ t ¼ 39900m;

a ¼ 10 km; r ¼ rM rC ¼ 700 kgm3

we obtain

CI ¼ 84 gu

Solving for gA in Equation (43.1) we obtain

gA ¼ gI þ g 3:086hþ 0:419rCh CI ¼ 9795 748 gu

(b) For a Worden gravimeter the increment in gravity between two points (Dg) is propor-

tional to the increment in the values given by the instrument (DL) corrected by the

instrumental variations

g ¼ KL

gB gA ¼ KðLB LAÞ ) gB ¼ gA þ KðLB LAÞ

¼ gA 364 gu ¼ 9795 384 gu

where K is the constant of the gravimeter.

(c) The anomaly produced by a sphere buried at depth d under sea level at a point at height

h and at a horizontal distance x from the centre of the sphere is given by

Cam ¼G4

3pa3 rsph rC

ðhþ dÞ


ð43:2Þ

78 Gravity

For point A (Fig. 43) if r is the distance from the centre of the sphere to the point A,

x2 þ (h þd)2 ¼ r2 and solving for d in Equation (43.2) gives

d ¼Cam

r3

G4

3pa3 rsph rC

h

Substituting the values r ¼ 12 km, a ¼ 5 km, rsph – rC ¼ 1400 kg m3, h ¼ 2100 m,

C am ¼ 321 106 m s2, we obtain:

d ¼ 9245m

The horizontal distance is:

x ¼


r2 ðhþ dÞ2

q

¼ 3910m

44. In a gravity survey, two points A and B on the Earth’s surface gave the values 159

and 80 mGal for the free-air anomaly, and 51 and 25 mGal for the Bouguer

anomaly, respectively. Given that B is at an altitude 1000 m lower than A, and that the

density of the mantle is 25% greater than that of the crust, calculate:

(a) The value of gravity at A and B, and the densities of the crust and mantle.

(b) The isostatic anomaly according to the hypotheses of Airy (H ¼ 30 km) and Pratt

(D ¼ 100 km, r0 the value determined in the previous part) at point A. Take, for

compensation, cylinders of 10 km radius. g ¼ 980 Gal.

(a) The free-air anomaly at point A is given by

gFAA ¼ gA gþ CFAA ¼ gA gþ 3:086hA ð44:1Þ

r

d

x

a

rC

rsph

h

A

x

Fig. 43



gBA ¼ gA gþ CFAA CB

A ¼ gFAA 0:419rChA

Changing values from mGal to gu, we write for the Bouguer anomalies at points A and B

gBA ¼ 510 ¼ 1590 0:419rChA ) 0:419rChA ¼ 2100 gu

gBB ¼ 250 ¼ 800 0:419rChB ) 0:419rChB ¼ 1050 guð44:2Þ

Dividing both equations, we find

hA

hB¼ 2

Knowing that the difference in height between A and B is 1000 m, we obtain for the

heights of both points,

hB ¼ hA 1000 ) hA ¼ hB þ 1000 ¼ 2hB ) hB ¼ 1000m ) hA ¼ 2000m

The density of the crust can be obtained from Equation (44.2):

0:419rChB ¼ 105 105 ms2 ¼ 0:419rC 1000 ) rC ¼ 2:505 g cm3

The density of the mantle is 25% more than that of the crust, so

rM ¼ rC 1þ 0:25ð Þ ¼ 1:2 2:505 ¼ 3:131g cm3

The gravity at A and B is obtained using Equation (44.1):

gA ¼ gFAA 3:086hA þ g ¼ 1590 3:086 2000þ 9800 000 ¼ 9 795 418 gu

gB ¼ 800 3:086 1000þ 9800 000 ¼ 9797 714 gu

(b) For the isostatic anomaly at point A, according to the Airy hypothesis, we first

calculate the value of the root t corresponding to its height:

t ¼rChA

1:25rC rC¼

hA

0:25¼ 8000m

For the isostatic correction we use, as in other problems, a cylinder



a2 þ c bð Þ2q


a2 þ c2p


b ¼ t ¼ 8000m; c ¼ hþ H þ t ¼ 40 000m;

a ¼ 10 km; r ¼ rM rC ¼ 626 kgm3

we obtain

CI ¼ 77 gu

80 Gravity

The isostatic anomaly is

gIA ¼ gBA þ CI ¼ 510þ 77 ¼ 433 gu

If we use the Pratt hypothesis, we first calculate the density corresponding to the material

under point A:

r ¼Dr0Dþ h

¼100 2:505

100þ 2¼ 2:456 g cm3

and the contrast of density

r ¼ r0 r ¼ 2:505 2:456 ¼ 0:049 g cm3

The isostatic correction is determined using a cylinder,



a2 þ c bð Þ2q


a2 þ c2p

where Dr ¼ 0.049 g cm3, b ¼ D ¼ 100 km, a ¼ 10 km, c ¼ D þ h ¼ 102 km.

Then, we obtain

CI ¼ 158 gu

The isostatic anomaly will be the Bouguer anomaly plus the isostatic correction

gIA ¼ gBA þ CI ¼ 510þ 158 ¼ 352 gu

In both cases the anomaly is negative, but using the Airy model the value is greater than

using the Pratt model.

45. At a point P on the Earth’s surface, the observed value of gravity is 9.795

636 m s2, and the Bouguer anomaly is 26 mGal. Assuming the Airy hypothesis

(rC ¼ 2.7 g cm3, rM ¼ 3.3 g cm3, H ¼ 30 km), calculate:

(a) The height of the point.

(b) The isostatic anomaly.

(c) The value of gravity that would be observed at the point if beneath it were a

sphere at a depth of 10 km below sea level, with a density of 2.5 g cm3 and a

radius of 5 km, such that the compensation was total.

Compensation with cylinders of 5 km radius; g ¼ 9.8 m s2.

(a) We calculate the height of point P from the Bouguer anomaly:

gB ¼ gP gþ 3:086h 0:419rCh ) h ¼gB gP þ g

3:086 0:419rC

so

h ¼260 9 795 636þ 9 800 000

3:086 0:419 2:7¼ 2099:8m ffi 2100m


(b) To calculate the isostatic anomaly, using the Airy hypothesis, we first calculate the

value of the root t corresponding to the height of the point:

t ¼rC

rM rCh ¼

2:7

3:3 2:7 2100 ¼ 9450m

Using a cylinder, the isostatic correction is given by

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð45:1Þ

where we substitute the values

a ¼ 5 km; b ¼ t ¼ 9450m;

c ¼ hþ H þ t ¼ 2100þ 30 000þ 9450 ¼ 41 550m;

r ¼ 3:3 2:7 ¼ 0:6 g cm3

and obtain

CI ¼ 22 gu

The isostatic anomaly is then

gI ¼ gB þ CI ¼ 260þ 22 ¼ 238 gu

(c) If the compensation is total then the isostatic anomaly must be zero. But now we have

to include the gravitational effect Cam produced by the presence of the anomalous mass

of the sphere.

gI ¼ 0 ¼ gP gþ 3:086h 0:419rCh Cam þ CI

Solving for gP:

gP ¼ g 3:086hþ 0:419rChþ Cam CI ð45:2Þ

where the effect of the sphere is given by

Cam ¼G4

3pa3 rsph rC

hþ dð Þ2¼ 48 gu

We calculate the isostatic correction according to the Airy hypothesis. First we calculate

the value of root t, but now we add the effect of the sphere on point P (Fig. 45):

rCpa2hþ rCpa

2H þ4

3pa3 rsph rC

þ rCpa2t ¼ rCpa

2H þ rMpa2t

Solving for t, we obtain

t ¼rChþ

4

3a rsph rC

rM rC¼ 7228m

82 Gravity

We substitute this value of t in Equation (45.1) together with the other values

a ¼ 5 km; b ¼ 7 ¼ 7228m;

c ¼ hþ H þ t ¼ 39 328m; r ¼ 3:3 2:7 ¼ 0:5 kgm3

and obtain

CI ¼ 18 gu

By substitution in (45.2) we find the value of the gravity at P under the given conditions:

gP ¼ 9800000 3:086 2100þ 0:419 2:7 2100 48 18 ¼ 9795830 gu

46. Consider a point on the surface of the Earth in an overcompensated region

at which the values of the free-air and the Bouguer anomalies are 1300 gu and

1200 gu, respectively.

(a) Is this a mountainous or an oceanic zone? Give reasons.

(b) Calculate the altitude and the value of gravity at the point given that the density

of the crust is 2.72 g cm3.

(c) If the isostatic anomaly is -1062 gu calculate, according to the Airy hypothesis

(rC ¼ 2.72 g cm3, rM ¼ 3.30 g cm3, H ¼ 30 km), the value of the root

responsible for this anomaly. Compare it with the value that it would have if

the region were in isostatic equilibrium.

H

t

h

d

P

rC

rM

rsph

ax

Fig. 45


Compensation with cylinders of 10 km radius; g ¼ 9.8 m s2.

(a) Since the free-air anomaly is positive and the Bouguer anomaly is negative, this

indicates that this is a mountainous region.

(b) From the free-air and Bouguer anomalies we can easily calculate the height of the point:

gFA ¼ gP gþ 3:086h

gB ¼ gP gþ 3:086h 0:419rChð46:1Þ

and solving for h,

h ¼gFA gB

0:419rC¼ 2193m

The observed gravity at P can be obtained from either of the two equations (46.1):

gP ¼ g 3:086hþgFA ¼ 9794 532 gu

(c) The isostatic correction is found from the known Bouguer and isostatic anomalies:

gI ¼ gB þ CI ) CI ¼ 1062þ 1200 ¼ 138 gu

The isostatic correction, using the Airy hypothesis, is given, as in previous problems, as a

function of the root t, by

CI ¼ 138 ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

where we substitute

r ¼ 3:3 2:72 ¼ 0:58 g cm3; a ¼ 10 km;

b ¼ t; c ¼ hþ H þ t ¼ 2:193þ 30þ t km

and obtain for t,

t ¼ 19 984 m

If the region is in equilibrium the root due to the height h would be

t ¼rCh

rM rC¼

2:72 2193

3:3 2:72¼ 10 284m

Since we have already obtained a larger value (t ¼ 19 984 m), this indicates that the region

is overcompensated.

47. In an oceanic region, gravity is measured at a point on the surface of the sea,

obtaining a value of 979.7950 Gal. Calculate, using the Airy hypothesis (H ¼ 30 km,

rC ¼ 2.9 g cm3, rM ¼ 3.2 g cm3, rW ¼ 1.04 g cm3):

(a) The isostatic anomaly if the thickness of the crust is 8.4 km.

84 Gravity

(b) The thickness that the water layer would have to have if 15 km vertically below

the point there was centred a sphere of 10 km radius such that the anti-root is null

and the compensation total. Also calculate the density of the sphere.

For the compensation, take cylinders of 10 km radius. g ¼ 9.8 m s2.

(a) We are in an oceanic region, therefore in the calculation of the root for the isostatic

compensation according to the Airy hypothesis we have to consider the layer of water

of density rW. The value of the root is now given by

t0 ¼

rC rWrM rC

h0 ¼2:9 1:04

3:2 2:9h0 ¼ 6:2h0 ð47:1Þ

According to Fig. 47a, we have the following relation: H h0 t0 ¼ e) h0 ¼ H t

0 e,

where e is the thickness of the crust at the oceanic region, H the thickness of

the normal (sea level) continental crust, h0 the thickness of the water layer, and t0 the

negative root.

Substituting the values of t 0 from Equation (47.1) we obtain for h0

h0 ¼ 30 6:2h0 8:4 ) h0 ¼ 3 km ¼ 3000m

t0 ¼ 6:2 3000 ¼ 18 600m

From the value of the root we calculate the isostatic correction using a cylinder of 10 km

radius,

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

where Dr ¼ 3.2 2.9 ¼ 0.3 g cm3 ¼ 300 kg m3, b ¼ t0, c ¼ H ¼ 30 000 m, and so

CI ¼ 269 gu

To calculate the isostatic anomaly we first have to apply the Bouguer correction which in

this case consists of two terms: the first to eliminate the attraction of the water layer

rM

rC

rW

t

h

eH

P

Fig. 47a


(2pGrWh0) and the second to replace this layer by one of density equal to the crustal

density (þ2pGrCh0). Since the point is at sea level the free-air correction is null:

DgI ¼ g gþ CB CI

CB ¼ 0:419ðrC rWÞh0 ¼ 2338 gu

DgI ¼ 19 gu

(b) If the anti-root is null and there is total compensation, then we have

t0 ¼ 0

gI ¼ 0 ¼ g g 2pGrWh0 þ 2pGrCh

0 Camð47:2Þ

Since the isostatic anomaly must be null, then the anomalous spherical mass and the water

layer must compensate each other. The attraction of the anomalous mass is Cam ¼ GDM/d2

where d is the depth of its centre below sea level. Then we can write

g gþ 2pG rC rWð Þh0 GM

d2¼ 0

where the mass of the sphere is

M ¼4

3pa3rsph

If the point P is totally isostatically compensated and the anti-root is null, then (Fig. 47b)

pa2h0rW þ pa2ðH h0ÞrC þ4

3pa3ðrsph rCÞ ¼ pa2HrC

Solving for h0 gives

h0 ¼4a rC rsph

3 rW rCð Þð47:3Þ

rCa

rsph

d

rW

P

h

e

H

x

Fig. 47b

86 Gravity

Substituting this value in (47.2) we obtain

rsph ¼g g

4

3pG 2a

a3

d2

þ rC ¼ 3372 kgm3

and putting this value in (47.3), h 0 ¼ 3369 m.

48. At a point on the Earth’s surface, 500 m below sea level, a gravity value is

measured of 980.0991 Gal. If the region is in isostatic equilibrium calculate, using

the Airy hypothesis (H ¼ 30 km, rC ¼ 2.7 g cm3, rM ¼ 3.2 g cm3):

(a) The thickness of the crust.

(b) The isostatic anomaly in gu, with reasons for the sign of each correction. Take

compensating cylinders of 5 km radius; g ¼ 9.8 m s2.

(a) We calculate first the root, according to the Airy hypothesis which corresponds to

the depth of the point, applying the condition of isostatic equilibrium (Fig. 48)

rCH ¼ rCðH h0 t0Þ þ rMt

0

so

t0 ¼

rCrM rC

h0 ¼ 2700m

The thickness of the crust e at that point is

e ¼ H h0 t0 ¼ 30 000 2700 500 ¼ 26 800m

(b) The isostatic anomaly is given by

gI ¼ gP g 3:086h0 þ 0:419rCh0 CI

The free-air correction (3.086h0) is negative because the point is below sea level. For the

same reason the Bouguer correction (0.419 rC h0) is positive. The isostatic correction,

using the Airy hypothesis, is calculated as in previous problems using a cylinder under the

point of 5 km radius and density contrast D r ¼ 3.2 2.7 ¼ 0.5 g cm3. The value of the

anti-root t 0 has already been calculated, so

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

trM

rC

Ph

eH

Fig. 48


Substituting c ¼ H – h0 and b ¼ t0, we obtain

CI ¼ 9 gu

Then the isostatic anomaly is DgI ¼ 5 gu.

49. In an oceanic region where the density of the crust is 2.90 g cm3 and that of the

mantle 3.27 g cm3, the value of gravity measured at a point P on the sea floor at

depth 4000 m is 9.806 341 m s2.

Calculate, according to the Airy hypothesis:

(a) The thickness of the crust.

(b) The isostatic anomaly in gravimetric units.

Data: rw ¼ 1.04 g cm3, H ¼ 30 km, g ¼ 9.8 m s2. Take, for compensation, cylinders

of 10 km radius.

(a) First we calculate the value of the root according to the Airy hypothesis

t0 ¼

rC rWrM rC

h0 ¼ 20 108m

The thickness of the crust under the point is found by (Fig. 49)

e ¼ H h0 t0 ¼ 30000 40000 2018 ¼ 5892m

(b) Because the point is located at the bottom of the sea, to reduce the observed value of

gravity to the surface of the geoid (sea level) we eliminate first the attraction of the

water layer. Then we apply the free-air and the Bouguer corrections, to take into

account the attraction of a layer of crustal material which replaces the water. Finally we

apply the isostatic correction:

gI ¼ gP gþ 0:419rWh0 3:086h0 þ 0:419rCh

0 CI

The isostatic correction is calculated using a cylinder of 10 km radius,

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

h

e

t rM

rW P

rCH

Fig. 49

88 Gravity

where b ¼ t0 ¼ 20 108 m, c ¼ H h0 ¼ 26 000 m, Dr ¼ 3.27 2.9 ¼ 0.37 g cm3 ¼

370 kg m3

By substitution we obtain

CI ¼ 598 gu

The isostatic anomaly is:

gI ¼ 9 806 341 9 800 000 3:086 4000þ 0:419 1:04þ 2:9ð Þ 4000 598

¼ 4 gu

50. At a point with coordinates 42.78º N, 0.5º E and height 1572 m, the observed value

of gravity is 980.0317 Gal.


(b) If cylinders of 10 km radius beneath that point are used for the isostatic compen-

sation, calculate the gravimetric attraction of the mass defect corresponding to

the altitude of the point according to the Airy and Pratt hypotheses. Take, for

the crust, H ¼ 30 km, rC ¼ 2.67 g cm3, for the mantle, rM ¼ 3.27 g cm3, and

D ¼ 100 km for the Pratt level of compensation.

(c) How deep should the root of the Airy model be for the compensation to be

total?

(a) We calculate first the normal gravity at the point where gravity has been observed

using the expression

g ¼ 9:780 32 1þ 0:005 3025 sin2 ’

¼ 9:804 243m s2

where ’ is the latitude.

The free-air anomaly is given by

gFA ¼ gP gþ CFA ¼ 9800 317 9804 243þ 3:086 1572 ¼ 925 gu

and the Bouguer anomaly by

gB ¼ gP gþ CFA þ CB ¼ 9800 317 9804 243þ 1:967 1572 ¼ 834 gu

(b) If we approximate the isostatic compensation by means of a cylinder of radius a

under the point, we use the expression

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

ð50:1Þ

where Dr is the density contrast, b the height of the cylinder, and c the distance from the

base of the cylinder to the observation point.

Airy: We calculate first the root given by the equation

t ¼ 4:45 h ¼ 4:45 1572 ¼ 6995 km

For the isostatic correction we substitute in (50.1) the values


a ¼ 10 km; b ¼ t ¼ 6995 km; c ¼ t þ H þ h ¼ 38 567 km;

r ¼ rM rC ¼ 3:27 2:67 ¼ 0:6 g cm3

and obtain

CI ¼ 2 3:1416 6:67 1011 m3 s2 kg1 0:5 103 kgm3 A

where:

A ¼ 6:995þ


102 þ 38:567 6:995ð Þ2q


102 þ 38:5672p

103 ¼ 270m

so

CI ¼ 68 gu

Pratt: The contrast of densities is now given by

r ¼h

Dþ hr0 ¼

1575

100 000þ 1572 2:67 ¼ 0:04 g cm3

and substituting in Equation (50.1) with the values

a ¼ 10 km, b ¼ 100 km, c ¼ D + h ¼ 101 572 m, we have

CI ¼ 135 gu

(c) If the isostatic compensation is total (isostatic anomaly null) the isostatic correc-

tion, according to the Airy hypothesis, coincides with the value of the Bouguer

anomaly (834 gu):

gB ¼ 2pGr t þ


a2 ðH þ hÞ2q


a2 þ t þ H þ hð Þ2q

so

t þ


a2 ðH þ hÞ2q



¼gB

2pGr

In this expression we solve for the value of the root t:



¼ t þ


a2 ðH þ hÞ2q

gB

2pGr¼ t þ N

a2 þ t þ H þ hð Þ2 ¼ t2 þ N 2 þ 2tN

a2 þ t2 þ ðH þ hÞ2 þ 2tðH þ hÞ ¼ t

2 þ N2 þ 2tN

so

t ¼N 2 a2 ðH þ hÞ2

2ðH þ h NÞ

90 Gravity

By substitution of the values of N, a, H, and h we obtain

t ¼ 58 875m

Because this root has a negative value greater than the thickness of the crust, total

compensation is not possible.

51. Calculate the free-air anomaly observed on a mountain of height 2000 m which

is fully compensated by a root of depth t ¼ 10 km. The compensation is by a

cylinder of radius 20 km, the density of the crust is 2.67g cm3, and that of the

mantle is 3.27g cm3.

The free-air anomaly is given by

gFA ¼ g gþ 3:086h

Since the point is isostatically compensated, we calculate the isostatic correction using a

cylinder as in previous problems:

CI ¼ 2pGr bþ


a2 þ c bð Þ2q


a2 þ c2p

where we substitute

b ¼ t ¼ 10 km, c ¼ h þH þ t ¼ 42 km, a ¼ 20 km, Dr ¼ 600 kg m3

and obtain

CI ¼ 306 gu

Since the point is totally compensated the isostatic anomaly must be zero:

gI ¼ g gþ CFA CB þ CI ¼ 0

The free-air anomaly can now be written as

gFA ¼ g gþ CFA ¼ CB CI ð51:1Þ

We can calculate the Bouguer correction:

CB ¼ 1:119h ¼ 1:119 2000 ¼ 2238 gu

and substituting in (51.1) we obtain, for the free-air anomaly,

gFA ¼ 2238 301:5 ¼ 1932 gu

52. Calculate for a point P at height 2100 m, latitude 40ºN, and observed gravity

979.7166 Gal the refined Bouguer anomaly in gravimetric units. Consider a surplus

mass compartment of 3000 m, mean height, 3520 m inner radius, 5240 m outer

radius, with n ¼ 16. The density of the crust is 2.5 g cm3.

The refined Bouguer anomaly is obtained using, besides the free-air and Bouguer correc-

tions, the correction for the topography or topographic or terrain correction (T):

gB ¼ g gþ CFA CB þ T


The normal gravity at latitude 40º N is given by

g ¼ 9:780 32 1þ 0:0053025 sin2 ’

¼ 9:801 747m s2

the free-air correction by

CFA ¼ 3:086h ¼ 6481 gu

and the Bouguer correction by

CB ¼ 0:419rCh ¼ 2200 gu

The topographic correction is introduced in order to correct for the topographic masses not

included in the Bouguer correction, that is, in this case those above the height h (Fig. 52).

Remember that the Bouguer correction corrects for an infinite layer or plate of thickness h

and doesn’t consider the additional masses above h or the lack of masses below h. The

topographic correction is always positive because the masses above height h produce on

point P an attraction of negative sign which must be added and the lack of mass under h

must also be taken into account with a positive sign, since it has been subtracted in the

Bouguer correction.

To calculate the attraction of the mass above height h we use the attraction of

concentric cylinders (in our case two) with axis passing through point P and with

height equal to the difference between the height h of the point P and the height of

the mass of the topography hm above h. The cylinder is divided into n sectors with

radius a1 and a2 to approximate the topography (Fig. 52). Then the topographic

correction is given by

T ¼2pGrC

n


a22 þ c bð Þ2q


a22 þ c2q


a21 þ c bð Þ2q

þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a21 þ c2q

In our case we substitute the values

a2 ¼ 5240m; a1 ¼ 3520m; b ¼ hm h ¼ 900m; c ¼ 0; n ¼ 16

P

hm

rC

a1

a1

a2

a2

P x

h

Fig. 52

92 Gravity

and obtain

T ¼ 2 gu

The refined Bouguer anomaly is

gB ¼ g gþ CFA CB þ T ¼ 298 gu

53. Calculate the topographic correction for a terrestrial compartment of inner radius

a1 ¼ 5240 m, outer radius a2 ¼ 8440 m, n¼ 20, mean height 120 m, with 2000 m being

the height of the point P. Take r ¼ 2.65 g cm3.

In this problem we consider the topographic correction for the case of the lack of mass in

the topography at heights below that of the point P. Since in the Bouguer correction we

have subtracted an infinite layer of thickness h, we have to correct for the places where the

mass was not present (Fig. 53).

The topographic or terrain correction T in this case is calculated in the same way as in

the previous problem. Thus we take n sectors of cylinders with axis at point P and height

equal to the difference between h and hm (Fig. 52b). The correction is then given by

T ¼2pGrC

n


a22 þ c bð Þ2q


a22 þ c2q


a21 þ c bð Þ2q


a21 þ c2q

where we substitute a2 ¼ 8440 m, a1 ¼ 5240 m, b ¼ c ¼ h – hm, n ¼ 20 to obtain

T ¼ 0:67mGal

54. Calculate the topographic correction for an oceanic sector or compartment

of inner radius a1 ¼ 5240 m, outer radius a2 ¼ 8440 m, n ¼ 20, mean depth

525 m, with 600 m being the height of the point P. Take rC ¼ 2.67 g cm3, rW ¼

1.03 g cm3.

In this problem we have to correct for the lack of mass in the oceanic area near the point P,

between the sea level and height h (column 1 in Fig. 54). Also we have to take into account

the attraction produced by the water layer between sea level and the bottom of the sea

(column 2 in Fig. 54).

hmrC

a1

a2

P

h

Fig. 53


To calculate the necessary topographic correction we proceed as in Problems 52 and 53,

using cylindrical sectors:

T ¼2pGr

n


a22 þ c bð Þ2q


a22 þ c2q


a21 þ c bð Þ2q


a21 þ c2q

where

a2 ¼ 8440m; a1 ¼ 5240m; n ¼ 20

For the correction corresponding to the attraction of column 1, between height h and sea

level, we substitute the values:

b ¼ h ¼ c; r ¼ rC

and obtain

T1 ¼ 0:07mGal

For the correction of the attraction of column 2 between sea level and the bottom of the sea

we use the difference between the densities of the crust and of water:

b ¼ p ¼ 525m

c ¼ pþ hs ¼ 1125m

c b ¼ hs

r ¼ rC rW

T2 ¼ 0:11mGal

The total topographic correction is

T ¼ T1 þ T2 ¼ 0:18mGal

rw

a2

a1 P

P

h h1

2

Fig. 54

94 Gravity

Tides

55. Two spherical planets A and B of radii 2a and a and masses 3m and m are

separated by a distance (centre to centre) of 6a. The only forces acting are gravita-

tional, and the system formed by the two planets rotates in the equatorial plane.

(a) Calculate the value of the components of the acceleration of the tides at the Pole of

each planet directly and using the tidal potential. On which planet are they greater?

(b) If each planet spins on its axis with the same angular velocity as the system, what,

for each planet, is the ratio between the centrifugal force and the maximum of the

tidal force at the equator? On which planet is this ratio greater?

(a) From Fig. 55a we can deduce that at the Pole of planet A, the radial component of

the acceleration of the tides produced by planet B is

gTr¼

Gm

q2cos a

where q is the distance from the Pole of planet A to the centre of planet B and a the angle

formed by q and the radius at the Pole of planet A.

By substitution of the required values we obtain

q ¼


6að Þ2 þ 2að Þ2q

¼ffiffiffiffiffi

40p

a

cos a ¼2affiffiffiffiffi

40p

a) a ¼ 71:6

gTr¼

Gm

40a2cos 71:6ð Þ ¼ 0:008

Gm

a2

The tangential component gTy is given by (Fig. 55a)

gTy ¼ Gm

36a2þGm

q2sin a ¼

Gm

36a2þ

Gm

40a2sin 71:6ð Þ

¼Gm

a20:028þ 0:024ð Þ ¼ 0:004

Gm

a2

2a

3m

6a

qa

A

q

b a

m

B

Fig. 55a

95 Tides

If we use the tidal potential,

c ¼GMr

2

2R33cos2# 1

where R is the distance between the centres of planets A and B (Fig. 55a), and # is the

angle the position vector r forms with the distance vector R (in this case it is equal to the

colatitude, # ¼ y)

c ¼GMr

2

2R33cos2y 1

The radial and tangential components of the acceleration are given by

gTr¼

@c

@r¼

@

@r

Gmr2

2 6að Þ33cos2y 1

¼Gmr

216a33cos2y 1

gTy ¼1

r

@c

@y¼

1

r

@

@y

Gmr2

2 6að Þ33cos2y 1

¼Gmr

216a33 cos y sin yð Þ

For planet A, at the Pole, r ¼ 2a and y ¼ 90º, so

gTr¼

Gm

108a2¼ 0:009

Gm

a2

gTy ¼ 0

For planet B, we proceed in a similar manner:

q02 ¼ 36a2 þ a2 ¼ 37a2

cos a ¼affiffiffiffiffi

37p

a) a ¼ 80:5

Therefore,

gTr¼

G3m

37a2cos 80:5ð Þ ¼ 0:013

Gm

a2

gTy ¼ G3m

36a2þG3m

37a2sin 80:5ð Þ ¼ 0:003

Gm

a2

Using the tidal potential, we obtain the acceleration components

c ¼G3mr2

2 6að Þ33cos2y 1

gTr¼

@c

@r¼

Gmr

72a33cos2y 1

gTy ¼ 1

r

@c

@y¼

9

216

Gmr

að Þ3cos y sin yð Þ

Substituting at the Pole of planet B, r ¼ a and y ¼ 90º, we have

gTr¼ 0:014

Gm

a2

gTy ¼ 0

96 Gravity

(b) First we calculate the centre of gravity of the system formed by the two planets

measured from the centre of planet A (Fig. 55b):

x ¼3m 0þ m 6a

3mþ m¼

3

2a

The rotation radius for planet A is 3/2a and for planet B

6a3

2a ¼

9

2a

In the rotating system the centrifugal force equals the force of gravitational attraction,

which at the equator (y ¼ 0º) is

fg ¼Gm3m

6að Þ2¼ fC ¼ o2

r ¼ mo2 9

2a

From this expression we obtain the value of the angular velocity o of the rotation system:

3Gm2

36a2¼ mo2 9

2a ) o2 ¼

Gm

54a3

Since the angular velocity of the spin of each planet is equal to that of the system, the spin

centrifugal force at the equator of planet A is

fC ¼ o2r ¼

Gm

54a32a ¼

Gm

27a2

The tidal force is

fT ¼ gTr¼

2Gmr

R3¼

2Gm2a

63a3

2a

3m

(3/2)a (9/2)a

6a

A

m

B

w

Fig. 55b

97 Tides

and their ratio

fC

fT¼

Gm

27a2

Gm

54a2

¼ 2

If we repeat these calculations for planet B, we obtain

fC ¼ o2r ¼

Gm

54a3a ¼

Gm

54a2

fT ¼2G3ma

63a3¼

Gm

36a2

and the ratio

fC

fT¼

Gm

54a2

Gm

36a2

¼ 0:666

The ratio is larger for planet A, as expected owing to its larger radius.

56. Two planets of mass M and radius a are separated by a centre-to-centre distance

of 8a. The planets spin on their own axes with an angular velocity such that the value

of the centrifugal force at the equator is equal to the maximum of the tidal force (the

equatorial plane is the plane in which the system formed by the two planets rotates

around an axis normal to that plane).

(a) Calculate the value of the components of the vector g as multiples of GM/a2 for a

point of l'¼ 60º and w ¼ 45º (with l¼ 0º, being the meridian in front of the other

planet) including all the forces that act.

(b) What is the relationship between the angular velocity of each planet and that of

the system?

(a) The tidal potential is given by

c ¼GMr

2

2R33cos2# 1

where R is the centre-to-centre distance between the planets, and # the angle formed by the

vector r to a point and R (Fig. 56a). From this potential we calculate the radial component

of the tidal force:

f Tr¼

@c

@r¼

@

@r

GMr2

2R33cos2# 1

¼GMr

R33cos2# 1

At the equator of one planet # ¼ 0º, r ¼ a, and R ¼ 8a, so

f Tr¼

GMa

8að Þ32 ¼

GM

256a2

98 Gravity

The spin centrifugal force is

fC ¼ o2a

Equating these two expressions we find the value of the spin angular velocity,

fC ¼ f Tr) o2a ¼

GM

256a2) o ¼

1

16


GM

a3

r

At a point on the surface of one of the planets the total potential is the sum of the

gravitational potential V, the spin potential F, and the tidal potential c:

U ¼ V þ Fþ c ¼GM

rþ1

2o2

r2cos2’þ

GMr2

2R33cos2# 1

For a point P at latitude ’ and longitude l (Fig. 56b)

cos# ¼ cos’ cos l

and the potential U is

U ¼GM

rþ1

2o2

r2cos2’þ

GMr2

2R33cos2’cos2l 1

The components of gravity including the three effects are,

gr ¼@U

@r¼

GM

r2

þ o2r cos2 ’þ

GMr

R33 cos2 ’ cos2 l 1

g’ ¼ 1

r

@U

@’¼ o2

r cos’ sin’þGMr

R33 cos2 l cos’ sin’

gl ¼1

r cos’

@U

@l¼

1

r cos’

GMr2

2R36 cos2 ’ cos l sin l

¼ GMr

R33 cos’ cos l sin l

ww

gr

gλ

gθ

θ

a

MMR

Ω

x

Fig. 56a

99 Tides

At the required point,

r ¼ a; ’ ¼ 45;

l ¼ 60 ) cos# ¼ cos 45 cos 60 ¼ 0:35 ) # ¼ 69:3

so

gr ¼ GM

a2þ

GM

256a3

a1

2þ

GMa

8að Þ331

2

1

4 1

¼ 0:9993GM

a2

g’ ¼GM

256a3a

1ffiffiffi

2p

1ffiffiffi

2p þ

GMa

8a3ð Þ31

4

1ffiffiffi

2p

1ffiffiffi

2p ¼ 0:0027

GM

a2

gl ¼ GMa

ð8aÞ33

ffiffiffi

2p

2

1

2

ffiffiffi

3p

2¼ 0:0018

GM

a2

(b) To obtain the angular velocity of the rotation of the system (O) we take into account

that the centrifugal force due to the rotation of the system at the equatorial plane is

equal to the gravitational attraction between the two planets:

M2p ¼ G

MM

ð8aÞ2

where p is the distance from the centre of one planet to the centre of gravity of the system.

Then we find

M24a ¼ G

MM

ð8aÞ2)

2 ¼GM

4 64a3¼

GM

256a3

and finally O ¼ o.

l

j

j

a

P

ϑ

ϑ

Fig. 56b

100 Gravity

57. Consider two planets of equal massm and radius a separated by a centre-to-centre

distance of 8a. Only gravitational forces act.

(a) Calculate the tidal force at the equator on one of the planets directly, using the

formula of the tidal potential (do so at l ¼ 0º, i.e. the point in front of the other

planet). Express the result in mGal given that Gm/a2 ¼ 980 000 mGal.

(b) Compare and comment on the reason for the difference between the results of the

direct calculation and using the tidal potential.

(c) What relationship must there be between the angular velocities of the planet’s

spin and of the system’s rotation for the centrifugal force due to the planet’s spin

to be equal to the tidal force at the equator and l ¼ 0º?

(a) The exact calculation of the tidal force at a point located at the equatorial plane in

front of the other planet is (Fig. 57)

fT ¼Gm

7að Þ2

Gm

8að Þ2¼

Gm

a21

49

1

64

¼ 46 875 gu

Using the tidal potential

c ¼Gmr2

2 Rð Þ33cos2# 1

where R is the centre-to-centre distance between the planets (8a), r the radius to the point

where the tide is evaluated (a), and # the angle between r and R (at the equator in front of

the other planet # ¼ 0º), the radial component of the tidal force can be derived from the

potential. At the equator this is the total tidal force

f Tr¼

@c

@r¼

@

@r

Gmr2

2 8að Þ33cos2# 1

!

¼2Gma

2 512a3ð Þ2 ¼ 38 281 gu

(b) The difference between the value obtained by the exact calculation and by using

the tidal potential is 8594 gu, that is, 18%. This is explained because the tidal

w w

a

a

m m

8a

Ω

x x

Fig. 57

101 Tides

potential is a first-order approximation corresponding to terms of the order of

(r/R)2. For a relatively large value of r/R (1/8) this approximation is not very good.

(c) Now we make the spin centrifugal force equal to the tidal force for a point at the

equator of one of the planets which is given by exact calculation

fT ¼Gm

a21

49

1

64

If the spin angular velocity is op, the spin centrifugal force at the equator is given by

fC ¼ o2pa

Equating these two expressions we obtain

fC ¼ fT ) o2pa ¼

Gm

a21

49

1

64

) o2p ¼

Gm

a31

49

1

64

ð57:1Þ

The centrifugal force due the rotation of the system with angular velocity os is equal to the

gravitational attraction between the two planets:

mo2s4a ¼

Gm

64a2) o2

s ¼Gm2

256a3ð57:2Þ

From (57.1) and (57.2) we obtain the relation between the two angular velocities:

o2p

o2s

¼

Gm

a31

49

1

64

Gm

256a3

¼ 1:22 )op

os

¼ 1:10

58. Two planets of equal mass m and radius a are separated by a distance R. The spin

angular velocity of each planet is such that the centrifugal force at the equator is

equal to the maximum of the tidal force. If the sum of the two forces at the equator

cancels the gravitational force, what is the distance R?

The tidal potential is given by

c ¼Gmr2

R3

1

23cos2# 1

where R is the centre-to-centre distance between the planets, r the radius to the point where

the tide is evaluated, and # the angle between r and R. The maximum value is at a point at

the equator in front of the other planet, # ¼ 0º and r ¼a. Then

fT ¼@c

@r¼

Gmr

R33cos2# 1

¼2Gma

R3ð58:1Þ

The spin centrifugal force for a point at the equator is

fC ¼ o2a ð58:2Þ

Equating (58.1) and (58.2) we obtain the spin angular velocity,

102 Gravity

2Gma

R3¼ o2a ) o2 ¼

2Gm

R3

If, at the point considered, the sum of the spin centrifugal force and the tidal force cancel

the gravitational force of the planet, then

F þ fC þ fT ¼ 0

where F ¼ Gm/a2.

The value of R must be

4Gma

R3¼

Gm

a2) R ¼

ffiffiffiffiffi

4a3p

59. Two spherical planets A and B of radii 2a and a and masses 5M and M spin on

their axes with equal angular velocities. They are separated by a centre-to-centre

distance of 8a, and form a system that rotates in the equatorial plane of both planets

with an angular velocity that is equal to that of the spin angular velocity of each one.

(a) Determine the total potential for points on planet A.

(b) Determine the expression for the three components of the total gravity, including

the tide, for a point on the surface of planet A at longitude 0º.

(c) If the Love number h on planet A is 0.5, determine the height of the terrestrial tide

as a multiple of a at the equator, at local noon with respect to planet B, in the case

that the system’s rotational angular velocity is the same as that of the spin of the

two planets about their axes.

(a) We calculate the centre of gravity of the system, measured from the centre of planet

A (Fig. 59a):

X ¼5M 0þM 8a

5M þM¼

4

3a

8a

A

(4/3)a

2a

5M

M

B

xx

Fig. 59a

103 Tides

The total potential U at a point on the surface of planet A at latitude ’ is given by the sum

of the gravitational potential V, plus the spin potential Ф, plus the tidal potential c

produced by planet B:

U ¼ V þ Fþ c

U ¼5GM

rþ1

2o2

r2cos2’þ

GMr2

8að Þ323cos2# 1 ð59:1Þ

where according to Fig. 59b

cos# ¼ cos’ cos t

where t is the local time of planet Bwith respect to planet A (hour-angle), at a point of l¼ 0º,

the geographical longitude at planet A. For t ¼ 0, the planet B is in front of the point, so

U ¼5GM

rþ1

2o2

r2cos2’þ

GMr2

2 8að Þ33cos2’cos2t 1

(b) The components of gravity are

gr ¼@U

@r¼

5GM

r2

þ o2rcos2’þ

GMr

8að Þ33cos2’cos2t 1

gy ¼ 1

r

@U

@’¼ o2

r cos’ sin’þGMr

8að Þ33 cos’ sin’cos2t

gl ¼1

r cos’

@U

@t¼

GMr

8að Þ33 cos’ cos t sin t

2a

l

j

ϑ

ϑ

j

P

Fig. 59b

104 Gravity

At a point on the surface of planet A, r ¼2a,

gr ¼ 5GM

4a2þ o22acos2’þ

GM2a

8að Þ33cos2’cos2t 1

gy ¼ o22a cos’ sin’þGM2

83a23 cos’ sin’cos2t

gl ¼GM2

83a23 cos’ cos t sin t

(c) At the equator ’ ¼ 0º, at 12 h with respect to B, t¼ 180º, h ¼ 1/2, and o ¼ O. The

height of the equilibrium terrestrial tide is given by

B ¼ hc

g

At the equator of planet A the tidal potential (59.1) is

c ¼2GM 2að Þ2

2 8að Þ3¼

GM

128a

If we approximate g by gr

gr ¼ 5GM

4a2þ o22aþ

GM2a

8að Þ32 ¼

5GM

4a2þ

GM

128a2þ o22a

¼ 159GM

128a2þ 2o2a

and the height of the equilibrium tide is

B ¼1

2

GM

128a159MG

128a2þ 2ao2

ð59:2Þ

We know that the angular velocity of the rotation of the system is equal to the spin angular

velocity of both planets, so the spin angular velocity is given by

5Mo2 4

3a ¼

G5M 2

8að Þ2) o2 ¼

3GM

256a2

By substitution in (59.2)

B ¼1

2

GM

128a159MG

128a2þ 2a

3GM

256a2

¼ a

312

60. The Earth is formed by a sphere of radius a and density r, and a core of radius a/2

and density 2r, in the northern hemisphere, centred on the axis of rotation and

105 Tides

tangent to the equatorial plane. The Moon has mass M/4 (whereM ¼ (4/3)pra3), is at

a distance (centre-to-centre) of 4a, and orbits in the equatorial plane. Determine:

(a) The total potential and the components of gravity including the tidal forces.

(b) The total deviation of the vertical from the radial at lunar noon, and the deviation

due to the tide at the same hour for latitude 45º N, with m ¼ 1/8.

(a) The total potential U is equal to the gravitational potential of the planet V1 with

uniform density plus that of the core V2 using the differential mass, the spin

potential F, and the tidal potential c. The gravitational potentials are given by

(Fig. 60):

V1 ¼GM

r

V2 ¼GM 0

q

where the differential mass of the core is

M 0 ¼4

3pð2r rÞ

a3

8¼

M

8) V2 ¼

GM

8q

and q is the distance to the centre of the core. Its inverse can be approximated by

1

q¼

1

r1þ

a

2rcos yþ

a

2r

2 1

23cos2y 1

The spin potential is

F ¼1

2o2

r2sin2y

and the total potential is

U ¼GM

r1þ

1

81þ

a

2rcos yþ

a2

2rð Þ21

23cos2y 1

!

þr

a

3 m

2sin2y

" #

þ c

r

2r

4a

a/2q P

ϑ

q r

j L

t – 180°

w

Fig. 60

106 Gravity

where m ¼ o2 a3/GM.

The tidal potential c due to the Moon is given by

c ¼GMLr

2

2R33cos2# 1

ð60:1Þ

According to Fig. 60

cos2# ¼ cos2’cos2 t 180ð Þ

By substitution in (60.1), since R (the centre-to-centre distance between the Earth and the

Moon) is 4a, ML ¼ M/4, and ’ ¼ 90º y, we obtain

c ¼GMr

2

512a33cos2’cos2 t 180ð Þ 1

The potential U is

U ¼ GM5

8rþ

a

16r2sin’þ

a2

64r33 sin2 ’ 1

þr2

a3m

2cos2 ’þ

r2

512a33 cos2 ’ cos2 t 1

The components of gravity are found by taking the derivatives of U with respect to r and ’:

gr ¼ GM 5

8r2

2a

16r3sin’

3a2

64r43 sin2 ’ 1

þr

a3m cos2 ’þ

2r

512a33 cos2 ’ cos2 t 1

gy ¼ 1

r

@U

@’

¼GM

r

a

16r2cos’

a2

64r36 cos’ sin’ð Þ

þr2

a3m

22 sin’ cos’þ

r2

512a36 sin’ cos’ cos2 t

By substituting r ¼ a, ’ ¼ 45º, m ¼ 1/8, and at 12 h lunar time, t ¼ 180º, we obtain

gr ¼ 1:128GM

a2

gy ¼ 0:023GM

a2

(b) The deviation of the vertical with respect to the radial direction at 12 h lunar time is

given by

tan i ¼gy

gr¼ 0:020 ) i ¼ 1:2

107 Tides

The part of the deviation due to the lunar tide is given by

tan i0 ¼gMygr

¼1

gr

1

r

@c

@’¼

1

gr

GMr2

r512a36 sin’ cos’cos2t

and by substitution of the same values

i0 ¼ 0:3

The greater part of the deviation is due to the core.

61. Two spherical planets, planet A of radius 2a and mass 3m and planet B of radius a

and mass m, are separated by a centre-to-centre distance of 6a. The system rotates in

the equatorial plane and each planet spins on its axis with the same angular velocity.

What, for each planet, is the ratio between the force of gravity and the maximum of

the tidal force at the equator in front of the other planet?

The centre of gravity of the system measured from the centre of planet A is (Fig. 61)

0 3mþ 6am

4m¼

3

2a

The angular velocity of the system is given by

3mGm

6að Þ2¼ 3mo2 3

2a ) o2 ¼

Gm

54a3

The tidal force (radial component) can be calculated from the tidal potential

c ¼GMr

2

2R33cos2# 1

) f Tr¼

@c

@r¼

GMr

R33cos2# 1

where R is the centre-to-centre distance between the planets. For a point on the equator in

front of the other planet, # ¼ 0º, and

f Tr¼

2GMr

6að Þ3

Planet A : f Tr¼

2Gm2a

216a3¼ 0:018

Gm

a2

Planet B : f Tr¼

2G3ma

216a3¼ 0:028

Gm

a2

Gravity without tides is the sum of the gravitational and centrifugal forces:

g ¼Gm

r2 o2

r

Planet A : g ¼G3m

4a2 o22a ¼

3Gm

4a2

Gm

54a32a ¼ 0:71

Gm

a2

Planet B : g ¼Gm

a2 o2a ¼

Gm

a2

Gm

54a3a ¼ 0:98

Gm

a2

108 Gravity

The ratios between the gravity and tidal forces are:

Planet A :g

f Tr

¼0:71

0:018¼ 39:44

Planet B :g

f Tr

¼0:98

0:028¼ 35:00

62. The Earth is of radius a and density r, with a core of radius a/2 and density 3r

on the axis of rotation in the southern hemisphere tangent to the equatorial

plane. The Moon has mass M/2 and its centre is at 4a from the centre of the Earth

(M ¼ 4/3pra3).

(a) Write down the total potential.

(b) What is the value of the angular velocity of the Earth if at the point 30º N, 30º E at

06:00 lunar time the radial component of gravity is equal to GM/a2?

(c) In this case, what is the ratio between the angular velocity of the Earth’s rotation

and that of the system?

(a) As in previous problems the total potential U is given by

U ¼ V1 þ V2 þ Fþ c

The differential mass of the core is:

M1 ¼4p

33r rð Þ

a3

8¼

M

4

A

2a

3m

(3/2)a (9/2)a

6am

a

B

xx

w

Fig. 61

109 Tides

Using the approximation for 1/q, where q is the distance from a point on the surface of the

Earth to the centre of the core, the gravitational potential due to the core is

V2 ¼GM

4q¼

GM

4

1

r

a

2r2cos yþ

a2

8r33cos2y 1

The potential due to the spin of the Earth is given by

F ¼1

2o2

r2sin2y ¼

GM

a3r2 m

2sin2y

where

m ¼o2a3

GM

and the tidal potential due to the Moon is

c ¼GM

2

r2

64a31

23cos2# 1

¼GMr

2

256a33sin2ycos2t 1

where t is the hour-angle of the Moon (Fig. 62)

Then, the total potential U is

U ¼ GM5

4r

a

8r2cos yþ

a2

32r33 cos2 y 1

þr3

a3m

2sin2 yþ

r2

256a33 sin2 y cos2 t 1

(b) The radial component of gravity is given by

gr ¼@U

@r

¼ GM 5

4r2þ

2a

8r3cos y

3a2

32r43 cos2 y 1

þrm sin2 y

a3þ

2r

256a33 sin2 y cos2 t 1

w

P

q

q

t

ϑ

r

a/2 x

3r

4a L90° – q

Fig. 62

110 Gravity

Substituting y ¼ 60º and at 6 h, t ¼ 90, we have

gr ¼GM

a2141

128þ m

3

4

1

128

¼GM

a2

Solving for m we obtain m ¼ 2.81, and the spin angular velocity is

o ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2:81GM

a3

r

rad s1

(c) The centre of gravity r0 of the Earth–Moon system, measured from the centre of the

Earth, is given by

r0 ¼

MTr1 þMLr2

MT þML

¼

M

24a

5

4þ1

2

M

¼8a

7

where the mass of the Earth MT includes that of the core,

MT ¼ M þM

4and ML ¼

M

2

and r1 ¼ 0 and r2 ¼ 4a.

To calculate the angular velocity of the system, O, we put the centripetal force at the

Moon equal to the gravitational force between the Earth and the Moon:

ML r2 r0ð Þ2 ¼ G

MTML

r2

Substituting and solving for O, we obtain

G

5

8M 2

16a2¼

M

2O2 4

8

7

a; so O ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

GM

a37

256

r

Then, the ratio between the angular velocities of the spin of the Earth and of the system is:

o

¼


2:81GM

a3

r

ffiffiffiffiffiffiffiffiffiffiffiffi

7GM

256a3

r ¼ 10:14

63. Consider two planets of equal mass m and radius a separated by a centre-to-

centre distance of 8a. The planets revolve around their centre of mass and

spin around their own axes. Their spin angular velocity is such that the value

of the centrifugal force is equal to the maximum tidal force of the two at the

equator.

111 Tides

(a) Calculate, for a point on the equator of one of the planets and longitude l¼ 90º, the

value of the vector g including all the forces acting at that point (l¼ 0º corresponds

to the point on the line joining the two centres of the planets) at t ¼ 0.

(b) What is the deviation of the vertical from the radial at the point w ¼ 45º, l ¼ 0º?

(a) If o is the spin angular velocity of the two planets, the centrifugal force at the

equator y ¼ 90, r ¼ a, only has radial component:

f Cr

¼ o2a ð63:1Þ

The radial component of the tidal force can be obtained from the tidal potential c which, in

the first-order approximation, is given by (Fig. 63)

c ¼Gmr2

2R33cos2# 1

where, if ’ is the latitude and l the longitude,

cos# ¼ cos’ cos t lð Þ

On the equator f ¼ 0º, so

c ¼Gmr2

2R33cos2ðt lÞ 1

The radial component of the tidal force is

f Tr¼

@c

@r¼

@

@r

Gmr2

2R33cos2ðt lÞ 1

¼Gmr

R33cos2ðt lÞ 1

Ω

w w

a a

90°

8ax x

m m

Fig. 63

112 Gravity

The maximum value is for t ¼ l and putting R ¼ 8a and r ¼ a, we obtain

f Tr¼

Gma

ð8aÞ32 ¼

Gm

256a2ð63:2Þ

Then, as the centrifugal force is equal to the tidal force, we put (63.1) equal to (63.2) and

solve for o:

o2a ¼Gm

256a2) o2 ¼

Gm

256a3ð63:3Þ

We know that m O2 4a ¼ Gm2/R2, so solving for O,

2 ¼

Gm

256a3

and then using (63.3), we obtain o/O ¼ 1.

The total potential U is the sum of the gravitational, spin, and tidal potentials, which for

t ¼ 0, is given by

U ¼ V þ Fþ c ¼Gm

rþ1

2o2

r2sin2yþ

Gmr2

2ð8aÞ33sin2ycos2l 1

The components of gravity including the tidal forces are

gr ¼@U

@r¼

Gm

r2þ o2

rsin2yþGmr

ð8aÞ33sin2ycos2l 1

gy ¼1

r

@U

@y¼ o2

r sin y cos yþGmr

ð8aÞ33 sin y cos ycos2l

gl ¼1

r sin y

@U

@l¼

Gmr

ð8aÞ33 sin y cos l sin l

ð63:4Þ

By substitution of r ¼ a, l ¼ 90º, and y ¼ 90º we have

gr ¼ Gm

a2þ o2a

Gm

512a2

gy ¼ 0

gl ¼ 0

(b) By substitution of r ¼ a, l ¼ 0, and y ¼ 45º in (63.4) we obtain

gr ¼ Gm

a2þ o2a

1

2þ

Gm

512a23

2 1

¼ Gm

a2þ

Gm

256a21

2þ

Gm

512a21

2¼

1021

1024

Gm

a2

gy ¼ o2a1

2þ

Gm

512a23

2¼

Gm

256a21

2þ

Gm

512a23

2¼

5

1024

Gm

a2

gl ¼ 0

113 Tides

The deviation of the vertical with respect to the radial direction is

tan i ¼gygr

¼ 0:28

64. Two spherical planets of radii 2a and a and masses 8M and M separated by a

centre-to-centre distance of 4a spin on their own axes and rotate in the equatorial

plane with the same angular velocity.

(a) Determine all the forces acting at a point on the smaller planet at geocentric

coordinates w ¼ 60º N, l ¼ 0º (00:00 h local time corresponds to passage of the

other planet through the zero meridian).

(b) For this same point, calculate the astronomical latitude and the tidal deviation of

the vertical.

(a) First we determine the centre of gravity of the system, putting the origin at the

centre of the small planet (Fig. 64):

x ¼0M þ 4a 8M

9M¼

32a

9

Because the spin angular velocity of each planet is equal to the angular velocity of the

system (o ¼ O), we can write, for the small planet, putting the gravitational attraction of

the two planets equal to the centripetal force:

G8MM

4að Þ2¼ Mo2 32a

9

and solving for o,

o2 ¼9GM

64a3

w

w

q ra

M

2a

4a

8M

q

x

Ω

x

Fig. 64

114 Gravity

On the small planet the gravitational force is

gr ¼ GM

r2

gy ¼ 0

and the force due to its spin is

fr ¼ o2rsin2y

fy ¼ o2r cos y sin y

For the point under consideration, r ¼ a, y ¼ 30º, we obtain

gGCr

¼ Gm

a2þ

9

64

GM

a21

4¼

247GM

256a2

gGCy ¼9

64

GM

a3a

ffiffiffi

3p

4¼

GM

a29ffiffiffi

3p

256

To add the tidal force we use the tidal potential in the first-order approximation,

c ¼G8Mr

2

2R33cos2# 1

where cos # ¼ sin y cos (t l).

The tidal force for the point considered, r ¼ a, y ¼ 30º, t ¼ l ¼ 0º, and R ¼ 4a, is

given by

f Tr¼

@c

@r¼

GM

a21

32

f Ty ¼1

r

@c

@y¼

GM

a23ffiffiffi

3p

32

The total force acting at the point is the sum of the three forces, gravitational, centrifugal,

and tidal:

gtotalr

¼ 255GM

256a2

gtotaly ¼GM

a212

ffiffiffi

3p

256

(b) The astronomical latitude is given by ’a ¼ ’ þ i, where i is the deviation of the

vertical without considering the tide:

tan i ¼gtotaly

gtotalr

¼9ffiffiffi

3p

247¼ 0:06 ) i ¼ 3:6 ) ’a ¼ 60þ 3:6 ¼ 63:6

The maximum deviation of the vertical due to the tide at the point considered is

i0 given by

115 Tides

tan i0 ¼f Tygtotalr

¼

3ffiffiffi

3p

32255

256

¼ 0:163 ) i0 ¼ 9:3

Gravity observations

65. Determine the values of gravity at the following series of points belonging to a

gravimetric survey with a Worden gravimeter, specifying the drift correction for each

of them.

The gravity at the base is 980.139 82 Gal, and the gravimeterconstant is 0.301 81 mGal/ru (ru: reading unit).

The instrument drift is given by

d ¼LAe LAb

tAe tAb

where LAb and LAe are the readings at the base A at the beginning and end of the

measurements taken at times tAb and tAe, respectively. By substitution we obtain

d ¼568:8 562:5

14:33 8:50¼ 1:08 ru=hour

The corrected reading for station j is given by

Ljc ¼ Lj – d (tj – tAb)

where Lj is the reading taken at time tj.

For a Worden gravimeter the increment in gravity between two points (D g)

is proportional to the increment in the readings corrected by the instrument drift

(D Lc):

D g ¼ K D Lc

where K is the instrument constant.

Thus, from the readings we obtain the following results.

Station Time Reading

A (base) 08:30 562.5

B 09:21 400.7

C 11:34 437.9

D 13:20 360.1

A 14:20 568.8

116 Gravity

66. Point A is at a geopotential level of 97.437 43 gpu. Point B is at a difference

of 15.213 m in height relative to A, and has a value of gravity of 9.712 611 m/s2.

Calculate:

(a) The value of gravity at point A, if the difference in readings of aWorden gravimeter

between B and A is 17.8 ru, and the gravimeter constant is 0.308 21 mGal/ru.

(b) The geopotential number, dynamic height, and Helmert height of the point B given

that the normal gravity at a point of latitude 45º on the ellipsoid is 980 629.40mGal.

(a) Using aWorden gravimeter the increment of gravity between pointsA andB is given by

DgBA ¼ KDLBA ¼ 5:5mGal

whereK is the instrument constant andDL is the difference between the readings at pointsAandB.

The gravity at point A is

gA ¼ gB þ DgBA ¼ 971 266:6mGal

(b) The geopotential number at B can be calculated from the value at A in the form

CB ¼ CA þgA þ gB

2

hBA ¼ 82:661 59 gpu

where gravity is given in Gal and increments in height in km, because the geopotential

units are, 1 gpu ¼ 1 kGal m ¼ 1 Gal km.

The dynamic height is given by

HBD ¼

CB

g45¼ 84:294m

The Helmert height can be calculated from the dynamic height by

H ¼C

g þ 0:0424H

where C is given in gpu, g in Gal, and H in km. Solving for H, we obtain

H ¼g


g2 þ 4 0:0424Cp

2 0:0424

Taking the positive solution because point B is above the geoid (CB > 0) we obtain

HB ¼ 85:107m

Station Corrected reading D g (mGal) g (mGal)

A (base) 562.5 980 139.82

B 399.8 49.10 980 090.72

C 434.6 10.50 980 101.22

D 354.9 24.05 980 077.17

A 562.5 62.66 980 139.83

117 Gravity observations

67. In a geometric survey with measurements of gravity using a Worden gravimeter,

the following values were obtained:

Calculate the gravimeter readings corrected for drift, and the gravimeter constant

The instrument drift is given by

d ¼LAe LAb

tAe tAb

where LAb and LAe are the readings at the base A at the beginning and end of the

measurements taken at times tAb and tAe, respectively. By substitution we obtain,

d ¼ 1:07 ru=hour

A reading corrected at station j is given by

Lcj ¼ Lj dðtj tAbÞ

where Lj is the reading at time tj.

For a Worden gravimeter the increment in gravity between two points (D g) is propor-

tional to the increment in the readings corrected by the instrument drift (D Lc):

D g ¼ K DLc

where K is the instrument constant. Thus, K can be calculated in the form

K ¼g

Lc

From each pair of observations we obtain a value of K. Finally we take the arithmetic mean

(Km) from all the values obtained. The results are given in the following table.

68. The following table is obtained from observations with a Lacoste–Romberg

gravimeter:

Station Gravimeter reading (ru) Time Gravity (gu) Height difference (m)

A (base) 1520.23 8 h 50 m 9 793 626.8

B 1759.15 9 h 15 m 9 794 363.9 30.410

C 1583.11 9 h 35 m 9 793 820.7 301.863

A 1521.30 9 h 50 m

Station Corrected reading (ru) Gravity (mGal) K (mGal/ru)

A (base) 1520.23 979 362.68

B 1758.70 979 436.39 0.3091

C 1582.31 979 382.07 0.3079

A 1520.23

Km ¼ 0.3085

118 Gravity

The gravimeter scale factor is 1.000 65, and the equivalence between reading units

and the relative value of gravity in mGal is given by

Given that the value of gravity at point A is 9.794 6312 m s2, calculate the values at

B and C.

First we correct the readings by the instrument drift:

d ¼LAe LAb

tAe tAb

where LAe and LAb are the readings at the base A at the end and the beginning of the survey

at times tAe and tAb. Then

d ¼ 0:0727 ru=hour

The corrected reading at each station j is given by

Lcj ¼ Lj dðtj tAbÞ

where Lj is the reading at time tj. The corrected readings are:

LcA ¼ 3614:351

LcB ¼ 3650:224

LcC ¼ 3610:600

These readings are converted into relative gravity values Rj using the conversion table.

The reading at station A is

LcA ¼ 3600þ 14:351

and the relative gravity value is

RA ¼ (3846.02 þ (14.351 1.07125)) 1.00065 ¼ 3863.90 mGal

For stations B and C,

Station Gravimeter reading Time

A 3614.351 10:10

B 3650.242 10:25

C 3610.633 10:37

A 3614.414 11:02

Reading Value in mGal Interval factor

3600 3846.02 1.071 25

3700 3953.15 1.071 40

119 Gravity observations

RB ¼ (3846.02 þ (50.224 1.07125)) 1.00065 ¼ 3902.36 mGal

RC ¼ (3846.02 þ (10.600 1.07125)) 1.00065 ¼ 3859.88 mGal

To convert the relative values into absolute values we need to know both values at one

station, in our case in station A:

gA ¼ 9.794 6312 m s2 ¼ 979 463.12 mGal

gB ¼ gA – RA þ RB ¼ 979 501.58 mGal

gC ¼ gA – RA þ RC ¼ 979 459.10 mGal

120 Gravity

3 Geomagnetism

Main field

69. Assume that the geomagnetic field of the Earth is a geocentric dipole with a North

Pole at 80 N, 45 E and a magnetic moment 8 1022 A m². Calculate for a point with

geographical coordinates 45 N, 30 W the components NS, EW, and Z of the Earth’s

magnetic field, the declination and inclination, and the geomagnetic longitude.

Earth’s radius: 6370 km and the constant C ¼ 107 H m1 (this value is used in all

problems).

We calculate first the geomagnetic latitude and longitude (f, l) from the geographical

coordinates (f, l) of the point and the geographical coordinates of the geomagnetic North

Pole (fB, lB) by the equation

sin’ ¼ sin’B sin’þ cos’B cos’ cosðl lBÞ

sin l ¼sinðl lBÞ cos’

cos’


fB ¼ 80 N

lB ¼ 45 E

f ¼ 45 N

l ¼ 30 W ¼ 330

we obtain

f ¼ 46:70

l ¼ 84:82

In the geocentric magnetic dipole model, the vertical (Z ) and horizontal (H ) components

of the magnetic field can be obtained from

Z ¼ 2B0 sinf

H ¼ B0 cosf

B0 ¼Cm

a3

ð69:1Þ

121

In these equations B0 is the geomagnetic constant, m the magnetic moment of the dipole,

a the Earth’s radius, and the constant C ¼107 H m1.

In this case we are given that

m ¼ 8 1022 A m2

a ¼ 6370 km ¼ 6.37 106 m

By substitution in Equations (69.1) we obtain:

B0 ¼ 30 951 nT

Z ¼ 45 051 nT

H ¼ 21 227 nT

The geomagnetic declination is given by


cosf

D ¼ 14:16

The NS (X ) and EW (Y ) components are

X ¼ H cosD ¼ 20 582 nT

Y ¼ H sinD ¼ 5193 nT

Finally, the geomagnetic inclination or dip (I ) at that point is given by

tan I ¼ 2 tanf ) I ¼ 64:77

70. Assume that the geomagnetic field is produced by a geocentric dipole of magnetic

moment 8 1022 Am², with North Pole at 80 N, 70 W, and that the Earth’s radius is

6370 km. Calculate for a point with geographical coordinates 60 N, 110 E:

(a) Its geomagnetic coordinates, the components of the Earth’s magnetic field (X ,

Y , Z , H ), the total field, the declination, and the inclination.

(b) The equation of the line of force passing through it.

(a) For this point the difference in longitude from the Geomagnetic North Pole

(GMNP) is 180º (Fig. 70), so both are on the same great circle. Then, the

geomagnetic coordinates are obtained from

f ¼ f ð90 fBÞ ¼ 50

l ¼ 180

The expressions for the geomagnetic vertical and horizontal components and for the total

geomagnetic field are

Z ¼ 2B0 sinf

H ¼ B0 cosf


H2 þ Z2p

B0 ¼Cm

a3

122 Geomagnetism

The values of the constants are

m ¼ 8 1022 A m2

a ¼ 6370 km ¼ 6.37 106 m

C ¼ 107 H m1

Substituting in the above equations we obtain

B0 ¼ 30 951 nT

Z ¼ 47 420 nT

H ¼ 19 895 nT

F ¼ 51 424 nT

For this point the geomagnetic declination D ¼ 0. So the NS (X ) and EW (Y )

components are

D ¼ 0

X ¼ H cosD ¼ H ¼ 19 895 nT

Y ¼ H sinD ¼ 0

The geomagnetic inclination (I ) is given by

tan I ¼ 2 tanf ) I ¼ 67:23

(b) The equation of the line of force passing through a point with geomagnetic

co-latitude y is

r ¼ r0 sin2 y

In this equation r0 is the distance from the Earth’s centre to a point on the line of force with

y ¼ 90º. The distance r0 is different for each line of force.

GMNP GNP

fB

f∗

l∗

l = 0

P

f

l

x

Fig. 70

123 Main field

For the point with geomagnetic latitude f ¼ 50º located in the Earth’s surface (r ¼ a),

y ¼ 90º – f ¼ 40º, so

r0 ¼a

sin2 y¼ 15 417km

71. Assume that the geomagnetic field is produced by a geocentric dipole of magnetic

moment 7.5 1022A m², with North Pole at 75 N, 65 W, and that the Earth’s radius

is 6372 km. Calculate:

(a) The NS and EW components for a point on the Earth’s surface at which the

inclination is 67 and the geomagnetic longitude is 120.

(b) The geographical coordinates of that point.

(c) The geomagnetic coordinates, field components, declination, and inclination of the

point on the geographical equator of zero geomagnetic longitude.

(a) The geomagnetic latitude f is obtained from

tan I ¼ 2 tanf ) f ¼ 49:7

The horizontal component, H , can be calculated from the geomagnetic constant, B0, and

the geomagnetic latitude:

B0 ¼Cm

a3¼ 28 989 nT

H ¼ B0 cosf ¼ 18 761 nT

To obtain the NS (X ) and EW (Y ) components it is necessary to calculate the declination (D )

from the spherical triangle with vertices at the Geographical North Pole (GNP), Geomagnetic

North Pole (GMNP), and the point P (Fig. 71a). But we need to calculate the geographic latitude

firstly by solving the spherical triangle. Applying the cosine law to the angle (90º – f):

cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ

þ sinð90 fBÞ sinð90 fÞ cosð180 lÞ

sinf ¼ sinfB sinf cosfB cosf

cos l ð71:1Þ

By substitution of the values,

f ¼ 55:1

To obtain the geomagnetic declination we apply the sine law in the spherical triangle of

Fig. 71a:

sinD

sinð90 fBÞ¼

sinð180 lÞ

sinð90 fÞ) sinD ¼

cosfB sin l

cosfð71:2Þ

and substituting the values we find

D ¼ 23:1

It is important to note that we have added a minus sign in the last equation in order for the

declination be positive toward the east.

124 Geomagnetism

The NS (X ) and EW (Y ) components are

X ¼ H cosD ¼ 17 262 nT

Y ¼ H sinD ¼ 7 349 nT

(b) The calculated geographical latitude is

f ¼ 55:1

The geographical longitude is obtained by applying the cosine law to the spherical triangle

of Fig. 71a:

cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ

þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ

cos y ¼ sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ

ð71:3Þ

so

cosðl lBÞ ¼sinf sinfB sinf

cosfB cosfð71:4Þ

Substituting the values, gives

l lB ¼ 101:6

To take the inverse cosine in the correct quadrant we bear in mind that l < 0 implies

that the point is to the west of the Geomagnetic North Pole, that is l – lB < 0. So we

obtain

l ¼ 166:6 W

(c) If this point is on the geographical equator (f ¼ 90) and has zero geomagnetic

longitude (l ¼ 0º), it is on the same geographical meridian as the Geomagnetic

North Pole. Then from Fig. 71b

90º– fB

GNP

GMNP

q = 90º– f∗

180º–l∗

l – lB

D∗

90º– f

P

Fig. 71a

125 Main field

f ¼ 90 fB ¼ 15:0

l ¼ 0

Z ¼ 2B0 sinf ¼ 15 006 nT

H ¼ B0 cosf ¼ 28 001 nT

D ¼ 0

X ¼ H

Y ¼ 0 nT

tan I ¼ 2 tanf ) I ¼ 28:2

72. Assume that the geomagnetic field is that of a dipole with North Pole at 75 N, 0 E.

What is the conjugate point of that of geographical coordinates 30 N, 30 E?

First, we calculate the geomagnetic coordinates (f , l ) (Problem 71; Fig. 71a):



cosf

ð72:1Þ

The values of the geographical coordinates are

fB ¼ 75 lB ¼ 0

f ¼ 30 l ¼ 30

f∗

P

fB

GMNPGNP

x

Fig. 71b

126 Geomagnetism

By substitution in Equations (72.1) we obtain:

f ¼ 42:6

l ¼ 36:0

A magnetic conjugate point is a point on the Earth’s surface that is located on the same line

of force and in the opposite hemisphere (Fig. 72, P and P1). Then, its geomagnetic

coordinates (f1, l1

) are

f1 ¼ f ¼ 42:6

l1 ¼ l ¼ 36:0

To calculate the geographical coordinates for this point (f1, l1) we use the spherical

triangle of Fig. 71a. We calculate the geographical latitude applying the cosine law:

cosð90 f1Þ ¼ cosð90 fBÞ cosð90 f

1Þ

þ sinð90 fBÞ sinð90 f

1Þ cosð180 l1Þ

sinf1 ¼ sinfB sinf1 cosfB cosf1

cos l1

f1 ¼ 53:9

To calculate the geographical longitude we apply the cosine law again:

cosð90 f1Þ ¼ cosð90 fBÞ cosð90

f1Þ

þ sinð90 fBÞ sinð90 f1Þ cosðl1 lBÞ

GMNP GNP

P

– f∗

f∗

x

P1

fB

x

Fig. 72

127 Main field

cosðl1 lBÞ ¼sinf

1 sinfB sinf

cosfB cosf

l1 lB ¼ 47:3

and taking the solution in the correct quadrant

l1 > 0 ) l1 lB > 0

l1 ¼ 47:3

73. Assume the centred dipole approximation, with the coordinates of the Geomag-

netic North Pole being 65 N, 0 E, and the magnetic moment of the dipole 8 1022

A m². Calculate, for a point on the Earth’s surface at geographical coordinates 30 N,

30 E:

(a) The geographical coordinates of the conjugate point.

(b) The declination, inclination, and vertical and horizontal components of the field

at both points. Compare and contrast the results.

Earth’s radius: 6370 km.

(a) First, we calculate the geomagnetic coordinates (f, l) for point P with geograph-

ical coordinates f ¼ 30 N, l ¼ 30 E using the equations (Problem 71; Fig. 71a)



cosf

ð73:1Þ

The geographical coordinates of the Geomagnetic North Pole are

fB ¼ 65; lB ¼ 0

Substituting in Equations (73.1) results in

f ¼ 50:4

l ¼ 42:7

The geomagnetic coordinates (f1, l1

) of the magnetic conjugate point P1 satisfy

(Problem 72):

f1 ¼ f ¼ 50:4

l1 ¼ l ¼ 42:7

To calculate the geographical coordinates for this point (f1, l1) we use the spherical

triangle of Fig. 71a. We calculate the geographical latitude applying the cosine law

cosð90 f1Þ ¼ cosð90 fBÞ cosð90 f

1Þ

þ sinð90 fBÞ sinð90 f

1Þ cosð180 l1Þ

sinf1 ¼ sinfB sinf1 cosfB cosf

1 cos l

1

f1 ¼ 63:7

128 Geomagnetism

To calculate the geographical longitude we apply the cosine law again and, solving for l1,

cosðl lBÞ ¼sinf

1 sinfB sinf1

cosfB cosf1

l1 lB ¼ 77:1

l1 > 0 ) l1 lB > 0

l1 ¼ 77:1

(b) First we calculate the geomagnetic constant B0

B0 ¼Cm

a3¼

8 1022

ð6379Þ3 109¼ 30 951 nT

We calculate the declination D , inclination I , and vertical Z and horizontal H

components of the field at both points. The results are shown in the table, where we notice

that except for the declinations, which are very different, all other values are equal for both

points except in sign.

74. Assume the centred dipole approximation, with the coordinates of the Geomag-

netic North Pole 78.5 N, 70.0 W, and the magnetic dipole moment being 8.25 1022

A m². Calculate, for a point on the surface with coordinates 60.0 S, 170.0 W:

(a) Its geomagnetic coordinates, declination, inclination, and vertical and horizontal

components of the field.

(b) The potential at that point.

(c) The declination and inclination at the point diametrically opposite to it.


(a) We calculate the geomagnetic coordinates (f, l) using the equations (Problem 71,

Fig. 71a)

sinf ¼ cos y ¼ sinfB sinfþ cosfB cosf cosðl lBÞ


cosf

ð74:1Þ

The values of the geographical coordinates are

fB ¼ 78:5 lB ¼ 70:0

f ¼ 60:0 l ¼ 170:0

P P1


cosf sinD1 ¼

cosfB sinðl1 lBÞ

cosf1D ¼ 19.33 D

1 ¼ 40:2

tan I ¼ 2 tan f ) I ¼ 67.5 tan I1 ¼ 2 tanf1 ) I1 ¼ I1 ¼ 67:5

Z ¼ 2B0 sinf ¼ 47 662 nT Z

1 ¼ 2B0 sinf1 ¼ Z ¼ 47 662 nT

H ¼ B0 cosf ¼ 19 750 nT H

1 ¼ B0 cosf1 ¼ H ¼ 19 750 nT

129 Main field

Substitution in Equations (74.1) gives

f ¼ 60:0

l ¼ 80:0

The geomagnetic declination is given by (Problem 71, Fig. 71a)


cosf

D ¼ 23:1

The inclination (I ) at that point is given by

tan I ¼ 2 tanf ) I ¼ 73:9

The vertical (Z ) and horizontal (H ) components of the magnetic field can be obtained

from

Z ¼ 2B0 sinf

H ¼ B0 cosf

B0 ¼Cm

a3

Substituting the values given we obtain:

B0 ¼ 31 918 nT

Z ¼ 2B0 sinf ¼ 55 279 nT

H ¼ B0 cosf ¼ 15 963 nT

(b) The potential at a point on the Earth’s surface (r ¼ a) at geomagnetic latitude f is

given by

F ¼Cm cos y

a2¼

Cm sinf

a2¼ 176 Tm

(c) We can observe in Fig. 74 that at the point diametrically opposite the geographical

and geomagnetic coordinates are

f1 ¼ f ¼ 60:0 N

l1 ¼ lþ 180 ¼ 10:0 E

f1 ¼ f ¼ 60:0

l1 ¼ l þ 180 ¼ 100:0

The geomagnetic declination at that point satisfies (Fig. 71a)

sinD1 ¼

cosfB sinðl1 lBÞ

cosf1

¼ cosfB sinðlþ 180 lBÞ

cosf

¼cosfB sinðl lBÞ

cosf ¼ sinD

130 Geomagnetism

Then, D1 ¼ D.

The geomagnetic inclination is given by

tan I1 ¼ 2 tanf1 ¼ 2 tanf ) I1 ¼ I

So, we can notice that the two points P and P1 (Fig. 74) that are diametrically opposite

are not magnetic conjugate points because the geomagnetic longitudes are different

by 180º.

75. Consider a point P on the Earth’s surface at coordinates 30 S, 10 W at which

the NS component of the geomagnetic field is 27 050 nT and the EW component is

5036 nT, with the geomagnetic inclination being negative. Assuming the centred

dipole hypothesis with magnetic moment 7.8 1022 Am², calculate:

(a) The geographical coordinates of the Geomagnetic North Pole.

(b) The geomagnetic coordinates of P’s conjugate point.

(a) We calculate first the geomagnetic constant B0:

B0 ¼Cm

a3¼ 30 177 nT

The geomagnetic declination D is obtained from the NS (X ) and EW (Y ) components

of the geomagnetic field:

tanD ¼Y

X ¼

5036

27 050) D ¼ 10:5

GMNPGNP

f1

fB

l∗

f∗f

P

P1

Fig. 74

131 Main field

The geomagnetic latitude f is calculated from the horizontal component H :

H ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X 2 þ Y 2p

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

270502 þ ð5036Þ2q

¼ 27 515 nT

H ¼ B0 cosf ) f ¼

H

B0

¼ 24:3

We then have two solutions for the geomagnetic latitude. To choose the correct one we bear

in mind that a negative value of the geomagnetic inclination implies a negative value of the

geomagnetic latitude:

tan I ¼ 2 tanf

I < 0 ) f < 0

f ¼ 24:3

With these results we calculate the geographical coordinates of the Geomagnetic North

Pole (fB, lB) using the spherical triangle in Fig. 71a. Applying the cosine rule for the angle

90º – fB:

cosð90 fBÞ ¼ cosð90 fÞ cosð90 fÞ þ sinð90 fÞ sinð90 fÞ cosD

sinfB ¼ sinf sinfþ cosf cosf cosD

fB ¼ 79:0

To calculate the longitude lB of the Geomagnetic North Pole, we apply the cosine law for

the angle 90º – f :

cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ þ sinð90 fBÞ sinð90

fÞ cosðl lBÞ



cosfB cosf) l lB ¼ 61:0

To choose the correct sign for the longitude we notice that the declination is negative and

then the point must be to the east of the Geomagnetic North Pole:

D < 0 ) l lB > 0

lB ¼ 61 10 ¼ 71:0 W

(b) The geomagnetic coordinates (f1, l1

) of P’s conjugate point verify that

f1 ¼ f ¼ 24:3

l1 ¼ l

We calculate the geomagnetic longitude l by


cosf

l ¼ 56:2 ¼ l1

132 Geomagnetism

76. At a point P on the Earth’s surface with coordinates 45 N, 30 W, the value of the

total geomagnetic field is 49 801 nT, the horizontal component is 21 227 nT, and the

EW component is 5171 nT, with the magnetic inclination being positive. Calculate:

(a) The geographical coordinates of the Geomagnetic North Pole.

(b) The value of the geomagnetic potential at P.

(c) The distance from the Earth’s centre to the point at which the line of force passing

through P intersects the geomagnetic equator.


(a) We calculate first the geomagnetic inclination, latitude, and declination by

cos I ¼H

F) I ¼ 64:8

tan I ¼ 2 tanf ) f ¼ 46:7

sinD ¼Y

H) D ¼ 14:1

With these results we calculate the geographical coordinates of the Geomagnetic North

Pole solving the spherical triangle (Fig. 71a) in the same way as in Problem 71:


fB ¼ 80:0



cosfB cosf

l lB ¼ 75:2

D > 0 ) l lB < 0

lB ¼ 45:2 E

(b) The geomagnetic potential at point Pon theEarth’s surface (r¼ a¼ 6370km) is givenby

F ¼Cm cos y

a2¼ B0a sinf

ð76:1Þ

We calculate the geomagnetic constant B0 from the horizontal component H :

H ¼ B0 cosf ) B0 ¼

21 227

cosð46:7Þ¼ 30 951 nT

Substituting in the potential equation (76.1) we obtain

F ¼ 143 Tm

(c) The equation of the line of force passing through a point with geomagnetic co-

latitude y is

r ¼ r0 sin2 y

133 Main field

In this equation r0 is the distance from the Earth’s centre to the point at which the line of

force passing through P intersects the geomagnetic equator. Substituting r ¼ a ¼ 6370 km

gives

r0 ¼a

sin2 y¼

a

cos2 f ¼ 13 543 km

77. Assume the centred dipole approximation, with the coordinates of the Geo-

magnetic North Pole being 75 N, 65 W, and the magnetic moment of the dipole

7.5 1022 Am2. For a point on the Earth’s surface at which the inclination is 67 and

the geomagnetic longitude is 120, calculate:

(a) The NS and EW components.

(b) Its geographical coordinates.

The Earth’s radius: 6372 km.

(a) We calculate first the geomagnetic constant B0, latitude f , and horizontal H

component:

B0 ¼Cm

a3¼ 28 989 nT

tan I ¼ 2 tanf ) f ¼ 49:7

H ¼ B0 cosf ¼ 18 761 nT

To calculate the NS (X ) and EW (Y ) components it is necessary to obtain first the

geographic latitude (f ) and the geomagnetic declination (D ). Applying the cosine rule

for the angle 90º – f (Fig. 71a),

sinf ¼ sinfB sinf cosfB cosf

cos l

f ¼ 55:1

The geomagnetic declination is given by

sinD ¼ cosfB sin l

cosf

D ¼ 23:1

From this value we obtain the NS and EW components:

X ¼ H cosD ¼ 17 262 nT


(b) The geographic latitude was already obtained,

f ¼ 55:1

134 Geomagnetism

To calculate the geographical longitude we apply the cosine law for the angle 90º – f

(Fig. 71a):



cosfB cosf

l lB ¼ 101:7

To choose the correct solution we notice that the declination is positive and then the point

must be to the west of the Geomagnetic North Pole

D > 0 ) l lB < 0

l ¼ 101:7 65 ¼ 166:7W

78. Assume a spherical Earth of radius 6370 km, with magnetic field produced by a

centred dipole whose northern magnetic pole is at 70 N, 60 W. Given that for a point

on the surface with coordinates 50 S, 80 W the horizontal component is 24 890 nT,

calculate:

(a) The magnetic dipole moment.

(b) The geographical coordinates of the conjugate point.

(a) We calculate first the geomagnetic latitude by (Fig. 71a)


f ¼ 30:9

To obtain the magnetic dipole moment m we need the geomagnetic constant B0, which is

related with the horizontal component H by

B0 ¼H

cosf ¼ 29 007 nT

B0 ¼Cm

a3) m ¼

B0a3

C¼ 7:5 1022 Am2

(b) Let us obtain first the geomagnetic longitude by (Problem 71, Fig. 71a)


cosf

l ¼ 14:8

The geomagnetic coordinates of the conjugate point (f1, l1

) are (Fig. 78)

f1 ¼ f ¼ 30:9

l1 ¼ l ¼ 14:8

135 Main field

Solving again the spherical triangle (Fig. 71a) we calculate the geographical coordinates

(f1, l1):

sinf1 ¼ sinfB sinf1 cosfB cosf1

cos l1

f1 ¼ 11:4

cosðl lBÞ ¼sinf1

sinfB sinf1

cosfB cosf1

l1 lB ¼ 13:0

l < 0 ) l1 lB < 0

l1 ¼ 73:0

79. If the Earth’s geomagnetic field is produced by a centred dipole, tilted 15 away

from the axis of rotation, of magnetic moment 7.6 1022 Am2, and the Geomagnetic

North Pole is at longitude 65 W, calculate:

(a) The geomagnetic constant in nT.

(b) The geographical coordinates of a point on the Earth’s surface at which the

declination is D ¼ 14 15.50 and the inclination is I ¼ 65 23.50. Discuss the

possible solutions.

(c) The geographical and geomagnetic longitude of the agonic line.

Assume a spherical Earth of radius 6370 km.

GMNPGNP

fB

P1

f∗1

f∗

l∗

P

Fig. 78

136 Geomagnetism

(a) We calculate the geomagnetic constant from the magnetic moment m, the Earth’s

radius a, and the constant C¼107 Hm1, by

B0 ¼Cm

a3¼ 29 403 nT

(b) Let us obtain first the geomagnetic latitude from the inclination I by

tan I ¼ 2 tanf ) f ¼ 47:5

If the dipole is tilted 15 away from the axis of rotation the latitude of the Geomagnetic

North Pole will be

fB ¼ 90 15 ¼ 75:0

With these results we calculate the geographical latitude f solving the spherical triangle

(Fig. 71a). Applying the cosine rule,


sinfB ¼ sinf sinfþ cosf cosf cosD ð79:1Þ

To obtain the geographical latitude from this equation we can carry out a change of

variables, introducing two new variables (m, N ) such that

sinf ¼ m cosN

cosf cosD ¼ m sinNð79:2Þ

From these equations we can calculate P and N:

tanN ¼cosf cosD

sinf ¼cosD

tanf ) N ¼ 41:6

P ¼cosN

sinf ¼ 1:02

Substituting Equations (79.2) in Equations (79.1) we obtain

sinfB ¼ P cosN sinfþ P cosN cosf ¼ P sinðfþ NÞ

sinðfþ NÞ ¼sinfB

m¼

sinfB cosN

sinf

fþ N ¼ 78:4 ) f ¼ 36:8

But another solution is also possible:

fþ N ¼ 180 ð78:4Þ ¼ 258:4 ) f ¼ 60:0

The two solutions are correct and we don’t have any additional information to choose one

or the other.

(c) The agonic line is the line where the declination is zero and this implies that the

point is on the great circle that contains the Geographic North Pole and the

Geomagnetic North Pole. So the geomagnetic longitude l is zero or 180º:

l ¼ 0 ) l ¼ lB ¼ 65 W

l ¼ 180 ) l ¼ lB þ 180 ¼ 115 E

137 Main field

80. The Earth’s magnetic field is produced by two dipoles of equal moment

(M ¼ Cm ¼ 9.43 109 nT m3) and polarity, forming angles of 30 and 45 with the

axis of rotation, and contained in the plane corresponding to the 0 meridian. Find

the potential of the total field and the coordinates of the resulting magnetic North

pole, taking the Earth’s radius to be 6000 km.

The total potential at a point is the sum of the potentials of the two dipoles. If M is the

magnetic moment (M ¼ Cm), r is the distance from the dipole’s centre, and y1 and y2are the geomagnetic co-latitude relative to each dipole (Fig. 80), the total potential F is

given by

F ¼ F1 þ F2 ¼M cos y1

r2

M cos y2

r2

¼Mðcos y1 þ cos y2Þ

r2

ð80:1Þ

We calculate the angles y1 and y2 (Fig. 71a) by

cos y1 ¼ sinfB1 sinfþ cosfB1 cosf cosðl lB1Þ

cos y2 ¼ sinfB2 sinfþ cosfB2 cosf cosðl lB2Þð80:2Þ

The geographical coordinates of the two Geomagnetic North Poles are given by

fB1 ¼ 90 30 ¼ 60; lB1 ¼ 0

fB2 ¼ 90 45 ¼ 45; lB2 ¼ 180

GMNP1

GMNP2

GNP45° 30°

q2

q1

P

f

Fig. 80

138 Geomagnetism

Substituting these values in Equations (80.2):

cos y1 ¼ sinfB1 sinfþ cosfB1 cosf cos l

cos y2 ¼ sinfB2 sinf cosfB2 cosf cos l

Adding the two equations gives

cos y1 þ cos y2 ¼ ðsinfB1 þ sinfB2Þ sinfþ ðcosfB1 cosfB2Þ cosf cos l

and substituting in the equation of the potential (80.1)

F ¼M ½ðsinfB1 þ sinfB2Þ sinfþ ðcosfB1 cosfB2Þ cosf cos l

r2

F ¼M

ffiffiffi

3p

þffiffiffi

2p

sinfþ 1ffiffiffi

2p

cosf cos l

2r2

If we call # the geographic co-latitude, # ¼ 90 f, then

F ¼M

ffiffiffi

3p

þffiffiffi

2p

cos#þ 1ffiffiffi

2p

sin# cos l

2r2

The resulting magnetic North Pole, the point where the inclination I ¼ 90º, due to the

combined effect of the two dipoles is given by

tan I ¼Z

H

and therefore at the magnetic Pole, H ¼ 0.

We derive the component H by taking the gradient of the potential F

X ¼ B# ¼1

r

@F

@#¼

M ffiffiffi

3p

þffiffiffi

2p

sin#þ 1ffiffiffi

2p

cos# cos l

2r3

Y ¼ Bl ¼1

r sin#

@F

@l¼

M 1ffiffiffi

2p

sin l

2r3

H ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X 2 þ Y 2p

Since the magnetic North Pole is contained in the plane corresponding to the 0 geograph-

ical meridian, then its longitude is either 0º or 180º.

If the longitude is 0º

l ¼ 0 ) H ¼M

2r3

ffiffiffi

3p

þffiffiffi

2p

sin#þ 1ffiffiffi

2p

cos#h i

¼ 0

# ¼ 8 ¼ 172

But this result doesn’t correspond to the north hemisphere. Then we must take the

geographical longitude 180º:

l ¼ 180 ) H ¼M

2r3

ffiffiffi

3p

þffiffiffi

2p

sin# 1ffiffiffi

2p

cos#h i

¼ 0

# ¼ 8 ) fB ¼ 82

139 Main field

This is the correct result and the coordinates of the magnetic North Pole are

fB ¼ 82; lB ¼ 180

81. The Earth’s magnetic field is produced by one dipole in the direction of the axis

of rotation (negative pole in the northern hemisphere) and another with the same

moment in the equatorial plane which rotates with differential angular velocity v

with respect to the points on the surface of the Earth (consider that the Earth doesn’t

rotate). Its negative pole passes through the 45 E meridian at time t ¼ 0 and

completes a rotation with respect to that point in 24 hours. Consider a point of

geographical coordinates 45 N, 45 E.

(a) Calculate the magnetic field components (Br, Bu, Bl) at that point.

(b) Illustrate graphically how each of them varies with local time.

(a) The total potential at a point on the surface of the Earth is the sum of the potentials

of the two dipoles (Problem 80, Equation 80.1):


r2

M cos y2

r2


r2

Dipole 1 is in the direction of the axis of rotation and so the geomagnetic co-latitude of the

point with respect to this dipole (Fig. 81a) is equal to the geographical co-latitude,

y1 ¼ 90 f

cos y1 ¼ sinf

GMNP1

GMNP2

GNP

f

q1

q2

P

Fig. 81a

140 Geomagnetism

Dipole 2 is on the equatorial plane (fB2 ¼ 0) and rotates with respect to the points of

the surface. Owing to this rotation its geographical longitude lB2 changes with time t in

the form

lB2 ¼ ot þ 45

where o is the angular velocity, o ¼ 360/T, T being the rotation period of 24 h.

The co-latitude y2 is


Substituting the geographical coordinates of the negative geomagnetic equatorial Pole

(fB2, lB2):

cos y2 ¼ cos’ cosðl ot 45Þ

Substituting in the equation for the potential

F ¼ F1 þ F2 ¼M sinfþ cosf cosðl ot 45Þ½

r2

If we consider the geographical co-latitude # ¼ 90 f, the potential is given by

F ¼M cos#þ sin# cosðl ot 45Þ½

r2

We obtain the magnetic field components (Br, By, Bl) at the point (#, l) by taking the

gradient in spherical coordinates of the potential F:

Br ¼ @F

@r¼

2M cos#þ sin# cosðl ot 45Þ½

r3

B# ¼ 1

r

@F

@ #¼

M sin#þ cos# cosðl ot 45Þ½

r3

Bl ¼ 1

r sin#

@F

@l¼

M sinðl ot 45Þ

r3

Substituting the values # ¼ 45, l ¼ 45, B0 ¼ M/a3, gives

Br ¼ 2B0ffiffiffi

2p 1þ cosotð Þ

B# ¼ B0ffiffiffi

2p 1 cosotð Þ

Bl ¼ B0 sinot

(b) The variation of each component with local time is shown in Fig. 81b

141 Main field

Magnetic anomalies

82. Calculate the magnetic anomaly created by a magnetic dipole buried at depth d,

arbitrarily oriented, at an angle to the vertical of a. The negative pole is upwards.

Consider a point P with coordinates (x, z), where x is measured along the horizontal from

the projection of the centre of the dipole and z is the vertical from the reference level (the

Earth’s surface). The position vector r forms an angle b to the vertical (Fig. 82). The

anomalous magnetic potential created by the dipole for this point is

F ¼Cm cosðaþ bÞ

r2¼

Cmðsin b cos aþ cos b sin aÞ

r2ð82:1Þ

where (Fig. 82)

cos b ¼zþ d


x2 þ ðzþ dÞ2q

sin b ¼x


x2 þ ðzþ dÞ2q

0 5 10 15 20

t (h)

–2

–1

0

1

Bθ

Br

Bl

Br,

q,l

B0

Fig. 81b

142 Geomagnetism


F ¼Cm ðzþ dÞ cos a x sin a½


To calculate the magnetic anomaly DB:

B ¼ rðFÞ

The vertical component of the magnetic field anomaly, taking the z-coordinate positive

downward, is

Z ¼@ðFÞ

@z¼

Cm x2 þ ðzþ dÞ2

cos a 3ðzþ dÞ ðzþ dÞ cos a x sin a½ h i


For points on the Earth’s surface (z ¼ 0)

Z ¼Cm ðx2 2d2Þ cos aþ 3dx sin a½

x2 þ d2½ 5=2ð82:2Þ

The component of the magnetic anomaly in an arbitrary horizontal direction x for the

Earth’s surface points (z ¼ 0) is given by

X ¼ @ðFÞ

@x¼

Cm ð2x2 d2Þ sin a 3dx cos a½

x2 þ d2½ 5=2ð82:3Þ

z

Z

rd

b

a

Px

X

+

–

Fig. 82

143 Magnetic anomalies

83. Calculate the magnetic anomaly produced at a point with geographical

coordinates 38 N, 30 W by a horizontal dipole buried at a depth of 10 m with

Cm ¼ 5 105 Tm3 which is in the vertical plane of geographical east, and with the

negative pole to the west. Also calculate the total values of the field in the NS, EW, and

vertical directions, and total field F, as well as the variations in the magnetic declin-

ation and inclination due to the existence of the dipole. Consider the Earth’s magnetic

field to be produced by a centred dipole with North Pole at 72 N, 30 W, and with

B0 ¼ 32 000 nT.

We calculate the horizontal and vertical components of the magnetic anomaly from

Equations (82.2) and (82.3), taking a ¼ 90º because the dipole is horizontal and x ¼ 0

because the dipole’s centre is beneath the point (Fig. 83a). We call DX and DY the

horizontal components in the NS and EW directions, respectively. Because the dipole is

on the vertical east–west plane, the north–south component DX ¼ 0,

Z ¼ 0

Y ¼Cm ð2x2 d2Þ sin a 3dx cos a½

x2 þ d2½ 5=2¼

Cm

d3

Substituting

Cm ¼ 5 105 Tm3

d ¼ 10m

we obtain

Y ¼ 50 nT

jBj ¼ Y ¼ 50 nT

X ¼ 0

Z

+–

a

d

PE

Fig. 83a

144 Geomagnetism

To calculate the components of the magnetic anomaly in the direction of the Earth’s

magnetic field, F, and their horizontal component, H, we need to determine the magnetic

declination and inclination at the point:

H ¼ X cosD þY sinD

F ¼ H cos I þZ sin Ið83:1Þ

Because the point has the same longitude as the Geomagnetic North Pole (Fig. 83b),

f ¼ 90 ðfB fÞ ¼ 56:0

D ¼ 0

The inclination is given by

tan I ¼ 2 tanf ) I ¼ 71:4

Substituting these values in Equations (83.1) we obtain

H ¼ 0

F ¼ 0

The total value of the field in the NS, EW, and vertical directions, and total field F are

XT ¼ X þX

YT ¼ Y þY

ZT ¼ Z þZ

FT ¼ F þF

f∗

f

P

fB

GMNPGNP

Fig. 83b


The vertical Z and horizontal H components of the geomagnetic field are given by

Z ¼ 2B0 sinf ¼ 53 058 nT

H ¼ B0 cosf ¼ 17 894 nT

The NS (X ) and horizontal EW (Y ) components of the geomagnetic field and its

magnitude F are given by

X ¼ H cosD ¼ H ¼ 17 894 nT

Y ¼ H sinD ¼ 0


H2 þ Z2p

¼ 55 994 nT

We finally obtain that the total field components are

XT ¼ 17 894 nT

YT ¼ 50 nT

ZT ¼ 53 058 nT

FT ¼ 55 994 nT

The variations in magnetic declination and inclination due to the presence of the buried

dipole are

tanD0 ¼YT

XT

) D0 ¼ 0:02

D D0 ¼ 0:02

tan I 0 ¼ZT

HT

) I 0 71:4 ¼ I

84. Buried at a point with magnetic latitude 30 N and a depth of 50 m is a horizontal

magnetic dipole with Cm ¼ 107 nT m3 with the positive pole to the geographical

north.

(a) Calculate DF if B0 ¼ 30 000 nT and the declination at that point is 15. Find the

ratio DF /F.

(b) How far from the dipole’s centre along the north–south line will the dipole field

strength be in the same direction as that of the Earth (take D ¼ 0 ).

(a) The component of the magnetic anomaly DF in the direction of the Earth’s

magnetic field (the total field anomaly) is given by

F ¼ H cos I þZ sin I ð84:1Þ

We first calculate the components in the geographical directions of the magnetic anomaly

produced by the buried dipole using Equations (82.2) and (82.3) of Problem 82, substitut-

ing a ¼ 90º because the dipole is horizontal, and x ¼ 0 because the dipole’s centre is

beneath the point. In this problem DY ¼ 0 because the dipole is on the geographical north–

south vertical plane (Fig. 84). Then

146 Geomagnetism

X ¼ Cm

d3

Y ¼ 0

Z ¼ 0


Cm ¼ 107 nTm3

d ¼ 50m

we obtain

X ¼ 80 nT

Substituting D ¼ 15º, the component of the magnetic anomaly in the direction of the

horizontal component H of the Earth’s magnetic field is

H ¼ X cosD ¼ 77 nT

At a point of magnetic latitude f ¼ 30 the magnetic inclination is

tan I ¼ 2 tanf ) I ¼ 49:1

Substituting in Equation (84.1), the total field anomaly is

F ¼ 50 nT

To calculate the geomagnetic field F we first obtain the components H and Z :

Z ¼ 2B0 sinf ¼ 30 000 nT

H ¼ B0 cosf ¼ 25 981 nT


H2 þ Z2p

¼ 39 686 nT

P

Z

N

a

d

– +

Fig. 84


The ratio of the total field anomaly and the Earth’s total magnetic field is

F

F¼ 1:26 103

(b) If the dipole field strength is in the same direction as that of the Earth then the

inclination I 0 due to the dipole is equal to that of the Earth’s field I , where

tan I 0 ¼Z

H

tan I ¼Z

H

Assuming D ¼ 0º, then

H ¼ E

If we substitute in Equations (82.2) and (82.3) of Problem 82, the angle a ¼ 90º because

the dipole is horizontal, we obtain

Z ¼3Cmdx

x2 þ d2 5=2

X ¼Cmð2x2 d2Þ

x2 þ d2 5=2

We have changed the sign of the vertical component because the negative pole is toward

the south.

Applying the condition, tan I 0 ¼ tan I , we obtain

Z

H¼

Z

X¼

Z

H

3Cmdx

x2 þ d25=2

Cm

2x2 d2Þ

x2 þ d25=2

¼3dx

2x2 d2¼

Z

H

2Zx2 3dHx Zd2 ¼ 0


d ¼ 50 m

Z ¼ 30 000 nT

H ¼ 25 981 nT

and solving the equation, we obtain

x1 ¼ 80m

x2 ¼ 15m

We have two solutions: a point 80 m to the north from the surface projection of the dipole’s

centre and another 15 m to the south.

148 Geomagnetism

85. Located at a point with geocentric geographical coordinates 45 N, 30 W, at a

depth of 100 m, is a dipole of magnetic moment Cm¼ 1 T m3, tilted 45 from the

horizontal to true north, with the negative pole to the north and downwards. At this

point on the surface, the following magnetic field values were observed (in nT):

F ¼ 55 101; H ¼ 12 413; DF ¼ 1268; DH ¼ 547.

Determine:

(a) At the indicated point, the main field components X , Y , Z .

(b) At the indicated point, the deviation of the compass with respect to geomagnetic

north.

(c) The geocentric geographical coordinates of the North Pole of the Earth’s dipole.

Precision 1 nT.

(a) We calculate first the magnetic anomaly produced by the dipole, applying Equa-

tions (82.2) and (82.3) of Problem 82, substituting a ¼ 225º and x ¼ 0. The

horizontal component is in the NS direction (DX) (Fig. 85a)

Z ¼2Cm cos a

d3¼ 1414 nT

X ¼Cm sin a

d3¼ 707 nT

Y ¼ 0

ð85:1Þ

To calculate the declination we use the equation

H ¼ X cosD ) cosD ¼H

X

D ¼ 39:3

PN

Z

d

a

+

45°

–

Fig. 85a


To obtain the Earth’s main field we eliminate the buried dipole contribution from the

observed values:

F ¼ F F ¼ 56 369 nT

H ¼ H H ¼ 11 866 nT

Z ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðFÞ2 ðHÞ2q

¼ 55 106 nT

and the X and Y components

X ¼ H cosD ¼ 9182 nT


(b) The observed declination is given by

tanD0 ¼Y

X¼

Y þY

X þX) D0 ¼ 37:2

The deviation of the compass due to the buried dipole with respect to geomagnetic north is

D0 D ¼ 2:1

(c) We calculate first the geomagnetic latitude of the point from the vertical and

horizontal components:

Z ¼ 2B0 sinf

H ¼ B0 cosf

tanf ¼Z

2H

f ¼ 66:7

With this value, the declination D and the geographical coordinates of the point (f, l), we

can solve the spherical triangle (Fig. 85b) and obtain the geographical coordinates of the

Geomagnetic North Pole:

GMNP

90º – f

q = 90º– f∗

90º – fB

l – lB

180º – l∗

D∗

GNP

P

Fig. 85b

150 Geomagnetism



fB ¼ 60:0

cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ



cosfB cosf

l lB ¼ 30:0

The correct solution is the negative one because a positive value of the declination implies

that the point is to the west of the Geomagnetic North Pole:

D > 0 ) l lB < 0

lB ¼ 0

86. Located at a point with geographical coordinates 45 N, 30 W, at a depth of 100 m,

is a dipole of magnetic moment Cm ¼ 1 T m3, inclined 45 to the vertical towards the

south, with the positive pole upwards, and in the geographical north–south vertical

plane. The Earth’s dipole has its north pole at 60 N, 0 E and B0 ¼ 30 000 nT.

Calculate:

(a) The values of Z, H, F at the given point.

(b) Where does the compass point to at that same point?

(a) We calculate first the geomagnetic latitude corresponding to the point by


f ¼ 66:7

From this value we obtain the geomagnetic components Z , H and the total main field F :

Z ¼ 2B0 sinf ¼ 55 107 nT

H ¼ B0 cosf ¼ 11 866 nT

F ¼


ðHÞ2 þ ðZÞ2q

¼ 56 370 nT

The geomagnetic declination D is given by


cosf

D ¼ 39:2

and the geomagnetic inclination I by

tan I ¼ 2 tanf ) I ¼ 77:8


The magnetic anomaly created by the dipole buried at depth d is given by Equations (85.1)

of Problem 85. Substituting Cm ¼ 1 T m3, d ¼ 100 m, and a ¼ 45º, we obtain

Z ¼2Cm cos a

d3¼ 1414 nT

X ¼Cm sin a

d3¼ 707 nT

Y ¼ 0

The field anomalies DH and DF are given by

H ¼ X cosD ¼ 548 nT

F ¼ H cos I þZ sin I ¼ 1266 nT

Finally, the observed values are

Z ¼ Z þZ ¼ 53 693 nT

F ¼ F þF ¼ 55 104 nT

H ¼ H þH ¼ 12 414 nT

(b) To calculate in what direction the compass points we need the value of the observed

declination D’ including the effects of the geomagnetic field and the buried dipole:

tanD0 ¼Y

X¼

Y þY

X þX



D0 ¼ 37:1

87. Located at a point on the Earth with geographical coordinates 45 N, 30 E, at a

depth of 100 m, is a dipole of magnetic moment Cm ¼ 107 nT m3, tilted 45 to the

vertical towards the south, with the positive pole downwards, and contained in the

plane of true north. The Earth’s field is produced by a centred dipole tilted 30 from

the axis of rotation in the plane of the 0 meridian, with B0 ¼ 30 000 nT. Calculate

the total values of F, Z, andH observed at the point of the surface above the centre of the

buried dipole.

We first calculate the geographical coordinates of the Geomagnetic North Pole and the

geomagnetic latitude

fB ¼ 90 30 ¼ 60

lB ¼ 0


f ¼ 66:7

152 Geomagnetism

The geomagnetic field components Z , H and the total main field F are given by

Z ¼ 2B0 sinf ¼ 55 107 nT

H ¼ B0 cosf ¼ 11 866 nT

F ¼


ðHÞ2 þ ðZÞ2q

¼ 56 370 nT

We calculate the geomagnetic declination D by


cosf

D ¼ 39:2

The inclination I is given by

tan I ¼ 2 tanf ) I ¼ 77:8

We obtain the magnetic anomaly produced by the buried dipole applying Equations (82.2)

and (82.3) of Problem 82, substituting a ¼ 45º and x ¼ 0. The horizontal component is in

the NS direction (DX):

Z ¼2Cm cos a

d3¼ 14 nT

X ¼Cm sin a

d3¼ 7 nT

Y ¼ 0

The field anomalies DH and DF are given by

H ¼ X cosD ¼ 5 nT


Finally the observed values are

Z ¼ Z þZ ¼ 55 121 nT

F ¼ F þF ¼ 56 383 nT

H ¼ HþH ¼ 11 861 nT

88. Buried at a point with geographical latitude 20 N and the same longitude as the

geomagnetic pole, at a depth of 200 m, is a sphere of 50 m radius of material with

magnetic susceptibility 0.01. The Earth’s field is produced by a centred dipole tilted

10 from the axis of rotation and magnetic moment M ¼ 1030 g cm3 (Earth’s radius:

6000 km). Calculate:

(a) The anomaly produced by induced magnetization in the sphere at a point on the

Earth’s surface above the centre of the sphere. Give the vertical and horizontal

components in units of nT.

(b) The total anomaly for a point on the Earth’s surface 100 m south of the above

point.


(a) We first calculate the geomagnetic co-latitude (y) and latitude (f ) of the point,

knowing that it is in the same meridian as the Geomagnetic North Pole (Fig. 88a):

y ¼ 90 f ¼ 90 10 20 ¼ 60

f ¼ 30

The geomagnetic field is given by

B0 ¼M

a3¼ 4630 nT

Z ¼ 2B0 sinf ¼ 4630 nT

H ¼ B0 cosf ¼ 4009 nT

F ¼


ðHÞ2 þ ðZÞ2q

¼ 6124 nT

The inclination is given by

tan I ¼ 2 tanf ) I ¼ 49

The magnetic anomaly created by a sphere is the same as the anomaly created by a

magnetic dipole oriented in the same direction as the geomagnetic field, that is, tilted

90 – I to the vertical and with the negative pole upwards (Fig. 88b). So we use Equations

(82.2) and (82.3) taking a ¼ 41º and x ¼ 0, but we change the sign of Equation (82.3)

because the negative pole is toward the north. The horizontal component is DX ¼ DH

because the dipole is in the magnetic north-vertical plane and DY ¼0:

Z ¼2Cm cos a

d3

H ¼Cm sin a

d3

GMNPGNP

10°

20°

q = 90° – f∗

P

Fig. 88a

154 Geomagnetism

To calculate Cm we use the magnetic susceptibility w and the volume V of the sphere:

Cm ¼ wFV ¼ 3:2 107 nT m3

Substituting this value in the equation of the components of the magnetic anomaly we

obtain

Z ¼2Cm cos a

d3¼

2 3:2 107 cos 41

8 106¼ 6 nT

X ¼Cm sin a

d3¼

3:2 107 sin 41

8 106¼ 3 nT

(b) The anomaly created by the sphere at a point at a distance x¼100 m to the south of

the above point is given by

Z ¼Cm

ðx2 2d2Þ cos aþ 3dx sin a

x2 þ d2 5=2

¼ 0:8 nT

H ¼Cm

ð2x2 d2Þ sin a 3dx cos a

x2 þ d2 5=2

¼ 3:3 nT

The total magnetic field anomaly is therefore

F ¼ H cos I þZ sin I ¼ 3:4 nT

P

Z

F∗

GN

MN

41°

49°

Fig. 88b


89. Buried at a point on the Earth at magnetic latitude 45 N, at a depth d, is a vertical

dipole of magnetic momentM (Cm) with negative pole upwards. IfM/d3 ¼ 10B0 (B0 is

the geomagnetic constant of the main field) calculate how far along the magnetic

meridian the direction of the buried dipole’s field will coincide with that of the Earth

(the terrestrial dipole field).

First we calculate the geomagnetic field components and the inclination by

Z ¼ 2B0 sinf ¼

ffiffiffi

2p

B0 nT

H ¼ B0 cosf ¼

ffiffiffi

2p

2B0 nT

tan I ¼Z

H¼ 2 ) I ¼ 63:4

The components of the magnetic field created by the dipole are given by Equations (82.2)

and (82.3) of Problem 82, putting a ¼ 0º:

Z ¼Cm

x2 2d2

x2 þ d25=2

H ¼Cm3dx

x2 þ d25=2

If the buried dipole’s field coincides with that of the Earth the magnetic inclinations due to

both have to be equal and so

Z

H¼ tan I ¼ 2

x2 þ 6xd 2d2 ¼ 0 ) x ¼ d 3ffiffiffiffiffi

28p

Of the two solutions, x ¼ 2.3d and x ¼ 8.3d, only the positive corresponds to the equal

direction of the two fields.

External magnetic field

90. The Earth’s magnetic field is produced by two dipoles of equal moment and

polarity that are at an angle of 60 to each other, with the bisector being the axis of

rotation. The dipoles are contained in the plane of the 0 geographical meridian.

(a) Calculate the potential on points of the Earth’s surface as a function of geograph-

ical coordinates f and l.

(b) At what points on the surface are the magnetic poles located?

(c) What form would the external field have in order to annul the internal field at the

magnetic equator?

156 Geomagnetism

(a) The total potential at a point is the sum of the potentials of the two dipoles (see

Equation 80.1 of Problem 80):


r2M cos y2

r2¼

Mðcos y1 þ cos y2Þ

r2

We calculate the geomagnetic co-latitudes by Equation (71.3) of Problem 71:


cos y2 ¼ sinfB2 sinfþ cosfB2 cosf cosðl lB2Þð90:1Þ

The geographical coordinates of the North Pole of each dipole are given by (Fig. 90)

fB1 ¼ fB2 ¼ 90 30 ¼ 60

lB1 ¼ 0

lB2 ¼ 180

Substituting these values in Equation (90.1):

cos y1 ¼

ffiffiffi

3p

2sinfþ

1

2cosf cos l

cos y2 ¼

ffiffiffi

3p

2sinf

1

2cosf cos l

Adding the two equations:

cos y1 þ cos y2 ¼ffiffiffi

3p

sinf

GMNP1GMNP2

GNP

P

30°

q1

q2

Fig. 90

157 External magnetic field

Substituting in the equation of the potential:

F ¼

ffiffiffi

3p

M sinf

r2

We note that this is the same potential as that produced by only a centred dipole with

magnetic momentffiffiffi

3p

M oriented in the direction of the rotation axis, because the geomag-

netic co-latitude is 90º f.

(b) Bearing in mind the last results, we calculate the inclination by

tan I ¼ 2 cotð90 fÞ ¼ 2 tanf ð90:2Þ

The magnetic poles are the points on the surface of the Earth where the value of the

inclination is equal to 90º:

I ¼ 90 )f ¼ 90

f ¼ 90

Therefore the magnetic poles coincide with the geographical poles.

(c) We derive themainfield by taking the gradient of the potentialF. The components are

Br ¼ @F

@r¼

2ffiffiffi

3p

M sinf

r3

Bf ¼1

r

@F

@f¼

ffiffiffi

3p

M cosf

r3

Bl ¼ 1

r cosf

@F

@l¼ 0

At the magnetic equator the inclination is null (I ¼ 0) and according to Equation (90.2) the

latitude is null too (f ¼ 0). Substituting in the last equations:

Br ¼ 0

Bf ¼

ffiffiffi

3p

M

r3

Bl ¼ 0

The external magnetic field to annul out the internal field is therefore

Be ¼ 0;

ffiffiffi

3p

M

r3

; 0

91. The Earth’s magnetic field is formed by a centred dipole with northern geomag-

netic pole at 60 N, 0 E and B0 ¼ 32 000 nTand a uniform external field from the Sun

of 10 000 nT parallel to the equatorial plane.

(a) For a point at coordinates 60 N, 60 W, calculate the components X, Y, Z of the

total field, and the values of D and I.

(b) How do D and Z of the total field vary throughout the day with local time t?

158 Geomagnetism

(a) We calculate the geomagnetic main field components Z , H obtaining the

geomagnetic latitude given by (71.3):


f ¼ 61

Z ¼ 2B0 sinf ¼ 55 976 nT

H ¼ B0 cosf ¼ 15 514 nT

To calculate the NS (X ) and EW (Y ) components we need the geomagnetic declination

D (71.2):


cosf

D ¼ 63


Y ¼ H sinD ¼ 13 823 nT

The external field is parallel to the equatorial plane and has a diurnal period (o ¼ 2p/24)

because it comes from the Sun. We assume that at local time t ¼ 0 the Sun is at the point’s

meridian. If we denote by N the modulus of the external field (N ¼ 10 000 nT) and bearing

in mind Fig. 91a (representation of the plane parallel to the equator that contains the point)

we have at time t 6¼ 0

Y ¼ Bel ¼ N sinot

The radial and tangential components (Fig. 91b) are

P

(a)

N

N coswt

Beλ

wt

Fig. 91a


Z ¼ Br ¼ N cosot cosf

X ¼ Bf ¼ N cosot sinf

The total magnetic field is the sum of the two contributions:

ZT ¼ ð55 976þ 5000 cosotÞ nT

XT ¼ ð7043þ 8660 cosotÞ nT

YT ¼ ð13 823þ 10 000 sinotÞ nT

The declination D and inclination I are given by

tanD ¼YT

XT

¼13 823þ 10 000 sinot

7043þ 8660 cosot

tan I ¼ZT

HT

¼ZTffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X 2T þ Y 2

T

p

(b) To see how D and Z vary during the day with local time t we substitute several

values for t, obtaining the values in the table:

GNP

f

r

z = –Br

X = Bf

PNcosωt

Fig. 91b

t (h) Z (nT) D (º)

0 60976 41

6 55976 28

12 50976 83

18 55976 73

160 Geomagnetism

92. The Earth’s magnetic field is formed by a centred dipole of moment m and

north pole 60 N, 0 E, and a uniform external field of magnitude N ¼ B0/4 (B0 is

the geomagnetic constant of the internal field) parallel to the axis of rotation.

Determine:

(a) The total potential at any point.

(b) The coordinates of the boreal magnetic pole.

(c) The magnetic declination at the point 45 N, 45 E.

(d) The angle along the meridian between that point and the magnetic equator.

(a) The total potential (F) is the sum of two contributions: the main (internal) field

(Fi) and the external field (Fe):

F ¼ Fi þ F

e

The main field is formed by a centred dipole of moment m so the potential is given by

Fi ¼

Cm cos y

r2

We calculate the geomagnetic co-latitude y ¼ 90 – f at a point with geographical

coordinates (f, l) by (71.3):

cos y ¼ sinfB sinfþ cosfB cosf cosðl lBÞ ¼

ffiffiffi

3p

2sinfþ

1

2cosf cos l

GNP

P

f

ze

Xe

Fe

Fig. 92a


The external field is parallel to the axis of rotation; therefore its components are in the

vertical and NS directions (Fig. 92a). If we call # ¼ 90 f the geographical co-latitude,

the components are given by

Ber¼ N cos# ¼ N sinf

Be# ¼ N sin# ¼ N cosf

Ze ¼ Ber¼ N sinf

X e ¼ Be# ¼ N cosf

Bearing in mind that Be ¼ ∇Fe the potential for the external field is

Fe ¼ Nr cos# ¼ Nr sinf

Therefore the total potential is given by

F ¼Cm

ffiffi

3p

2sinfþ 1

2cosf cos l

r2

þ Nr sinf

(b) At the magnetic boreal pole the inclination is I ¼ 90º and the horizontal field

components H, X, Y are given by

tan I ¼Z

H) H ¼ 0


X 2 þ Y 2p

)X ¼ 0

Y ¼ 0

We obtain the components X and Y by taking the gradient of the potential

Be ¼ rFe

We bear in mind the relations@F

@y¼

@F

@f

X ¼ By

and obtain that the components are given by

X ¼ 1

r

@F

@f¼

Cmffiffi

3p

2cosf 1

2sinf cos l

r3

N cosf ¼ 0

Y ¼ 1

r cosf

@F

@l¼

Cm

2r3sin l ¼ 0 ) l ¼ 0

ð92:1Þ

Substituting in the equation of component X (92.1) the values

l ¼ 0

r ¼ a

B0 ¼Cm

a3

N ¼ B0=4

162 Geomagnetism

we obtain the geographical latitude of the boreal magnetic pole

tanfBM ¼2ffiffiffi

3p

1

2) fBM ¼ 51

Therefore the geographical coordinates of the magnetic boreal pole are 51 N, 0. Notice

that this is different from the geomagnetic pole.

(c) The declination is given by

tanD ¼Y

X¼

B0

2sin l

B0

ffiffiffi

3p

2cosf

1

2sinf cos l

B0

4cosf

¼ sin l

ffiffiffi

3p

1

2

cosf sinf cos l

Substituting the geographical coordinates of the point (45 N, 45 E) we obtain

D ¼ 62

(d) We call feq the angle between the geographical and magnetic equators at longitude

l ¼ 45º (Fig. 92b). Then the angle to calculate will be feq þ 45.

Tocalculatefeqwe take into account that at themagnetic equator thevertical component isZ¼ 0:

Z ¼ Br ¼@F

@r¼

2CM

r3

ffiffiffi

3p

2sinfeq þ

1

2cosfeq cos l

þ N sinfeq ¼ 0

tanfeq ¼4 cos l

4ffiffiffi

3p

þ 1

GNP GMNP

fB

f

P

l

feq

Fig. 92b


Substituting l ¼ 45º in the equation we obtain

feq ¼ 20

Then the angle we are asked for is 45º – 20º ¼ 25º.

93. The internal magnetic field is formed by two orthogonal dipoles of equal moment

M, one of them in the direction of the axis of rotation and the other contained in the 0

meridian on the equator. There is also an external field of constant intensity N ¼ B0/4

(B0 is the geomagnetic constant of the internal field) and lines of force parallel to the

axis of rotation.

(a) Calculate the potential of the total field for points on the surface.

(b) At which latitude is Z maximum on the 0 meridian?

(a) The potential of the total field is the sum of the two potentials of the internal

dipoles and the potential of the external field:

F ¼ F1 þ F2 þ Fe ¼

M cos y1

r2

M cos y2

r2

þ Fe


r2

þ Fe

In this equation r is the distance from the dipole’s centre (the Earth’s centre), y1 and y2 are

the co-latitudes relative to each dipole, and M ¼ Cm.

Dipole 1 is on the direction of the axis of rotation and so the geomagnetic co-latitude of

the point with respect to this dipole is

y1 ¼ 90 f

cos y1 ¼ sinf

Dipole 2 is in the equatorial plane (fB2 ¼ 0) and contained in the 0 meridian so the

geomagnetic co-latitude y2 is given by (71.3)


fB2 ¼ 0

lB2 ¼ 0

cos y2 ¼ cosf cos l

The equation for the potential of the external field is the same as that of Problem 92:

Fe ¼ þNr sinf

Therefore the potential of the total field is given by

F ¼Mðsinfþ cosf cos lÞ

r2þ Nr sinf

164 Geomagnetism

(b) We obtain the component Z by taking the vertical component of the gradient of the

potential (B ¼ ∇F)

Z ¼ Br ¼@F

@r¼

2Mðsinfþ cosf cos lÞ

r3þ N sinf

To calculate the maximum of Z on the 0 meridian we substitute l ¼ 0 and apply the

condition that the first derivate with respect to the latitude is null:

@Z

@f¼

2M

r3ðcosf sinfÞ þ N cosf ¼ 0

At the Earth’s surface r ¼ a and we know that

B0 ¼M

a3

Substituting this constant and solving the equation we determine the latitude at which the

Z component is maximum:

2B0ðcosf sinfÞ þB0

4cosf ¼ 0 ) f ¼ 48

94. The internal field has its northern geomagnetic pole at the coordinates 60 N, 0 E,

and B0 ¼ 30 000 nT. At a point with coordinates 30 N, 45 W, one observes an

increase of 7.7 in the value of the declination from 00:00 h to 09:00 h. There is

known to be an external field parallel to the Earth’s axis of rotation in the direction

from N to S which is null at 00:00 h and maximum at 12:00 h local times. Calculate:

(a) The components of the internal and external fields.

(b) The difference in the inclination at 00:00 h and 09:00 h.

(c) The maximum value of the declination during the day.

(a) To calculate the geomagnetic main field intensity components we obtain first the

geomagnetic latitude (Equation 71.3):


f ¼ 47:7

The declination and inclination are given by (71.2):


cosf ) D ¼ 31:7

tan I ¼ 2 tanf ) I ¼ 65:5

So the geomagnetic main field components are

Z ¼ 2B0 sinf ¼ 44 378 nT

H ¼ B0 cosf ¼ 20 190 nT

X ¼ H cosD ¼ 17 178 nT

Y ¼ H sinD ¼ 10 609 nT


The external field is parallel to the axis of rotation so its components are in the vertical and

NS directions. This field is null at 00:00 h and maximum at 12:00 h local time (period T ¼

24 h). Its components are given by

Ze ¼N

2ð1 cosotÞ sinf

X e ¼ N

2ð1 cosotÞ cosf

Y e ¼ 0

He ¼ X e

o ¼2p

T¼

2p

24

(b) To calculate the difference in the inclination we obtain first the value of N, bearing

in mind the time variation of the declination. The observed declination as a

function of time is given by

tanD ¼Y

X¼

Y þ Y e

X þ X e¼

Y

X N

2ð1 cosotÞ cosf

For t ¼ 0 h:

tanD1 ¼Y

X ¼ tanD ) D1 ¼ D ¼ 31:7

Since we know the change in declination between 0 h and 9 h, we find the declination at

9 h, D2:

D2 D1 ¼ 7:7 ) D2 ¼ 39:4

We know that at t ¼ 9 h

tanD2 ¼Y

X N

2ð1 cosotÞ cosf

¼ 0:82

Solving for N we obtain

N ¼2

ð1 cosotÞ cosfX

Y

tanD2

¼ 5766 nT

The magnetic inclination is given by

tan I ¼Z

H¼

Z þ Ze

H þ H e¼

Z þN

2ð1 cosotÞ sinf

H N

2ð1 cosotÞ cosf

At t ¼ 0 h:

I1 ¼ I ¼ 65:5

At t ¼ 9 h:

I2 ¼ 71:2

166 Geomagnetism

The difference in the inclination is therefore

I2 I1 ¼ 5:7

(c) The declination is given by

tanD ¼Y

X N

2ð1 cosotÞ cosf

ð94:1Þ

The maximum value is at 12 h because at that time the external field has the maximum

value. Substituting ot ¼ p and the values obtained for X , Y , N, and f:

Dmax ¼ 41:0

95. At a point on the Earth with coordinates 45 N, 45 E, measurements are made of

the magnetic field components at 00:00 h and 12:00 h in nT with a 2 nT precision:

0 h X ¼ 20 732 Y ¼ 2500 Z ¼ 57 768

12 h X ¼ 24 267 Y ¼ 2500 Z ¼ 54 232

It is known that the modulus of the magnetic field intensity has a harmonic diurnal

variation, and that the geomagnetic pole is on the zero meridian. Calculate:

(a) The moment and coordinates of the main field dipole.

(b) Expressions for the potential and components of the external field.

(Earth’s radius a ¼ 6400 km, m0 ¼ 4p107

kg m s2

A2).

(a) To calculate themoment and coordinates of themainfield dipolewe need to obtain the

geomagnetic main field intensity components. The observed values are equal to the

sum of the geomagnetic main field (X , Y , Z ) and the external field (X e, Ye, Z e):

X ¼ X þ X e

Y ¼ Y þ Y e

Z ¼ Z þ Ze

The geomagnetic main field is constant but the external field changes with time. So if we

denote by (X0, Y0, Z0) and (X12, Y12, Z12) the observed values at 0 h and at 12 h respectively

then the differences are due to the variations of the external field:

X12 X0 ¼ 3535 nT ¼ X e12 X e

0

Y12 Y0 ¼ 0 nT ¼ Y e12 Y e

0

Z12 Z0 ¼ 3536 nT ¼ Ze12 Ze

0

We notice that the Y component doesn’t vary, which implies that the component Y e is zero,

and so the external field is parallel to the axis of rotation. We also notice that the NS

component increases in the time interval between 0 h and 12 h, while the vertical

component diminishes, which implies that the polarity of the external field is inverted

with respect to that of the main field.

The modulus of the magnetic field intensity has a harmonic diurnal variation and

increases with time. Therefore the components of the external field are (Fig. 95a)


X e ¼ Nð1 cosotÞ cosf

Ze ¼ Nð1 cosotÞ sinf

o ¼2p

T¼

2p

24h

We notice that at time t ¼ 0 the external field is null and so

X e12 ¼ 3535 nT

Y e12 ¼ 0 nT

Ze12 ¼ 3536 nT

Therefore we calculate the main field components by

X ¼ X12 X e12 ¼ 20 732 nT

Y ¼ Y12 Y e12 ¼ 2500 nT

Z ¼ Z12 Ze12 ¼ 57 768 nT

H¼


ðX Þ2 þ ðY Þ2q

¼ 20 882 nT

If we consider the centred magnetic dipole model, the vertical and horizontal field

components are given by the equations

Z ¼ 2B0 sinf

H ¼ B0 cosf

Fe

ZeX

e

P

f

GNP

Fig. 95a

168 Geomagnetism

From these equations we obtain the geomagnetic latitude (f ) and the geomagnetic

constant B0:

tanf ¼Z

2H) f ¼ 54:1

B0 ¼Z

2 sinf ¼ 35 657 nT

From B0 we calculate the magnetic moment:

B0 ¼m04p

m

a3) m ¼ B010

7a3 ¼ 9:3 1022 A m2

The longitude of the Geomagnetic North Pole is lB ¼ 0 and we calculate the latitude fB

from the spherical triangle (Fig. 95b), but obtaining first the declination from the X and

Y components:

tanD ¼Y

X ) D ¼ 6:9

Applying the cosine rule:



fB ¼ 79:9

(b) We obtain the radial and transverse components of the external field from the

vertical and NS components:

Ber¼ Ze ¼ Nð1 cosotÞ sinf ¼

@Fe

@r

Bef ¼ X e ¼ Nð1 cosotÞ cosf ¼

1

r

@Fe

@f

q = 90º – f∗

GMNP

90º – fB

GNP

l – lB

180º – l∗

D∗

P

90º – f

Fig. 95b


Therefore the potential of the external field is given by

Fe ¼ Nrð1 cosotÞ sinf

where

N ¼Ze12

2 sinf¼ 2500 nT

96. The Earth’s main magnetic field is that of a centred dipole of moment M (M ¼

Cm) in the direction of the axis of rotation, and the external field is produced by

electric currents of intensity J circulating in a clockwise sense in a ring in the plane of

the ecliptic at a distance of 10 Earth radii around the Earth.

(a) Express the potential of the total field and the components Br and Bø on the

Earth’s surface for l ¼ 0.

(b) If the inclination of the ecliptic is 30, and the external field strength is N ¼ M /

4R3, what is the latitude of the northern magnetic pole?

(a) The potential of the total field is the sum of the potentials of the dipole and of the

external field:

FT ¼ Fþ F

e ¼M cos y

r2þ F

e

where r is the distance from the dipole’s centre (Earth’s centre) and y is the geomagnetic

co-latitude. The dipole is in the direction of the axis of rotation and so the geomagnetic co-

latitude is equal to the geographic co-latitude:

y ¼ 90 f

cos y ¼ sinf

To calculate the potential of the external field we know that it is produced by electric

currents of intensity J circulating in a clockwise sense at a distance of 10 Earth radii around

the Earth. These electric currents produce, at remote points, a magnetic dipolar field whose

modulus is m0J/2R, J being the current intensity and R the radius of the circular currents;

the dipole is oriented in the direction of the ecliptic’s axes with the negative pole in the

southern hemisphere, because the currents are clockwise. So the potential of the external

field is given by

Fe ¼

Cme cos y2

r2

In this equation me ¼ Jp100a2, a is the Earth’s radius, and y2 is the angle between the axes

of the circular currents and the direction of the point from the negative pole (Fig. 96)

Therefore

y2 ¼ 180 ðyþ eÞ

cos y2 ¼ cosðyþ eÞ

where e is the angle between the axes of the circular currents and the axes of rotation of the

Earth.

170 Geomagnetism

The potential of the external field is

Fe ¼

Cme cosðyþ eÞ

r2

and the total potential is given by

FT ¼

M cos y

r2þCme cosðyþ eÞ

r2

We calculate the components of the main field intensity by taking the gradient of the

potential:

Br ¼ @F

@r¼

2M cos y

r3

By ¼ 1

r

@F

@y¼

M sin y

r3

The magnitude of the external field is given by

Be ¼m04

J

10a¼ N

The radial and tangential components are

Ber¼ N cosðyþ eÞ

Bey ¼ N sinðyþ eÞ

GNP

qP

q2

e

e

–

+

Fig. 96


The total field is the sum of the two contributions:

Br ¼ 2M cos y

r3þ N cosðyþ eÞ

By ¼ M sin y

r3

þ N sinðyþ eÞ

(b) We know that the Earth’s dipole is in the direction of the axis of rotation and y is

the geographical co-latitude (Fig. 96). At the magnetic North Pole the total field is

vertical and the tangential component is By ¼ 0:

By ¼ M sin y

r3þ N sinðyþ eÞ ¼ 0

The values of N and e are known:

N ¼M

4r3)

M

r3¼ 4N

e ¼ 30

From these values we calculate y, the geographical co-latitude of the magnetic North Pole,

By ¼ 4N sin yþ N sinðyþ 30Þ ¼ 0

4 sin yþ

ffiffiffi

3p

2sin yþ

1

2cos y ¼ 0 ) y ¼ 9

97. Two spherical planets of radius a and separated by a centre-to-centre distance of 4a

orbit around each other and spin in the equatorial plane. Each has a magnetic field

produced by a centred dipole in the direction of the axis of rotation, with the positive

pole in the northern hemisphere andB0¼ 10 000 nT. Determine the componentsX, Y, Z,

D, and I of the totalmagnetic field at theNorth Pole of one of the planets (precision 1 nT).

The total magnetic field in each planet is the sum of its main field and the external field

created by the other planet. To determine the main field of either of the planets we need the

geomagnetic latitude, which is positive toward the negative pole, in this case, the South

Pole. Therefore the geomagnetic latitude and the components of the main field at the North

Pole are

f ¼ 90

Z ¼ 2B0 sinf ¼ 20 000 nT

H ¼ B0 cosf ¼ 0

X ¼ Y ¼ 0

The external field at one of the planets is created by the main field of the other planet and

corresponds to that of a magnetic dipole. Its components are (Fig. 97)

Ber¼

2Cm cos y

r3

Bey ¼

Cm sin y

r3

172 Geomagnetism

At the North Pole r ¼ffiffiffiffiffi

17p

a ¼ 4:12a and we calculate the geomagnetic co-latitude y by

(Fig. 97)

tanðy 90Þ ¼a

4a¼

1

4

y ¼ 14 þ 90 ¼ 104

Substituting these values in the equations for the radial and tangential components with

respect to the planet producing the external field:

Ber¼

2Cm cos y

r3

¼Cm cos y

35a3¼

B0 cos y

35¼ 69 nT

Bey ¼

Cm sin y

r3

¼Cm sin y

70a3¼

B0 sin y

70¼ 139 nT

From this value we calculate the vertical and horizontal components (Fig. 97):

Ze ¼ Bercosð180 yÞ þ Be

y cosðy 90Þ ¼ Bercos yþ Be

y sin y ¼ 118 nT

H e ¼ Bersinð180 yÞ þ Be

y sinðy 90Þ ¼ Bersin y Be

y cos y ¼ 33 nT

X e ¼ H e ¼ 33 nT

Y e ¼ 0

The components of the total field are finally

Z ¼ Z þ Ze ¼ 19882 nT

H ¼ H þ H e ¼ 33 nT

X ¼ X þ X e ¼ 33 nT

Y ¼ Y þ Y e ¼ 0

tanD ¼Y

X) D ¼ 0

tan I ¼Z

H) I ¼ 89:9

4aθ

+

–

aa

r

q – 90°

Ber

Beθ

q – 90°

Fig. 97


Main (internal), external, and anomalous magnetic fields

98. At a point with geographical coordinates 40 N, 45 E, measurements are made of

the magnetic field components, obtaining the values (in nT):

At 06:00 h: X ¼ 19 204; Y ¼ 0; Z ¼ 38 195

At 12:00 h: X ¼ 11 544; Y ¼ 0; Z ¼ 44 623

Buried at a depth of 20 m below this point is a dipole of magnetic moment Cm ¼ 0.01

T m3, oriented in the NS plane at an angle of 45 with the vertical towards the south,

and the positive pole upwards. Given that the external field at 12:00 h is twice that at

06:00 h, determine:

(a) The geomagnetic constant Bo and the coordinates of the northern geomagnetic pole.

(b) The magnitude and direction of the external field. How does the magnitude of the

external field vary with time?

(a) The observed magnetic field is composed of three parts: the geomagnetic main

(internal) field, the anomalous field (magnetic anomaly) created by the buried

dipole, and the external field.

We determine first the magnetic anomaly produced by the buried dipole, applying Equa-

tions (82.2) and (82.3) in Problem 82, substituting a ¼ 225º and x ¼ 0. The horizontal

component is in the NS direction (DX), because the buried dipole is in the NS-vertical

plane (Fig. 98a):

Z ¼2Cm cos a

d3¼ 1768 nT

X ¼Cm sin a

d3¼ 884 nT

Y ¼ 0

P

45°

α

d

N

Z

+

–

Fig. 98a

174 Geomagnetism

The components of the total magnetic field at 06:00 h are given by

X1 ¼ X þX þ X e ¼ 19 204 nT

Y1 ¼ Y þY þ Y e ¼ 0

Z1 ¼ Z þZ þ Ze ¼ 38 195 nT

At 12:00 h given that the external field is twice that at 06:00 h:

X2 ¼ X þX þ 2X e ¼ 11 544 nT

Y2 ¼ Y þY þ 2Y e ¼ 0

Z2 ¼ Z þZ þ 2Ze ¼ 44 623 nT

If we subtract both sets of equations and obtain

X2 X1 ¼ X e ¼ 7660 nT

Y2 Y1 ¼ Y e ¼ 0

Z2 Z1 ¼ Ze ¼ 6428 nT

Now we can calculate the elements of the main field

X ¼ X1 X X e ¼ 25 980 nT

Y ¼ Y1 Y Y e ¼ 0

Z ¼ Z1 Z Ze ¼ 33 535 nT

H ¼


ðX Þ2 þ ðY Þ2q

¼ X ¼ 25 980 nT

tanD ¼Y

H) D ¼ 0

We calculate the geomagnetic latitude of the point f and the geomagnetic constant B0

from the vertical and horizontal geomagnetic main field components by

Z ¼ 2B0 sinf

H ¼ B0 cosf

tanf ¼Z

2H) f ¼ 32:8

B0 ¼Z

2 sinf ¼ 30 953 nT

We obtain the coordinates of the Geomagnetic North Pole by (Fig. 98b)

D ¼ 0 ) lB ¼ 180þ l ¼ 225 E ¼ 135 W

90 fB ¼ f f ) fB ¼ 82:8

(b) The magnitude of the external field at 06:00 h is

B6e ¼


X 2e þ Y 2

e þ Z2e

q

¼ 10 000 nT

175 Main (internal), external, and anomalous magnetic fields

At 12:00 h the magnitude is

B12e ¼ 2B6

e ¼ 20 000 nT

The direction of the external field is in the NS-vertical plane because the EW component is

null, forming with the horizontal an angle Ie (Fig. 98c). This direction is the same at 06:00

h and at 12:00 h. We calculate the angle Ie by

tan Ie ¼Ze

Xe

Ie ¼ 40

GNPGMNP

P

f90° – fB

f∗

Fig. 98b

Z

Ze

Ie

Xe

Be

N

Fig. 98c

176 Geomagnetism

Because at

06 :00 h ðt ¼ p=2Þ ! N ¼ 10 000

12 :00 hðt ¼ pÞ ! N ¼ 20 000

the variation of the magnitude of the external field with time is given by

N ¼ 10 000ð1 cos tÞ

99. Buried at a depth of 100 m at a point with geographical coordinates 45 N, 45 W is

a dipole anomaly of Cm¼ 0.1 T m3, inclined 45 from the horizontal northwards in the

vertical plane with the negative pole downwards. Measurements gave the following

results (in nT):

09:00 h X ¼ 27 759; Y ¼ 0; Z ¼ 30 141

12:00 h X ¼ 28 052; Y ¼ 0; Z ¼ 30 141

Find:

(a) The coordinates of the magnetic dipole’s North Pole.

(b) The value of B0.

(c) An expression for the variation Sq knowing that it is zero at 00:00 h and

maximum at 12:00 h.

(a) As in Problem 98, the observed field is the result of three parts: the main (internal)

field, the buried dipole field, and the external field. To calculate the coordinates of

the magnetic dipole’s North Pole we need to obtain the components of the

geomagnetic main field from the components of the total field. With this aim we

begin by calculating the magnetic anomaly created by the buried dipole, applying

Equations (82.2) and (82.3), and substituting a ¼ 225º and x ¼ 0. The horizontal

component is in the NS direction (DX) given that the dipole is on the NS-vertical

plane (Fig. 99a).

Z ¼2Cm cos a

d3¼ 141 nT

X ¼Cm sin a

d3¼ 71 nT

Y ¼ 0

The total observed field at 09:00 h is

X1 ¼ X þX þ X e1

Y1 ¼ Y þY þ Y e1 ¼ 0 ) Y ¼ Y e

1

Z1 ¼ Z þZ þ Ze1

The total field at 12:00 h is

X2 ¼ X þX þ X e2

Y2 ¼ Y þY þ Y e2

Z2 ¼ Z þZ þ Ze2


Subtracting both sets of equations we obtain

X2 X1 ¼ X e2 X e

1 ¼ 293 nT

Y2 Y1 ¼ Y e2 Y e

1 ¼ 0 ) Y e2 ¼ Y e

1 ¼ Y

Z2 Z1 ¼ Ze2 Ze

1 ¼ 0 ) Ze2 ¼ Ze

1

ð99:1Þ

We assume that the time variation of the observations is due to the diurnal Sq variation

which is zero at 00:00 h and maximum at 12:00 h. Therefore the only possible values

for the components Ye and Ze are zero because these components have the same values at

09:00 h and at 12:00 h:

Y e2 ¼ Y e

1 ¼ Y ¼ 0

Ze2 ¼ Ze

1 ¼ 0

Then, the intensity of the external field is given by

X e ¼ Nð1 cosotÞ

o ¼2p

24

The NS components of this field at 09:00 h (X1) and at 12:00 h (X2) are

X e1 ¼ N 1 cos

3p

4

¼ N 1þ

ffiffiffi

2p

2

X e2 ¼ Nð1 cos pÞ ¼ 2N

Subtracting the two values and using Equation (99.1) we obtain

X e2 X e

1 ¼ 1

ffiffiffi

2p

2

N ¼ 1

ffiffiffi

2p

2

X e2

2¼ 293 nT ) X e

2 ¼ 2001 nT

X e1 ¼ 1708 nT

45° a

Z

∆X

∆Z

d

–

+

NP

Fig. 99a

178 Geomagnetism

The components of the geomagnetic main field intensity are given by

X ¼ X1 X X e1 ¼ 26 122 nT

Y ¼ Y1 Y Y e1 ¼ 0

Z ¼ Z1 Z Ze ¼ 30 000 nT

H ¼


ðX Þ2 þ ðY Þ2q

¼ X ¼ 26 122 nT

tanD ¼Y

H) D ¼ 0

The geomagnetic latitude of the point f is determined from the vertical and horizontal

geomagnetic main field components:

Z ¼ 2B0 sinf

H ¼ B0 cosf

tanf ¼Z

2H) f ¼ 29:9

We obtain the coordinates of the Geomagnetic North Pole by (Fig. 99b)

D ¼ 0 ) lB ¼ 180þ l ¼ 135 E

90 fB ¼ f f ) fB ¼ 74:9

GMNPGNP

f

f∗

90° – fB

P

Fig. 99b


(b) The geomagnetic constant B0 is given by

H ¼ B0 cosf

B0 ¼H

cosf ¼ 30 133 nT

(c) We have obtained above that

Fe ¼ X e ¼ Nð1 cosotÞ

To calculate N we take into account that the Sq variation is maximum at 12:00 h:

X e2 ¼ 2N ) N ¼ 1000 nT

100. At a point with geographical coordinates 60 N, 45 E, a magnetic

observation results in the following values: FT ¼ 48 277 nT, DT ¼ 2.9, IT ¼ 63.7.

It is known that at 20 m below this point there is a horizontal magnetic dipole of

momentCm¼ 0.01 Tm3, with the positive pole oriented in the direction N 60 E. There

is also an external field parallel to the axis of rotation directed southwards of magni-

tude 1000 nT.

Determine the main (internal) field components, the constant B0, and the geocentric

geographical coordinates of the northern and southern geomagnetic poles (precision

1 nT).

We first calculate the components of the total field intensity from FT, DT, and IT(Fig. 100a) by

HT ¼ FT cos IT ¼ 21 390 nT

ZT ¼ FT sin IT ¼ 43 280 nT

XT ¼ HT cosDT ¼ 21 363 nT

YT ¼ HT sinDT ¼ 1082 nT

Y

Z

ZT

XYT

FT

IT

HT

DT

XT

Fig. 100a

180 Geomagnetism

The observed field is the result of three parts: the main field (X , Y , Z ), the buried dipole

field (DX, DY, DZ), and the external field (Xe, Ye, Ze):

XT ¼ X þX þ X e

YT ¼ Y þY þ Y e

ZT ¼ Z þZ þ Ze

We determine the magnetic anomaly created by the buried dipole from Equations (82.2)

and (82.3) substituting a¼ 90º and x¼ 0. If we call X60 the direction N 60º E the horizontal

component of this anomaly is DX60 (Fig. 100b):

Z ¼2Cm cos a

d3¼ 0 nT

X60 ¼Cm sin a

d3¼ 1250 nT

The NS and EW components will be given by (Fig. 100c)

X ¼ X60 cos 60 ¼ 625 nT

Y ¼ X60 sin 60 ¼ 1083 nT

The external field is parallel to the Earth’s axis of rotation in the southwards direction so its

components are in the vertical and NS direction and are given by (Fig. 100d)

Ze ¼ 1000 sinf ¼ 866 nT

X e ¼ 1000 cosf ¼ 500 nT

We calculate the main field components from these values:

X ¼ XT X X e ¼ 21 238 nT

Y ¼ YT Y Y e ¼ 1082 nT

Z ¼ ZT Z Ze ¼ 42 414 nT

H ¼


ðX Þ2 þ ðY Þ2q

¼ 21 266 nT

tanD ¼Y

X ) D ¼ 2:9

PX60

d

Z

a

– +

Fig. 100b


The geomagnetic latitude of the point f and the geomagnetic constant B0 are found from

the vertical and horizontal geomagnetic main field components by

Z ¼ 2B0 sinf

H ¼ B0 cosf

tanf ¼Z

2H) f ¼ 44:9

B0 ¼Z

2 sinf ¼ 30 953 nT ¼ 30 044 nT

N

E

60°

∆Y

∆X60

∆X

Fig. 100c

GNP

Xe

Fe

Ze

P

f

Fig. 100d

182 Geomagnetism

To calculate the geographical coordinates of the Geomagnetic North Pole we use the

corresponding spherical triangle (Fig. 100e). We obtain the latitude fB by applying the

cosine law for the angle 90º – fB:



fB ¼ 75:0

We obtain the longitude lB by applying the cosine rule for the angle 90º – f :

cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ



cosfB cosf

l lB ¼ 180:0

lB ¼ 135:0 W

Therefore the coordinates of the Geomagnetic North Pole are

fB ¼ 75 N lB ¼ 135 W

The coordinates of the Geomagnetic South Pole (the antipodal point) are:

fA ¼ fB ¼ 75 S lA ¼ 180 þ lB ¼ 45 E

101. At a point with geographical coordinates 30 N, 30 E, the observed geomagnetic

field components are (in nT): X ¼ 15 364, Y ¼ 7660, Z ¼ 48 980. The northern

geomagnetic pole is at 60 N, 0 E, andB0¼ 30 000 nT. There is also a constant external

magnetic field normal to the equatorial plane, with a southwards direction, of 1000 nT

intensity. Buried 10 m below the observation point is a magnetic dipole. Calculate:

(a) The magnetic anomalies DX, DY, DZ, DH, DF.

90º – fB

GNP

l – lB

180º – l∗

q = 90º – f∗

GMNP

D∗

90º – f

P

Fig. 100e


(b) The orientation and magnetic moment (Cm, in nT m3) of the buried dipole.

(a) The observed values are the sum of the geomagnetic main field, the magnetic field

due to the buried dipole, and the external field:

X ¼ X þX þ X e

Y ¼ Y þY þ Y e

Z ¼ Z þZ þ Ze

To obtain the magnetic anomalies from these equations we first calculate the geomagnetic

latitude and the vertical and horizontal components of the geomagnetic main field by (71.3):


f ¼ 54

Z ¼ 2B0 sinf ¼ 48 479 nT

H ¼ B0 cosf ¼ 17 676 nT

The declination and inclination are given by


cosf ) D ¼ 25

tan I ¼ 2 tanf ) I ¼ 70

The NS and EW components are

X ¼ H cosD ¼ 16 007 nT


The external field is parallel to the axis of rotation directed southwards so its components

are in the vertical and NS direction and are given by (Fig. 101a)

Ze ¼ 1000 sinf ¼ 500 nT

X e ¼ 1000 cosf ¼ 866 nT

Y e ¼ 0

From these values we calculate the magnetic anomalies DX, DY, DZ, DH, DF:

X ¼ X X X e ¼ 223 nT

Y ¼ Y Y Y e ¼ 162 nT

Z ¼ Z Z Ze ¼ 1 nT

H ¼ X cosD þY sinD ¼ 271 nT


(b) We call b

the angle between the geographical north and the buried dipole direc-

tions. Then using Equations (82.2) and (82.3) we obtain

Z ¼2Cm cos a

d3ð101:1Þ

184 Geomagnetism

X ¼Cm sin a

d3cos b ð101:2Þ

Y ¼Cm sin a

d3sinb ð101:3Þ

To solve this system of three equations in three unknowns (Cm, a, b) we divide Equation

(101.3) by (101.2):

tan b ¼Y

X) b ¼ 36 þ 180 ¼ 144

This value of the angle bimplies that the dipole is oriented in the N 144º E direction.

To calculate the angle a between the buried dipole and the vertical we divide Equation

(101.2) by (101.1):

X

Z¼

1

2tan a cos b

tan a ¼ 2

cos b

X

Z) a 90

Therefore the dipole is practically horizontal (Fig. 101b) and this explains the small value

of the vertical component DZ.

Finally we calculate the magnetic moment of the buried dipole from Equation (101.2):

Cm ¼ d3

sin a cos bX ¼ 2:8 105 nTm3

GNP

P

Xe

Ze

Fef

Fig. 101a


102. At a point with geographical coordinates 30 N, 30 E, magnetic measurements

give the following values: F ¼ 52 355 nT, I ¼ 70.5, and D ¼ -26.0. The terrestrial

dipole has a north pole at 60 N, 0 E and B0 ¼ 30 000 nT. There is also a constant

external field of 1000 nTwith lines of force contained in the plane of the 30 Emeridian

at an angle of 60 with the equatorial plane and directed southwards.

(a) Calculate the anomalies DX, DY, DZ.

(b) If the anomalies are produced by a dipole buried at 10 m below the point, what is

its orientation and its magnetic moment, Cm? Neglect values less than 10 nT.

(a) We first calculate the components of the total field from F, D, and I by

H ¼ F cos I ¼ 17 476 nT

Z ¼ F sin I ¼ 49 352 nT

X ¼ H cosD ¼ 15 707 nT


The magnetic anomalies are found by subtracting from the observed values the main and

external field contributions:

X ¼ X X X e

Y ¼ Y Y Y e

Z ¼ Z Z Ze

First we determine the geomagnetic latitude and declination using (71.3) and (71.2):


f ¼ 53:9


cosf ) D ¼ 25:1

The vertical and horizontal components are given by

Z ¼ 2B0 sinf ¼ 48 479 nT

H ¼ B0 cosf ¼ 17 676 nT

X ¼ H cosD ¼ 16 007 nT


– +

a

P

d

N 144° E

Fig. 101b

186 Geomagnetism

The external field is on the plane containing the vertical and NS directions (Figs 102a

and 102b):

Ze ¼ 1000 cosð60 fÞ ¼ 866 nT

X e ¼ 1000 sinð60 fÞ ¼ 500 nT

Y e ¼ 0

GNP

P60°

Xe

Ze

f Fe

Fig. 102a

Ze

Fe

Xe

f

60°

N

E

Fig. 102b


Finally, the magnetic anomalies are given by

X ¼ X X X e ¼ 200 nT

Y ¼ Y Y Y e ¼ 163 nT

Z ¼ Z Z Ze ¼ 7 nT

(b) We call b the angle between the positive pole of the buried dipole and the geographical

north. Then applying Equations (82.2) and (82.3) we obtain

Z ¼2Cm cos a

d3ð102:1Þ

X ¼Cm sin a

d3cos b ð102:2Þ

Y ¼Cm sin a

d3sin b ð102:3Þ

We divide Equation (102.3) by Equation (102.2) and obtain

tan b ¼Y

X) b ¼ 39:2 þ 180 ¼ 140:8

Therefore the dipole is oriented in the N 140.8º E direction.

To calculate the inclination of the dipole from the vertical we divide Equation (102.2) by

Equation (102.1):

X

Z¼

1

2tan a cos b

tan a ¼ 2

cos b

X

Z) a ¼ 89:2

This value implies that the dipole is nearly horizontal.

Finally we calculate Cm from the total anomalous field DB

B ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X 2 þY 2 þZ2p

¼ 258 nT

B ¼Cm

d3) Cm ¼ 258 000 nTm3

103. The Earth’s magnetic field is formed by a centred dipole with a geomagnetic pole

at 60 N, 0 E, and B0 ¼ 30 000 nT, and an external field of 10 000 nT parallel to the

equatorial plane and to the zero meridian.

(a) Calculate the components X, Y, Z of the total field at a point P with geographical

coordinates 60 N, 30 W.

(b) If at 30 m in the direction of the compass needle from P there is a vertical dipole of

moment Cm ¼ 4000 nT m3 buried 40 m deep, what would be the anomaly DZ

produced at P?

188 Geomagnetism

(a) The components of the total field are the sum of the geomagnetic main field and

the external field:

X ¼ X þ X e

Y ¼ Y þ Y e

Z ¼ Z þ Ze

Let us first calculate the geomagnetic latitude, declination, and inclination using (71.3)

and (71.2):


f ¼ 75:1


cosf ) D ¼ 76:5

tan I ¼ 2 tanf ) I ¼ 82:4


Z ¼ 2B0 sinf ¼ 57 983 nT

H ¼ B0 cosf ¼ 7714 nT



The external field is parallel to the equatorial plane and to the zero meridian. If its

magnitude is N ¼ 10 000 nT, its components, from Fig. 103a (plane through the point

parallel to the equator) and Fig. 103b (plane through the geographical meridian of the

point), are given by

Ze ¼ Ber¼ N cos l cos’ ¼ 4330 nT

X e ¼ Bef ¼ N cos l sin’ ¼ 7500 nT

Y e ¼ Bel ¼ N sin l ¼ 5000 nT

Therefore the components of the total field are

Z ¼ 62 313 nT

X ¼ 9301 nT

Y ¼ 12 501 nT

(b) The buried vertical dipole is in the direction of the compass, that is, in the direction

of the magnetic north (Fig. 103c). We calculate the anomaly DZ produced at

P using Equations (82.2) and (82.3) substituting a ¼ 0º and x ¼ 30:

Z ¼Cm½ðx2 2d2Þ cos aþ 3dx sin a

½x2 þ d25=2

Z ¼Cmðx2 2d2Þ

½x2 þ d25=2¼ 0:029 nT


Be

l

P

Ncosλ

Plane parallel

to the equator

Beλ= Nsinl

Fig. 103a

GNP

P

f

Ncosλ

Bef

Ber

Fig. 103b

190 Geomagnetism

104. The Geomagnetic North Pole is at 60 N, 150 W, with B0 ¼ 30 000 nT, and there

is an external magnetic field of intensity 3000 nT parallel to the axis of rotation

pointing away from the North Pole. Buried 10 m below a point with coordinates

30 N, 30 E there is a horizontal dipole with Cm ¼ 40 000 nT m3 and the negative

pole pointing in the direction N 45 E.

(a) What are the components X, Y, Z of the total field?

(b) Calculate the total field anomaly DF.

(c) What is the angle between the direction of the compass and geographic north?

(a) The components of the total field are the sum of the geomagnetic main field, the

magnetic field due to the buried dipole, and the external field:

X ¼ X þX þ X e

Y ¼ Y þY þ Y e

Z ¼ Z þZ þ Ze

We determine first the geomagnetic latitude, declination, and inclination using (71.3)

and (71.2):


f ¼ 0


cosf ) D ¼ 0

tan I ¼ 2 tanf ) I ¼ 0

P x

d

z

–

+

NM

Fig. 103c



Z ¼ 2B0 sinf ¼ 0 nT

H ¼ B0 cosf ¼ 30 000 nT

X ¼ H cosD ¼ 30 000 nT

Y ¼ H sinD ¼ 0 nT

We calculate the magnetic anomaly created by the buried dipole from Equations (82.2) and

(82.3), substituting a ¼ 270º and x ¼ 0. The vertical component DZ and the horizontal

component DX45 of the anomaly in the direction N 45 E are (Fig. 104)

Z ¼2Cm cos a

d3¼ 0 nT

X45 ¼Cm sin a

d3¼ 40 nT

The NS and EW components are given by

X ¼ 40 sin 45 ¼ 28:3 nT

Y ¼ 40 cos 45 ¼ 28:3 nT

The external field is parallel to the axis of rotation directed southwards, so its components

are in the vertical and NS direction and are given by

Nr ¼ Ze ¼ N cos 60 ¼ 1500 nT

N# ¼ X e ¼ N sin 60 ¼ 2598 nT

The components of the observed total magnetic field are

X ¼ 27 430 nT

Y ¼ 28 nT

Z ¼ 1500 nT

P

d

a

–+

z

X45

Fig. 104

192 Geomagnetism

(b) The total field anomaly DF is given by

F ¼ X ¼ 28 nT

(c) The direction of the compass is affected by the three fields. The angle D between

the direction of the compass and the geographic north is obtained from the

horizontal components of the total observed field

tanD ¼Y

X¼

28

27430) D ¼ 0:06

105. The geomagnetic field is that of a dipole in the direction of the axis of rotation

and B0 ¼ 30 000 nT. There is also a constant external field of 2500 nT normal to the

equatorial plane in the direction of the South Pole.

(a) Calculate the value of the inclination observed at a point P with coordinates 45 N,

45 E given that, at 10 m below it, there is a vertical dipole with the negative pole

upwards and moment Cm ¼ 40 000 nT m3.

(b) For a point 20 m north of P, calculate the observed inclination and declination,

and the total field anomaly DF.

(a) The components of the total observed field are the sum of the geomagnetic main

field, the magnetic field due to the buried dipole, and the external field:

X ¼ X þX þ X e

Y ¼ Y þY þ Y e

Z ¼ Z þZ þ Ze

The magnetic dipole is oriented in the direction of the axis of rotation and therefore

f ¼ f ¼ 45

D ¼ 0

Then the components of the geomagnetic main field are

Z ¼ 2B0 sinf ¼ 42 426 nT

H ¼ B0 cosf ¼ 21 213 nT

X ¼ H cosD ¼ 21 213 nT

Y ¼ H sinD ¼ 0

The magnetic anomaly created by the dipole is obtained from Equations (82.2) and (82.3)

substituting a ¼ 0º and x ¼ 0 (Fig. 105a):

Z ¼2Cm cos a

d3¼ 80 nT

X ¼Cm sin a

d3¼ 0

Y ¼ 0


The external field is parallel to the axis of rotation directed southwards so its components

are in the vertical and NS direction (Fig. 105b) and are given by

Ze ¼ 2500 sinf ¼ 1768 nT

X e ¼ 2500 cosf ¼ 1768 nT

Y e ¼ 0

P

d

z

X

–

+

Fig. 105a

Ze

GNP

f Fe

Xe

P

Fig. 105b

194 Geomagnetism

Therefore the components of the observed field are

X ¼ X þX þ X e ¼ 19 445 nT

Y ¼ Y þY þ Y e ¼ 0

H ¼ X

Z ¼ Z þZ þ Ze ¼ 44 274 nT

From these values we calculate the observed inclination

tan I ¼Z

H) I ¼ 66:3

(b) For a point Q located 20 m to the north of P we can assume that the main and

external fields have the same value as at point P and only the magnetic anomaly

created by the buried dipole is different. We calculate this anomaly from Equations

(82.2) and (82.3) substituting a ¼ 0º and x ¼ 20:

Z ¼Cm½ðx2 2d2Þ cos aþ 3dx sin a

½x2 þ d25=2¼

Cmðx2 2d2Þ

½x2 þ d25=2¼ 1 nT

X ¼Cm½ð2x2 d2Þ sin a 3dx cos a

½x2 þ d25=2¼

Cm3dx

½x2 þ d25=2¼ 4 nT

Y ¼ 0

Therefore, the components of the observed field at that point are

X ¼ X þX þ X e ¼ 19 441 nT

Y ¼ Y þY þ Y e ¼ 0

H ¼ X

Z ¼ Z þZ þ Ze ¼ 44 193 nT

From these values we calculate the observed inclination and declination by the expressions

tan I ¼Z

H) I ¼ 66:2

tanD ¼Y

X) D ¼ 0

The total field anomaly DF is given by

F ¼ X cos I þZ sin I ¼ 4 cos 66 1 sin 66 ¼ 3 nT

106. Consider a point with coordinates 30 N, 30 E under which is buried at a depth

of 100 m a horizontal dipole of moment m0 m/4p ¼ 1 T m3, with the positive pole in

the direction N 60 E. The terrestrial field is formed by a centred dipole in the

direction of the axis of rotation and a constant external field of 10 000 nT from the

Sun, B0 ¼ 30 000 nT.

(a) Calculate F, D, and I at that point on December 21 at 12 noon.


(b) How do D and I vary throughout the year?

(a) The components of the total field intensity are the sum of the geomagnetic main

field, the magnetic field due to the buried dipole, and the external field:

X ¼ X þXþ X e

Y ¼ Y þY þ Y e

Z ¼ Z þZ þ Ze

We first calculate the geomagnetic latitude and the declination and inclination. The

magnetic dipole is oriented in the direction of the axis of rotation and therefore

f ¼ f ¼ 30

D ¼ 0

Then the components of the geomagnetic main field are

Z ¼ 2B0 sinf ¼ 30 000 nT

H ¼ B0 cosf ¼ 25 981 nT

X ¼ H cosD ¼ 25 981 nT

Y ¼ H sinD ¼ 0

We calculate the magnetic anomaly produced by the buried dipole from Equations (82.2)

and (82.3) substituting a ¼ 90º and x ¼ 0. We call X60 the direction N 60º E and DX60 the

horizontal component of the anomaly (Fig. 106a):

Z ¼2Cm cos a

d3¼ 0 nT

X60 ¼Cm sin a

d3¼ 1000 nT

P

a

– +

d

z

X60

Fig. 106a

196 Geomagnetism

The NS and EW components are given by (Fig. 106b)

X ¼ X60 cos 60 ¼ 500 nT

Y ¼ X60 sin 60 ¼ 866 nT

The external field has a diurnal period (o ¼ 2p / 24 h) and we know that the Sun is at the

meridian point at 12:00 h (solar time). This field changes through the year because the Sun

moves on the ecliptic plane (apparent motion) which is tilted with respect to the equatorial

plane by an angle e¼ 23º. Therefore the solar declination (d), the angle from the Sun to the

celestial equator, changes through the year. On December 21 (winter solstice) this angle is

d¼e¼23º (Fig. 106c). If we call N the magnitude of the external field (N¼ 10 000nT)

its components on December 21 at 12:00 are (Fig. 106d)

Ze ¼ Br ¼ N cosðfþ eÞ ¼ 6018 nT

X e ¼ Bf ¼ N sinðfþ eÞ ¼ 7986 nT

Y e ¼ 0

60°

N

E∆Y

∆X

∆X60

Fig. 106b

GNP

P

e

f

f +e

N = Be

Sun

Fig. 106c


The components of the observed field are

X ¼ X þX þ X e ¼ 33 467 nT

Y ¼ Y þY þ Y e ¼ 866 nT

Z ¼ Z þZ þ Ze ¼ 36 018 nT

Therefore the total field F and the declination D are

F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X 2 þ Y 2 þ Z2p

¼ 49 166 nT

tanD ¼Y

X) D ¼ 1:5

sin I ¼Z

F) I ¼ 47:1

(b) Any other day at 12:00 h

X ¼ X þX þ X e ¼ 25 481 nTþ N sinðf dÞ

Y ¼ Y þY þ Y e ¼ 866 nT

Z ¼ Z þZ þ Ze ¼ 30 000 nTþ N cosðf dÞ

107. The internal field of the Earth corresponds to a centred dipole with the negative

pole in the northern hemisphere at coordinates 80 N, 130 W and B0 ¼ 30 000 nT.

There is a uniform external field from the Sun of 1000 nT. Buried at 500 m depth

under a point P with geocentric coordinates 40 N, 50 E there is a positive magnetic

pole of strength CP ¼ 0.5 T m². Calculate:

(a) The total field components X, Y, and Z, and the magnetic and geomagnetic

declination at P on March 21 at 12:00 h.

GNP

N

Z = – Br

X = Bf

r

f

P

f + e

Fig. 106d

198 Geomagnetism

(b) The same parameters for a point 200 m north of P, assuming that neither the

internal nor the external fields change (precision 1 nT).

(a) The components of the intensity of the total field are the sum of the geomagnetic

main field, the external field, and the magnetic field due to the buried pole.

To calculate the main field we determine first the geomagnetic latitude and the declination

by (71.3) and (71.2):


f ¼ 30


cosf ) D ¼ 0

From these values we obtain the vertical and horizontal components of the geomagnetic

main field:

Z ¼ 2B0 sinf ¼ 30 000 nT

H ¼ B0 cosf ¼ 25 981 nT

X ¼ H cosD ¼ 25 981 nT

Y ¼ H sinD ¼ 0

To calculate the external field we notice that it comes from the Sun which on March 21

(spring equinox) is on the equatorial plane so that the external field is parallel to this plane.

In addition this field changes during the day as a function of local time t with a diurnal

period (o ¼ 2p/24 h). At t ¼ 12 h, the external field is maximum given that at this time

the Sun is at the meridian point (Fig. 107a). Calling N its magnitude (N ¼ 1000 nT), the

components of the external field are given by

GNP

Xe

P

Ze

N

f

Fig. 107a


Ze ¼ Br ¼ N cosf ¼ 766 nT

X e ¼ Bf ¼ N sinf ¼ 643 nT

Y e ¼ Bel ¼ 0

The magnetic field anomaly created by the buried pole is derived from its potential DF,

given by

F ¼CP

r

Applying the gradient, we obtain

B ¼ F

but the only component is the vertical:

Z ¼ Br ¼@ðFÞ

@r¼

CP

r2

Substituting r ¼ d ¼ 500 m in this equation we obtain

Z ¼ 2000 nT

Therefore the components of the total field and the observed declination are given by

X ¼ X þ X e þX ¼ 26 624 nT

Y ¼ Y þ Y e þY ¼ 0

Z ¼ Z þ Ze þZ ¼ 28 766 nT

tanD ¼Y

X) D ¼ 0

(b) The radial component of the magnetic field anomaly created by the buried pole for

a point 200 m north of P (x ¼ 200 m) is given by

Br ¼ @ðFÞ

@r¼

CP

r2

From Fig. 107b the vertical and NS components are

Z ¼ Br cos a ¼CP

r2

d

r¼

CPd

ðx2 þ d2Þ3=2

X ¼ Br sin a ¼CP

r2

x

r¼

CPx

ðx2 þ d2Þ3=2

Y ¼ 0

Substituting the values given (d ¼ 500 m, x ¼ 200 m, CP ¼ 0.5 Tm2), we obtain

Z ¼ 1601 nT

X ¼ 640 nT

Y ¼ 0

200 Geomagnetism

The components of the observed total field and the declination are the sum of the three

contributions:

X ¼ X þ X e þX ¼ 27 264 nT

Y ¼ Y þ Y e þY ¼ 0

Z ¼ Z þ Ze þZ ¼ 29 165 nT

tanD ¼Y

X) D ¼ 0

Paleomagnetism

108. At a point with geographical coordinates 60 N, 60 Wa 1 cm3 sample was taken

of a rock with remanent magnetism, age 10 000 years, specific susceptibility

0.01 cm3. The magnetization components of the rock were:

X ¼ 40, Y ¼ 30, Z ¼ 50 nT (N, E, nadir).

The current field is B0 ¼ 30 000 nT and the geomagnetic pole coincides with the

geographical pole. Calculate:

(a) The coordinates of the virtual geomagnetic pole which corresponds to the

sample.

(b) The magnetic moment of the terrestrial dipole 10 000 years ago.

(c) The secular variation of F, D, and I in nT and minutes per year assuming that the

variation since that time has been constant.

Px

X

α

d

z

r a

+

Fig. 107b

201 Paleomagnetism

(a) First we determine the declination D and the geomagnetic co-latitude y, corres-

ponding to the virtual pole, from the magnetization components of the rock X, Y,

and Z:

tanD ¼Y

X) D ¼ 36:9


X 2 þ Y 2p

¼ 50 nT

tan I ¼Z

H) I ¼ 45

tan I ¼ 2 cot y ) y ¼ 63:4;fvirtual ¼ 26:6

Since at present the geomagnetic pole coincides with the geographical pole, the geograph-

ical latitude of the point coincides with the present geomagnetic latitude:

f ¼ fpresent ¼ 60N

To determine the coordinates of the virtual Geomagnetic North Pole (VP), corresponding

to the magnetization of the rock, we solve the spherical triangle of Fig. 108 for ’B and lBusing the obtained values of y and D. The latitude fB applying the cosine rule is given by

cosð90 fBÞ ¼ cos y cosð90 fÞ þ sin y sinð90 fÞ cosD

sinfB ¼ cos y sinfþ sin y cosf cosD

fB ¼ 48:2

To obtain the longitude lB we again apply the cosine rule:

cos y ¼ sinð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ

þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ

cos y ¼ sinfB sinfþ cosfB cosf cosðl lBÞ

90º – fB

GNP

VP

l – lB

90º – f∗

180º – l∗

D∗

P

90º – f

Fig. 108

202 Geomagnetism

cosðl lBÞ ¼cos y sinfB sinf

cosfB cosf

l lB ¼ 126:4

To choose between the positive and negative solution we bear in mind that the declination

is negative and so the point is to the east of the virtual magnetic North Pole:

D < 0 ) l lB > 0

lB ¼ 173:6 E

(b) To obtain the magnetic moment we first calculate the constant B0. The susceptibility

w relates the magnetization and the magnetic field. If we call F the magnitude of the

paleomagnetic field andF0 the remanent magnetization, the relation between them is

F 0 ¼ wF ð108:1Þ

w ¼ 0:01

We calculate F0 from its components

F 0 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X 2 þ Y 2 þ Z2p

¼ 71 nT

The field F of the virtual pole is given by

F ¼ B0


1þ 3 cos2 yp


F 0 ¼ wB0


1þ 3 cos2 yp

B0 ¼F 0

wffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ 3 cos2 yp ¼ 5610 nT

From this value we calculate the magnetic moment of the virtual pole taking a ¼ 6370 km

for the Earth’s radius

B0 ¼Cm

a3) m ¼ 1:45 1022 Am2

(c) The magnetic field, the declination, and inclination 10 000 years ago were

F ¼F 0

w¼ 7100 nT

D ¼ 36:9

I ¼ 45

At present the values of these parameters are

Fp ¼ Ba0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ 3 sin2 fpresent

q

¼ 54 083 nT

Dp ¼ 0

tan Ip ¼ 2 tanf ) Ip ¼ 73:9

203 Paleomagnetism

The secular variation of F, D, and I, for this period of time, is

Fp F

10 000¼ 4:7 nT=yr

Dp D

10 000¼ 0:220=yr

Ip I

10 000¼ 0:170=yr

109. The following table gives the demagnetization data for a sample that was

subjected to stepwise thermal demagnetization of its natural remanent magnetization

(NRM).

Calculate the direction of each stable component identified by the demagnetization

curve.

First, construct a vector component diagram of the demagnetization data. Decompose

each observation into its north (X), east (Y), and vertical (Z) components:

H ¼ J cos I

X ¼ H cosD

Y ¼ H sinD

Z ¼ J sin I

In Fig. 109 plotting X versus Y gives the projection of the demagnetization vector

onto the horizontal plane, while plotting X versus Z gives the projection onto the vertical

plane.

A stable component of NRM is represented by collinear points on the vector component

diagrams, so that two stable components can be identified in the range 20–300 C and in

the range 400–700 C.

The declination of a stable component is determined by measuring or by calculating the

angle between the north axis and the trajectory of the stable component in the horizontal

plane.

Demagnetization

temperature (°C) Declination (D, °E) Inclination (I, °) NRM Intensity (J, mA/m1)

20 32 33 0.056

100 36 22 0.056

200 38 12 0.057

300 39 4 0.058

400 41 5 0.058

500 41 5 0.050

600 41 5 0.016

650 41 5 0.009

700 300 55 0.000

204 Geomagnetism

For the 20–300 C component:

a ¼ tan1 X300 X20

Y300 Y20

¼ tan1 0:0051

0:0115

¼ 23:9

D ¼ 180þ 23:9 ¼ 203:9


D ¼ b ¼ tan1 Y400

X400

¼ tan1 0:0379

0:0436

¼ 41:0

Note: this value can be obtained directly from the declination of the observations between

400 and 650 C.

The apparent inclination, Iap, of a stable component is determined by measuring or by

calculating the angle between the north axis and the trajectory of the stable component in

the vertical plane. Iap is related to the true inclination, I, by:

tan I ¼ tan Iapj cosDj


Iap ¼ g ¼ tan1 Z20 Z300

X300 X20

¼ tan1 0:0265

0:0051

¼ 79:1

I ¼ tan1ðtanð79:1Þj cosð203:9ÞjÞ ¼ 78:1


Iap ¼ d ¼ tan1 Z400

X400

¼ tan1 0:0051

0:0436

¼ 6:7

I ¼ tan1ðtanð6:7Þj cosð41ÞjÞ ¼ 5:0

Note: this value can be obtained directly from the inclination of the observations between

400 and 650 C.

Therefore the stable component isolated in the range 20–300 C has D ¼ 203.9 and

I ¼ 78.1 and the stable component isolated in the range 400–700 C has D ¼ 41.0 and

I ¼ 5.0.

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.0350.00

0.01

0.02

0.03

0.04

0.05

0.06

X X

YZ

a

b d

γ

–0.01 0.00 0.01 0.02 0.03

0.01

0.00

0.02

0.03

0.04

0.05

0.06

Fig. 109

205 Paleomagnetism

110. A palaeomagnetic study of a late Jurassic limestone outcrop near Alhama de

Granada (37 N, 4 W) in southern Spain yielded a well-defined primary remanent

magnetization whose directions are given in the table below. Calculate the mean

direction of the primary remanence of the seven samples. Compare this direction

with that defined by the reference late Jurassic palaeomagnetic pole for the stable

Iberian tectonic plate (252 E, 58 N). How much vertical axis rotation has the studied

outcrop suffered with respect to stable Iberia?

Use unit vector addition to calculate the mean direction of the primary remanence.

Calculate the direction cosines of each direction, the resultant total field vector, F, and

then the mean direction using:

X ¼ cos I cosD

Y ¼ cos I sinD

Z ¼ sin I

F ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X

N

i¼1

Xi

!2

þX

N

i¼1

Yi

!2

þX

N

i¼1

Zi

!2v

u

u

t ¼ 6:98542

Xmean ¼

P

N

i¼1

Xi

F¼ 0:6446;

Ymean ¼

P

N

i¼1

Yi

F¼ 0:37186;

Zmean ¼

P

N

i¼1

Zi

F¼ 0:66799

Dmean ¼ tan1 Ymean

Xmean

¼ 30:0

Imean ¼ sin1ðZmeanÞ ¼ 41:9

The mean direction of the primary remanence has a declination of 30.0 and an inclination

of 41.9.

Declination (D, °E) Inclination (I, °)

30 43

28 39

34 44

25 45

32 38

35 44

26 40

206 Geomagnetism

Next, calculate the expected field direction at the site using the reference palaeomagnetic

pole. The first step is to determine y ¼ 90 f, (Equation 71.3), from the pole (fp, lp) to

the site (fs, ls) using spherical triangles:

sinf ¼ sin’p sin’s þ cos’p cos’s cosðls lpÞ ) f ¼ 24:1

The expected inclination can then be calculated using:

tan Iexp ¼ 2 tanf ) Iexp ¼ 41:8

The expected declination can be calculated by (71.2):

sinDexp ¼ cos’p sinðls lpÞ

cosf ) Dexp ¼ 34:5 W ¼ 325:5

rotation about the vertical axis should give rise to a difference between the observed and

expected declinations, defined as positive for an observed declination clockwise from the

expected declination.

Therefore the outcrop has suffered 64.5 of clockwise rotation with respect to stable

Iberia.

207 Paleomagnetism

4 Seismology

Elasticity

111. Determine the principal stresses and principal axes of the stress tensor:

2 1 1

1 0 1

1 1 2

0

@

1

A

Find the invariants I1, I2, I3, the deviator tensor, its eigenvalues, and the invariants

J2 and J3.

To calculate the principal stresses (s1, s2, s3) and principal axes (n1i, n2i, n3i), we

calculate the eigenvalues and eigenvectors of the matrix. They are found through the

equation

ðtij sdijÞni ¼ 0 ð111:1Þ

The eigenvalues are the roots of the cubic equation for s resulting from putting the

determinant of the matrix in (111.1) equal to zero:

2 s 1 1

1 s 1

1 1 2 s

¼ 0

) 2 sð Þ sð Þ 2 sð Þ 1 1þ s 2 sð Þ 2 sð Þ ¼ 0

s3 4s2 þ sþ 6 ¼ 0

The three roots of the equation are the principal stresses

s1 ¼ 1

s2 ¼ 2

s3 ¼ 3

From these values we obtain s0 ¼13s1 þ s2 þ s3ð Þ ¼ 4

3.

The invariants of the matrix are the coefficients of the characteristic equation

s3 I1s2 þ I2s I3 ¼ 0

208

which, in terms of the roots of the equation, are

I1 ¼ 4 ¼ s1 þ s2 þ s3

I2 ¼ 1 ¼ s1s2 þ s1s3 þ s2s3

I3 ¼ 6 ¼ s1s2s3

The principal axes of stress are the eigenvectors ni associated with the three eigenvalues.

For s1 ¼ 1

3 1 1

1 1 1

1 1 3

0

@

1

A

n1n2n3

0

@

1

A ¼ 0 ) n11; n12; n

13

¼ 1;2; 1ð Þ

For s2 ¼ 2

0 1 1

1 2 1

1 1 0

0

@

1

A

n1n2n3

0

@

1

A ¼ 0 ) n21; n22; n

23

¼ 1; 1; 1ð Þ

For s3 ¼ 3

1 1 1

1 3 1

1 1 1

0

@

1

A

n1n2n3

0

@

1

A ¼ 0 ) n31; n32; n

33

¼ 1; 0; 1ð Þ

The deviatoric stress tensor is defined as

t0ij ¼ tij s0dij

where in our problem, s0 ¼ 4/3.

The three components of the principal diagonal of the deviatoric tensor are

t011 ¼2

3

t022 ¼ 4

3

t033 ¼2

3

To calculate its eigenvalues we proceed as we did before:

2

3 s 1 1

1 4

3 s 1

1 12

3 s

0

B

B

B

B

B

@

1

C

C

C

C

C

A

¼ 0 ) s3 13

3sþ

38

27¼ 0

Comparing with the characteristic equation

s3 J1s2 þ J2s J3 ¼ 0

209 Elasticity

the invariants are

J2 ¼ 13

3

J3 ¼ 38

27

Solving the cubic equation we obtain s1 ¼ 2.22, s2 ¼ 0.33, s3 ¼ 1.89.

112. Given the stress tensor

1

42

5

4

1

2

32

25

4

1

42

1

2

32

1

2

32

21

2

32 3

2

0

B

B

B

B

B

B

B

B

B

@

1

C

C

C

C

C

C

C

C

C

A

calculate:

(a) The principal stresses.

(b) The angles formed by the greatest of these stresses with the axes 1, 2, 3.

(a) As in the previous problem to find the principal stresses we calculate the eigen-

values of the stress matrix

1

4 s

5

4

1

2

3=2

5

4

1

4 s

1

2

3=2

1

2

3=2

1

2

3=23

2 s

0

B

B

B

B

B

B

B

B

B

@

1

C

C

C

C

C

C

C

C

C

A

¼ 0 ) s3 2s2 sþ 2 ¼ 0

Solving the cubic equation, its three roots are

s1 ¼ 2

s2 ¼ 1

s3 ¼ 1

The largest is s1. The associated eigenvector corresponds to the axis of greatest stress

whose direction cosines are (n1, n2, n3). They are found by solving the equation

7

45

4

1

2

3=2

5

47

4

1

2

3=2

1

2

3=2

1

2

3=21

2

0

B

B

B

B

B

B

B

B

@

1

C

C

C

C

C

C

C

C

A

n1n2n3

0

@

1

A ¼ 0

210 Seismology

Solving this equation with the condition that n21 þ n22 þ n23 ¼ 1, we obtain,

n1 ¼ n2 ¼1

2

n3 ¼1ffiffiffi

2p

(b) From these values we obtain #, the angle with the vertical axis (x3) and ’, the angle

which forms its projection on the horizontal plane with x1:

n1 ¼ sin# cos’ ¼1

2

n2 ¼ sin# sin’ ¼ 1

2

n3 ¼ cos# ¼1ffiffiffi

2p

9

>

>

>

>

>

=

>

>

>

>

>

;

) ’ ¼ 315; # ¼ 45

113. The stress tensor tij in a continuous medium is

3x1x2 5x22 0

5x22 0 2x30 2x3 0

0

@

1

A

Determine the stress vector Tin acting at the point (2, 1, √3) through the plane

tangential to the cylindrical surface x22 1 x23 at that point.

First we calculate the value of the stress tensor at the given point:

tijð2; 1;ffiffiffi

3p

Þ ¼6 5 0

5 0 2ffiffiffi

3p

0 2ffiffiffi

3p

0

0

@

1

A

A unit vector normal to the surface f ¼ x22 þ x23 4 ¼ 0 at the given point is

ni ¼grad f

grad fj j¼

@f

@x1@f

@x1

;

@f

@x2@f

@x2

;

@f

@x3@f

@x3

0

B

B

@

1

C

C

A

¼ 0;1

2;

ffiffiffi

3p

2

Then, the stress vector acting at the point through that surface is given by

T ni ¼ tijnj ¼

6 5 0

5 0 2ffiffiffi

3p

0 2ffiffiffi

3p

0

0

@

1

A

01

2ffiffiffi

3p

2

0

B

B

B

B

@

1

C

C

C

C

A

¼5

2; 3;

ffiffiffi

3p

Wave propagation. Potentials and displacements

114. The amplitudes of P- and S-waves of frequency 4/2p Hz are

uP ¼4

3ffiffiffi

2p ;

4

3ffiffiffi

2p ;

4ffiffiffi

3p

uS ¼ ffiffiffi

3p

;ffiffiffi

3p

;ffiffiffi

2p

211 Wave propagation. Potentials and displacements

and their speeds of propagation are 6 km s1 and 4 km s1, respectively. Find the

scalar and vector potentials. Displacements are always given in µm.

The displacements of P-waves can be deduced from a scalar potential function ’ such that

uPi ¼ ðr’Þi. The general form of the potential function for P-waves for harmonic motion is

’ ¼ A exp ikaðnixi atÞ ð114:1Þ

where A is the amplitude, ni the direction cosines of the ray or propagation direction, a the

velocity of propagation, and ka the wavenumber. If uj is given in µm and ka in km1, then A

is given in 103 m2. Taking the derivatives in (114.1) we obtain for the components of the

displacement

uPj ¼@’

@xj¼ Aikanj exp ika nkxk atð Þ

Their amplitude is

uPj ¼ Akanj ð114:2Þ

and the wavenumber is

ka ¼o

a¼

2p4

2p6

¼2

3km1

By substitution in (114.2), we obtain for the two horizontal components

uP1 ¼ Akan1 ¼4

3ffiffiffi

2p mm

uP2 ¼ Akan2 ¼4

3ffiffiffi

2p mm

Dividing these two expressions and writing the direction cosines of the ray in terms of the

incident angle i and azimuth az,

n1 ¼ sin i cos az

v2 ¼ sin i sin az

n3 ¼ cos i

ð114:3Þ

we have

uP1uP2

¼ 1 ¼n1

n2¼

sin i cos az

sin i sin az) a ¼ 45

Using the uP3 and uP1 components,

uP3uP1

¼Akan3

Akan1¼

cos i

sin i cos az¼

ffiffiffi

3p

) i ¼ 30

From the values of the direction cosines and the amplitude of the displacement we calculate

the amplitude of the potential A:

A ¼uP1kan1

) A ¼ 4 103 m2

212 Seismology

Finally, the expression for the scalar potential of P-waves is

’ ¼ 4 exp2

3i

ffiffiffi

2p

4x1 þ

ffiffiffi

2p

4x2 þ

ffiffiffi

3p

2x3 6t

Displacements of the S-wave are obtained from a vector potential ci of null divergence,

whose general form is

ci ¼ Bi exp ikb njxj bt

where b is the velocity of propagation and kb the wavenumber. The displacements are

given by

uS ¼ r c ð114:4Þ

The wavenumber is kb ¼ o/b ¼ 1 km1. According to (114.4) the relation between the

components of the displacement and of the amplitude of the potential is

uS1 ¼ B3

ffiffiffi

2p

4 B2

ffiffiffi

3p

2¼

ffiffiffi

3p

mm

uS2 ¼ B1

ffiffiffi

3p

2 B3

ffiffiffi

2p

4¼

ffiffiffi

3p

mm

uS3 ¼ B2

ffiffiffi

2p

4 B1

ffiffiffi

2p

4¼

ffiffiffi

2p

mm

The potential must have null divergence,

r c ¼

ffiffiffi

2p

4B1 þ

ffiffiffi

2p

4B2 þ

ffiffiffi

3p

2B3 ¼ 0

From these equations we obtain, in units of 103 m2,

B1 ¼ 2

B2 ¼ 2

B3 ¼ 0

The S-wave vector potential is

cj ¼ ð2; 2; 0Þ exp i

ffiffiffi

2p

4x1 þ

ffiffiffi

2p

4x2 þ

ffiffiffi

3p

2x3 4t

Note: These units will be used for all problems but not explicitly given.

115. The components of the S-wave with respect to the axes (x1, x2, x3) are (6, 3.25, 3)

where x2 is the vertical axis, the azimuth is 60, and the angle of incidence is 30.

Determine the amplitude and direction cosines of the SV and SH components.

From the azimuth and incident angles we calculate the direction cosines (x2 is the

vertical axis)


n1 ¼ sin i cos az ¼1

4

n2 ¼ cos i ¼

ffiffiffi

3p

2

n3 ¼ sin i sin az ¼

ffiffiffi

3p

4

Since the SV and SH components are on a plane normal to the direction of the ray r

(Fig. 115a) unit vectors in the direction of SV (a1, a2, a3) and of SH (b1, 0, b3) must satisfy

the equations

SV r ¼ 0 )a1

4þa2

ffiffiffi

3p

2þa3

ffiffiffi

3p

4¼ 0

SH r ¼ 0 )b1

4þb3

ffiffiffi

3p

4¼ 0

SH SV ¼ 0 ) a1b1 þ a3b3 ¼ 0

ð115:1Þ

The projections on the horizontal plane R of the ray r and SH are perpendicular

(Fig. 115b). Then SH forms an angle of 180 – 30 with the x1 axis. The direction

cosines of SH are

b1 ¼

ffiffiffi

3p

2

b3 ¼1

2

SV

SVH

SH

RX3

X2

X1

90°

60°

30°

r

Fig. 115a

214 Seismology

SV forms an angle of 60 with the vertical axis x2 (Fig. 115a). Then a2 ¼ sin i ¼ 12

From a2 using Equations (115.1) and a21 þ a22 þ a23 ¼ 1, we calculate a1 and a2:

a1 ¼

ffiffiffi

3p

4

a3 ¼3

4

116.Given the potentialci ¼

ffiffiffi

7p

ffiffiffi

5p ;

5ffiffiffi

7p

ffiffiffi

5p ;6

exp 4i 1ffiffiffi

5p x1 þ

1ffiffiffi

3p x2 þ

ffiffiffi

7p

ffiffiffiffiffi

15p x3 4t

,

calculate the polarization angle.

First we calculate the amplitudes of the components of the displacement of the S-wave

from the vector potential

uiS ¼ r ci )

uS1 ¼ c3;2 c2;3 ¼ 413ffiffiffi

3p ¼ 30:02

uS2 ¼ c1;3 c3;1 ¼ 47

5ffiffiffi

3p þ

6ffiffiffi

5p

¼ 13:97

uS3 ¼ c2;1 c1;2 ¼ 4ffiffiffi

7p

ffiffiffi

7p

ffiffiffiffiffi

15p

¼ 7:85

8

>

>

>

>

>

>

>

>

<

>

>

>

>

>

>

>

>

:

The modulus is

uS ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u21 þ u22 þ u23

q

¼ 34:03

From the vertical component uS3 we calculate the SV component (Fig. 116) knowing that

n3 ¼ cos i ¼

ffiffiffi

7p

ffiffiffiffiffi

15p ) sin i ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

17

15¼

rffiffiffi

8p

ffiffiffiffiffi

15p

Then, uS3 ¼ uSVcos 90 ið Þ ) uSV ¼ 10:75.

SH

R

SVH

X3

X130°

60°

30°

Fig. 115b


To find the SH component we use the relation

uS ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðuSVÞ2 þ ðuSHÞ2q

) uSH ¼ 32:29

From the SV and SH components we obtain the polarization angle e:

tan e ¼uSH

uSV) e ¼ 71:6

117. Given the amplitudes of the P- and S-waves (ka ¼ 1):

uP1 ¼ 4 uS1 ¼ 8

uP2 ¼ 4 uS2 ¼ 2ffiffiffi

2p

uP3 ¼ 8 uS3 ¼ 4þffiffiffi

2p

determine the angle of incidence i, azimuth az, polarization angle « of the S-wave,

and apparent polarization angle g.

Given that the displacements of the P-wave are on the incident plane, in the direction of the

ray r, the angle of incidence i can be obtained from the modulus and the vertical component:

uP ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

42 þ 42 þ 82p

¼ 4ffiffiffi

6p

cos i ¼uP3uP

¼8

4ffiffiffi

6p ) i ¼ 35:3

The azimuth, the angle between the horizontal projection of the ray and the north (x1), is

obtained from the horizontal components, uP1 and uP2 :

tan az ¼uP2uP1

) az ¼ 45

SV

R

X3

r

i

90° – i

Fig. 116

216 Seismology

We calculate the SV component from the vertical component uS3 , as in the previous

problem:

uSV ¼uS3

cos 90 i0ð Þ¼

uS3sin i0

¼ 9:37

The SH component is found from the horizontal components (Fig. 117)

uSH ¼ uS2 cos az uS1 sin az ¼ 3:66

From SV and SH we find the polarization angle e (Fig. 117):

tan e ¼uSH

uSV¼

3:66

9:37) e ¼ 21:3

To calculate the apparent polarization angle g, the angle between the horizontal component

of S, (uSH), and the radial direction R (Fig. 117), we use the relation

tan e ¼ cos i0 tan g ) g ¼ 25:5

118. Given the values az ¼ 60, « ¼ 30, g ¼ 45, and ub

¼ 5, calculate the

amplitudes of the components 1, 2, and 3 of the S-wave.

From the modulus of the displacement of S-waves and the polarization angle, we calculate

the SV and SH components (Fig. 118a):

uSV ¼ uS cos e ¼5ffiffiffi

3p

2

uSH ¼ uS sin e ¼5

2

SH

R

X1

X2

az

az

γ

uS2

uS1

uSH

90° – az

Fig. 117


From the angles e and g we obtain the incidence angle i:

tan e ¼ tan g cos i ) cos i ¼1ffiffiffi

3p ) sin i ¼

ffiffiffiffiffiffiffiffiffiffiffi

11

3

r

¼

ffiffiffi

2p

ffiffiffi

3p

The vertical component u3 of the S-wave is obtained from the value of SV (Fig. 118b):

uS3 ¼ uSV cos 90 ið Þ ¼5ffiffiffi

3p

2

ffiffiffi

2p

ffiffiffi

3p ¼ 3:53

To calculate the horizontal components we have to take into account the horizontal

component of SV (Fig. 118b):

uSVH ¼ uSV cos i ¼5

2

SH

SV

e

Fig. 118a

SV

R

X3

r

i

i

uS3

uSVH

Fig. 118b

218 Seismology

From SH and the horizontal component of SV we find the horizontal components

of S (Fig. 118c):

uS1 ¼ uSH sin az þ uSVH cos az

¼ 5

41þ

ffiffiffi

3p

¼ 3:42

uS2 ¼ uSH cos az uSVH sin az ¼5

41

ffiffiffi

3p

¼ 0:92

119. For a scalar potential w and a vector potential ci, it is known that A ¼ 3, Bi ¼

(2, 2, 0), ka ¼ 2/3, T ¼ p/2, and Poisson’s ratio is s ¼ 1/4. Calculate the amplitudes

on the free surface (x3 ¼ 0) of the components u1, u2, and u3 of the P- and SV-waves.

The direction cosines are1

2ffiffiffi

2p ;

1

2ffiffiffi

2p ;

ffiffiffi

3p

2

.

The general expressions for the scalar and vector potentials are

’ ¼ A exp ika n1x1 þ n2x2 þ n3x3 atð Þ

ci ¼ Bi exp ikb n1x1 þ n2x2 þ n3x3 btð Þ

Since ka ¼ o/a then a ¼ o/ka ¼ 6 km s1.

If Poisson’s ratio is 0.25 then

s ¼1

4¼

l

2 lþ mð Þ) l ¼ m

Substituting this condition in the equation for the P-wave velocity a, we find for the

velocity b of S-waves is given by

a ¼


lþ 2m

r

s

¼

ffiffiffiffiffiffi

3m

r

s

¼ bffiffiffi

3p

) b ¼affiffiffi

3p ¼ 2

ffiffiffi

3p

km s1

Then the wavenumber of S-waves is

kb ¼o

b¼

2ffiffiffi

3p

RX2

X1

az

az

az

u SH

SH

uSH

uSVH

Fig. 118c


Then we can write the complete expressions for the potentials:

’ ¼ 3 exp i2

3

1

2ffiffiffi

2p x1 þ

1

2ffiffiffi

2p x2 þ

ffiffiffi

3p

2x3 6t

ci ¼ 2; 2; 0ð Þ exp i2ffiffiffi

3p

1

2ffiffiffi

2p x1 þ

1

2ffiffiffi

2p x2 þ

ffiffiffi

3p

2x3 2

ffiffiffi

3p

t

ð119:1Þ

To determine the displacements of the P- and S-waves we use the relations

uP ¼ r’

uS ¼ r c

obtaining

uP1 ¼1ffiffiffi

2p uS1 ¼ 2

uP2 ¼1ffiffiffi

2p uS2 ¼ 2

uP3 ¼ffiffiffi

3p

uS3 ¼2ffiffiffi

2p

ffiffiffi

3p

From the components of the displacement of the S-wave we can obtain the SV component.

The values of the angles of incidence and azimuth are found from the direction cosines,


2ffiffiffi

2p

n2 ¼ sin i sin az ¼1

2ffiffiffi

2p ) i ¼ 30; az ¼ 45

n3 ¼ cos i ¼

ffiffiffi

3p

2

and from the angle i we obtain the SV component from uS3:

uS3 ¼ uSV cos 90 ið Þ ) uSV ¼4ffiffiffi

2p

ffiffiffi

3p ¼ 3:27

From the horizontal components of the S-waves and the azimuth we calculate the SH

component:

uSH ¼ uS2 cos az uS1 sin az ¼ 0

120. In an elastic medium of density r¼ 3 g cm3

and Poisson ratio 1/3 there

propagate P- and S-waves of frequency 1 Hz in the direction1

3;1ffiffiffi

2p ;

ffiffiffi

7p

3ffiffiffi

2p

. Given

that the pressure of the P-wave is 5000 dyn cm2, the magnitude of its displacement,

10 mm, is twice that of the S-wave, and the angle g ¼ 45, find all the parameters

involved in the expression of the potentials ’ and ci. (It is not necessary to solve the

equations to obtain the coefficients Bi)

220 Seismology

Given that Poisson’s ratio is 1/3 the relation between the elastic coefficients l and m is,

s ¼l

2 lþ mð Þ¼

1

3) l ¼ 2m

and between the velocities of P (a) and S (b)-waves is

a ¼


lþ 2m

r

s

¼

ffiffiffiffiffiffi

4m

r

s

) a ¼ 2b

The bulk modulus K is

K ¼ lþ2

3m ¼

8

3m

Expressing m in terms of b and a, we get for K:

b ¼

ffiffiffi

m

r

r

) m ¼ rb2 ) K ¼8

3rb2 ¼

8

3ra2

4¼

2

3ra2

Taking into account that the bulk modulus K is defined as the applied pressure divided by

the change in volume per unit volume y,

P ¼ Ky ¼2

3ra2y

y ¼ r2’ ¼ Ak2a

y ¼3P

2ra2¼ Ak2a ) a2 ¼

3P

2rAk2a

ð120:1Þ

The expression for the scalar potential is

’ ¼ A exp ika n1x1 þ n2x2 þ n3x3 atð Þ ð120:2Þ

where the wavenumber for P-waves is

ka ¼o

a¼

2pf

a

uP

¼ r’j j ¼ kaA ) A ¼uPj j

ka¼

uPj ja

2pf

By substitution in (120.1) we obtain the values of a and b:

a2 ¼3P

2rAk2a¼

3Pa

2r uPj j2pf) a ¼

3P

2r uPj j2pf

where

P ¼ 5000 dyn cm2

r ¼ 3 g cm3

f ¼ 1 Hz

uP ¼ 10 mm


We obtain a ¼ 3.98 km s1 and b ¼ 1.99 km s1.

Since we know A, a, and ka we can write the complete expression for the scalar

potential

’ ¼ 10 exp i1:581

3x1 þ

1ffiffiffi

2p x2 þ

ffiffiffi

7p

3ffiffiffi

2p x3 3:98t

The vector potential of the S-waves is given by

ci ¼ Bi exp ikb n1x1 þ n2x2 þ n3x3 btð Þ ð120:3Þ

We calculate kb:

kb ¼o

b¼ 3:16 km1

The displacements are given by

uS ¼ r c

and

uS1 ¼@c3

@x2@c2

@x3¼ kb B3n2 B2n3ð Þ

uS2 ¼@c1

@x3@c3


uS3 ¼@c2

@x1@c1


ð120:4Þ

The incidence angle i is found from n3 and, using tane¼ cosi tang, we find the polarization

angle e:

n3 ¼ cos i ¼

ffiffiffi

7p

3ffiffiffi

2p ) i ¼ 31:95 ) e ¼ 31:95

The azimuth is

az ¼ tan1 n2

n1¼ 64:76

Since the amplitude of the S-wave displacement is 5 mm, knowing the value of e we can

find the values of the SV and SH components:

uSV ¼ uS cos e ¼ 4:24 mm

uSH ¼ uS sin e ¼ 2:65 mm

From uSV we calculate its vertical and horizontal components:

uS3 ¼ uSV cos 90 ið Þ ¼ 2:25 mm

uSVH ¼ uSV cos i ¼ 3:61 mm

222 Seismology

Using uHSV and uSH we find the two horizontal components:

uS1 ¼ uSH sin az þ uSVH cos az

¼ 3:94 mm

uS2 ¼ uSH cos az uSVH sin az ¼ 2:14 mm

Using the values found for the displacements and Equations (120.4) and∇ ·c ¼ 0 (n1 B1þ

n2 B2 þ n3 B3 ¼ 0) we find the values of B1, B2, B3. Substituting all the values in (120.3) we

obtain for the vector potential

ci ¼ ð1:17; 2:5;3:44Þ exp 3:16i1

3x1 þ

1ffiffiffi

2p x2 þ

ffiffiffi

7p

3ffiffiffi

2p x3 1:99t

121. At the origin in an infinite medium in which s, Poisson’s ratio, is 0.25, and the

density is 3 g cm3, there is an emitter of elastic plane waves of frequency 0.5 cps.

Calculate:

(a) The equation of the P- and S-waves in exponential form and with arbitrary

amplitudes for the wave arriving at the point A(500, 300, 141) km.

(b) The arrival time.

(a) First we calculate the distance to point A and the direction cosines of the direction

of the ray (r) (Fig. 121):

r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

5002 þ 3002 þ 1412p

¼ 599:90 600 km

n1 ¼500

600¼

5

6

n2 ¼300

600¼

1

2

n3 ¼141

600¼

ffiffiffi

2p

6

X2

X3

X1

141

300

500

r

Fig. 121


The S-wave velocity is

b ¼

ffiffiffi

m

r

r

¼ffiffiffiffiffi

10p

km s1

Given that Poisson’s ratio is 0.25,

s ¼ 0:25 ) l ¼ m ) a ¼ bffiffiffi

3p

¼ffiffiffiffiffi

30p

km s1

The wavenumbers of the P- and S-waves are

ka ¼2pf

a¼

2p 0:5

5:5¼

pffiffiffiffiffi

30p km1

kb ¼2pf

b¼

2p 0:5ffiffiffiffiffi

10p ¼

pffiffiffiffiffi

10p km1

The general expressions for the scalar and vector potentials are


ci ¼ Bi exp ikb n1x1 þ n2x2 þ n3x3 btð Þ

Leaving the amplitudes A and Bi in arbitrary form and substituting the obtained values we

have for the potentials,

’ ¼ A exp ipffiffiffiffiffi

30p

5

6x1 þ

1

2x2 þ

ffiffiffi

2p

6x3

ffiffiffiffiffi

30p

t

ci ¼ Bi exp ipffiffiffiffiffi

10p

5

6x1 þ

1

2x2 þ

ffiffiffi

2p

6x3

ffiffiffiffiffi

10p

t

(b) The travel times for P- and S-waves from the origin to the given point are

ta ¼

r

a¼

600ffiffiffiffiffi

30p ¼ 109:5 s

tb ¼

r

b¼

600ffiffiffiffiffi

10p ¼ 189:7 s

Reflection and refraction

122. A P-wave represented by the potential

w ¼ 4 exp 0:25ix1ffiffiffi

6p þ

x2ffiffiffi

3p þ

x3ffiffiffi

2p 4t

is incident on the surface x3 ¼ 0 separating two liquids of densities 3 g cm3

and

4 g cm3. If the speed of propagation in the second medium is 2 km s1, write the

expressions for the potentials of the reflected and transmitted waves.

224 Seismology

The potentials of the reflected and transmitted waves are given by

’refl ¼ A exp ik tan e x3 þ x1 ctð Þ

’trans ¼ A0 exp ik tan e0 x3 þ x1 ctð Þð122:1Þ

where e ¼ 90 – i is the emergence angle and i the incidence angle, and k ¼ ka cos e is the

wavenumber corresponding to the apparent horizontal velocity, c ¼ a /cos e. These

expressions are written for rays contained on the incidence plane (x1, x3). Then, we have

to rotate the given potential to refer it to the incidence plane. First, from the direction

cosines we calculate the incidence angle i and the azimuth az:

n1 ¼ sin i cos az ¼1ffiffiffi

6p

n2 ¼ sin i sin az ¼1ffiffiffi

3p

n3 ¼ cos i ¼1ffiffiffi

2p ) i ¼ 45 ¼ e

cos az ¼1ffiffiffi

3p ) sin az ¼

ffiffiffi

2p

ffiffiffi

3p

Using the rotation matrix we obtain the direction cosines on the plane of incidence (x1, x3):

n01n02n03

0

@

1

A ¼cos az sin az 0

sin az cos az 0

0 0 1

0

@

1

A

n1n2n3

0

@

1

A)

1ffiffiffi

2p

01ffiffiffi

2p

0

B

B

B

@

1

C

C

C

A

The values of c and k are

c ¼a

cos e¼

a0

cos e0¼

8ffiffiffi

2p ¼ 4

ffiffiffi

2p

km s1

k ¼ ka cos e ¼ ka0 cos e0 ¼

ffiffiffi

2p

8km1

Then the potential of the incident wave is now given by

’inc ¼ 4 exp i1

4ffiffiffi

2p x3 þ x1 4

ffiffiffi

2p

t

The angle i0 of the transmitted or refracted ray is found from Snell’s law:

sin i

a¼

sin i0

a0) sin i0 ¼

ffiffiffi

2p

4¼ cos e0

) cos i0 ¼ sin e0 ¼

ffiffiffiffiffi

14p

4) tan e0 ¼

ffiffiffi

7p

Using the expressions for the reflection and refraction coefficients, V and W, we can

calculate the amplitude of the reflected and refracted potentials:

V ¼A

A0

¼r0 tan e r tan e0

r0 tan eþ r tan e0¼

4 3ffiffiffi

7p

4þ 3ffiffiffi

7p ) A ¼

16 12ffiffiffi

7p

4þ 3ffiffiffi

7p ¼ 1:07

W ¼A0

A0

¼2rtge0

r0 tan eþ r tan e0¼

6

4þ 3ffiffiffi

7p ) A0 ¼

24

4þ 3ffiffiffi

7p ¼ 2:01

225 Reflection and refraction


’refl ¼ 1:07 exp

ffiffiffi

2p

8i x3 þ x1

8ffiffiffi

2p t

’trans ¼ 2:01 exp

ffiffiffi

2p

8i

ffiffiffi

7p

x3 þ x1 8ffiffiffi

2p t

123. A P-wave of amplitude 5ffiffiffi

2p

; 5ffiffiffi

6p

; 10ffiffiffi

2p

and frequency v ¼ 12 rad s1 in a

semi-infinite medium of speed of propagation a ¼ 6 km s1 and Poisson’s ratio 0.25 is

incident on the free surface. Calculate:

(a) The potential of the incident P-wave.

(b) The potential of the reflected S-wave.

(c) The components u1, u2, u3 of the reflected S-wave.

(a) The displacements of the P-wave can be deduced from its scalar potential:


uP ¼ r’ð123:1Þ

where A is the amplitude, ka is the wavenumber (P), ni are the direction cosines, and a is the

P-wave velocity. The wavenumber is found from the given angular frequency and velocity:

ka ¼o

a¼

12

6¼ 2 km1

Since we know the amplitudes of the components of the displacements we can find the

incidence angle i and the azimuth az:

uP1 ¼@’

@x1¼ Akan1 ¼ A2 sin i cos az ¼ 5

ffiffiffi

2p

uP2 ¼@’

@x2¼ Akan2 ¼ A2 sin i sin az ¼ 5

ffiffiffi

6p

uP3 ¼@’

@x3¼ Akan3 ¼ A2 cos i ¼ 10

ffiffiffi

2p

Dividing the two first equations we obtainffiffiffi

3p

¼ tan az ) az ¼ 60, and dividing the

last two,

5ffiffiffi

2p

10ffiffiffi

2p ¼ tan i cos az ¼

1

2tan i ) i ¼ 45

The amplitude A is given by

uP

¼


uP1 2

þ uP2 2

þ uP3 2

q

¼ Aka ) A ¼uPj j

ka¼ 102 m2

Finally the potential is given by

’inc ¼ 102exp i2

ffiffiffi

2p

4x1 þ

ffiffiffi

6p

4x2 þ

ffiffiffi

2p

2x3 6t

m2

226 Seismology

If we express the potential referred to the plane (x1, x3) as the incidence plane, as we did in

Problem 122, then

’ ¼ A exp ik x1 þ tan e x3 ctð Þ

where

e ¼ 90 i ¼ 45; k ¼ ka cos e ¼ 21ffiffiffi

2p ¼

ffiffiffi

2p

c ¼a

cos e¼

6

1ffiffiffi

2p

¼ 6ffiffiffi

2p

km s1

and the potential is

’inc ¼ 102 exp iffiffiffi

2p

x1 þ x3 6ffiffiffi

2p

t

m2 ð123:2Þ

(b) Since Poisson’s ratio is 1/4, l ¼ m, and

b ¼affiffiffi

3p ¼ 2

ffiffiffi

3p

km s1

The angle f of the reflected S-wave is obtained by Snell’s law:

cos e

a¼

cos f

b) cos f ¼

b

acos e ¼

ffiffiffi

2p

2ffiffiffi

3p ¼

1ffiffiffi

6p ) sin f ¼

ffiffiffi

5

6

r

From the values of e and f we calculate the P-to-S reflection coefficient VPS, using equation

VPS ¼4a 1þ 3a2ð Þ

4abþ 1þ 3a2ð Þ2

where we substitute

a ¼ tan e ¼ 1 and b ¼ tan f ¼ffiffiffi

5p

so

VPS ¼4 1þ 3ð Þ

4ffiffiffi

5p

þ 1þ 3ð Þ2¼ 0:64

From this coefficient we calculate the proportion of the incident P-wave which is reflected

as an S-wave (only with SV component; the negative sign indicates the opposite sense of

the reflected ray):

B ¼ AVPS ¼ 10 0:64 ¼ 6:4 103 m2

When the ray is contained in the (x1, x3) plane we use a scalar potential for the S-wave

which in this case is given by

c ¼ B exp ik x1 tan f x3 ctð Þ

¼ 6:4 103 exp iffiffiffi

2p

x1 ffiffiffi

5p

x3 6ffiffiffi

2p

t

m2


(c) To calculate the amplitudes of the total displacements in terms of the two scalar

potentials, we remember that for this orientation of the axes the displacements are

given by

u1 ¼@’

@x1

@c

@x3¼ uP1 þ uSV1

u3 ¼@’

@x3þ

@c

@x1¼ uP3 þ uSV3

The displacements of the SV reflected wave in this case are

uSV1 ¼ @c

@x3¼ 6:4

ffiffiffi

5p

uSV3 ¼@c

@x1¼ 6:4

ffiffiffi

2p

If we want to determine the components 1 and 2, referred to the original system of axes, we

project uSV1 ¼ uSVR using the azimuth 60:

uSV1 ¼ uSVR cos az ¼ 3:2ffiffiffi

5p

uSV2 ¼ uSV2 sin az ¼ 3:2ffiffiffiffiffi

15p

124. An S-wave of vector potential

c i ¼ 10ffiffiffi

3p

; 2; 4

exp 5ix1

4þ

ffiffiffi

3p

4x2 þ

ffiffiffi

3p

2x3 4t

is incident on the free surface x3 ¼ 0. Find the SV and SH components of the reflected

S-wave referred to the same coordinate system as the incident wave, and the coeffi-

cient of reflection. Poisson’s ratio is 3/8.

According to the value of Poisson’s ratio the relation between l and m is

s ¼l

2 lþ mð Þ¼

3

8) l ¼ 3m

and the relation between the velocities of the P-waves and S-waves is

a ¼


lþ 2m

r¼

s ffiffiffiffiffiffi

5m

r

s

¼ffiffiffi

5p

b

The incidence angle i and the azimuth az are obtained from the direction cosines:

n3 ¼ cos i ¼

ffiffiffi

3p

2) i ¼ 30


4¼ sin 30 cos az ) az ¼ 60

Using Snell’s law we find the value of the critical angle

sin ic

b¼

1

a) sin ic ¼

b

a¼

1ffiffiffi

5p ) ic ¼ 26:5

228 Seismology

Since i > ic, there is no reflected P-wave. The components of the incident S-wave are

obtained from the potential

uSi ¼ r c )uS1 ¼ 0

uS2 ¼ 80

uS3 ¼ 40

8

<

:

The modulus of the displacement is uS

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

80ð Þ2þ 40ð Þ2q

¼ 40ffiffiffi

5p

:

The SV component is given by (Fig. 124a)

uSV ¼u3

cos 90 i0ð Þ¼ 80

and the SH component is

uSH ¼


uSð Þ2 uSVð Þ2q

¼ 40

The minus sign corresponds to that of uS2The amplitude of the reflected SH wave is equal to that of the incident SH wave. We find

the 1 and 2 components using the azimuth 60 (Fig. 124b):

uSHi ¼ 40

ffiffiffi

3p

2;1

2; 0

exp 5i1

4x1 þ

ffiffiffi

3p

4x2

ffiffiffi

3p

2x3 4t

For total reflection the amplitude of the reflected SV is equal to that of the incident one, but

with a phase shift d. The components are given by (Figs 124a and 124b)

uSV1 ¼ uSV cos i cos az

uSV2 ¼ uSV cos i sin az

uSV3 ¼ uSV sin i

X3

i

r

60°

R

uSV

Fig. 124a


The components of the reflected SV wave are

uSVi ¼ 80

ffiffiffi

3p

4;

3

4;1

2

exp 5i1

4x1 þ

ffiffiffi

3p

4x2

ffiffiffi

3p

2x3 4t

þ id

To determine the phase shift d we have to determine the reflection coefficients for a free

surface. Textbooks usually give those for Poisson’s ratio s = 1/4, but since in this problem

s ¼ 3/8 we have to calculate them. On a free surface the boundary conditions are that

stresses are null, which in terms of the scalar potentials ’ and c, are given by

t31 ¼ 0 ¼ m u3;1 þ u1;3

¼ 2’;31 c;11 þ c;33

t33 ¼ 0 ¼ l u1;1 þ u3;3

þ 2mu3;3 ¼ 3’;11 5’;33 2c;13

ð124:1Þ

where, using (x1, x3) as the plane of incidence, the scalar potential ’ of the reflected

P-waves is

’ ¼ A exp ik ax3 þ x1 ctð Þ

and the scalar potential of the incident and reflected S-wave is

c ¼ B0 exp ik bx3 þ x1 ctð Þ þ B exp ik bx3 þ x1 ctð Þ

Substituting in (124.1) we obtain for the coefficient of the reflected S-waves,

VSS ¼B

B0

¼i4ab 1 b2ð Þ 3þ 5a2ð Þ

i4abþ 1 b2ð Þ 3þ 5a2ð Þ

The phase shift is given by

d ¼ tan1 4ab

1 b2ð Þ 3 5a2ð Þ

60°

30°R

X2

X1

uHSV

uSH

Fig. 124b

230 Seismology

By substitution of the values of the problem,

b ¼ tan f ¼


c2

b2 1

s

¼1ffiffiffi

3p

a ¼


1c2

a2

r

¼

ffiffiffiffiffi

22

30

r

and finally we obtain d ¼ 77.33.

125. An S-wave represented by the potential

c ¼ 10 exp i31

2x1 þ

ffiffiffi

3p

2x3 4

ffiffiffi

2p

t

is incident from an elastic medium with l¼ 0 onto a liquid with velocity a0 ¼ 4 km s1

(the two media have the same density). Derive the equations relating the amplitudes of

the potentials of the incident, reflected, and transmitted waves.

Given that thewave is incident froman elasticmedium onto a liquidmedium, there are reflected

S- and P-waves in the elastic medium and transmitted P-waves in the liquid (Fig. 125).

If l ¼ 0, the P-wave velocity in the elastic medium is

a ¼


lþ 2m

r

s

¼

ffiffiffiffiffiffi

2m

r

s

¼ bffiffiffi

2p

¼ 4ffiffiffi

2p ffiffiffi

2p

¼ 8 km s1

Assuming (x1, x3) is the incidence plane, we use the scalar S-wave potential which, for the

incident and reflected waves in the solid medium, is given by

c ¼B0 exp ikb x1 cos f þ x3 sin f btð Þ

þ B exp ikb x1 cos f x3 sin f btð Þ

M

M ff

X3

X1

P

P

e

e

SS

a= 4

a = 8

b= 0

b = 4√2

Fig. 125


From the potential of the incident wave given in the problem,

cos f ¼1

2) f ¼ 60; B0 ¼ 10; kb ¼ 3

The potential can also be written in the form (Problem 122)

c ¼ B0 exp ik x1 þ tan f x3 ctð Þ þ B exp ik x1 tan f x3 ctð Þ

where

k ¼ kb cos f ¼3

2km1

c ¼b

cos f¼ 8

ffiffiffi

2p

km s1

and the potential is given by

c ¼ 10 exp i3

2x1 þ x3

ffiffiffi

3p

8ffiffiffi

2p

t

þ B exp i3

2x1 x3

ffiffiffi

3p

8ffiffiffi

2p

t

Applying Snell’s law we determine the angle e of the reflected P-wave in the solid medium

and the angle e0 of the transmitted P-wave onto the liquid (Fig. 125):

cos f

b¼

cos e

a)

cos 60

4ffiffiffi

2p ¼

cos e

8) e ¼ 45

cos e0

a0¼

cos f

b) cos e0 ¼

1

2ffiffiffi

2p ) e0 ¼ 69

sin e0 ¼


11

8

r

¼

ffiffiffi

7p

2ffiffiffi

2p ) tan e0 ¼

ffiffiffi

7p

The potential of the reflected P-wave in the solid medium (M) is

’ ¼ A exp ik x1 x3 tan e ctð Þ ¼ A exp i3

2x1 x3 8

ffiffiffi

2p

t

and that of the transmitted P-wave in the liquid (M0) is

’0 ¼ A0 exp ik x1 þ x3 tan e0 ctð Þ ¼ A0 exp i

3

2x1 þ x3

ffiffiffi

7p

8ffiffiffi

2p

t

The relation between the amplitude of the potential of the incident S-wave (B0 = 10) and

those of the reflected and refracted P-waves A, A0 and the reflected S-wave B can be

obtained from the conditions at the boundary between the two media (x3 ¼ 0), that is,

continuity of the normal component of the displacements (u3) and of the stress (t33) and

null tangential stress (t31):

u3 ¼ u03 ) ’;3 þ c;1 ¼ ’0;3

t31 ¼ 0 ) 2’;13 c;33 þ c;11 ¼ 0

t33 ¼ t033 ) l0 ’0;33 þ ’0

;11

¼ 2m ’;33 þ c;13

232 Seismology

By substitution of the potentials we obtain the equations (in units of 103 m2)

Aþ 10þ B ¼ A0ffiffiffi

7p

Aþ Bþ 10 ¼ 0

2A0 ¼ Aþ 10ffiffiffi

3p

ffiffiffi

3p

B

Solving the system of equations we obtain

A0 ¼40

ffiffiffi

3p

4þffiffiffi

7p

þffiffiffiffiffi

21p ¼ 6:2

A ¼20

ffiffiffiffiffi

21p

4þffiffiffi

7p

þffiffiffiffiffi

21p ¼ 8:2

B ¼10 4

ffiffiffi

7p

þffiffiffiffiffi

21p

4þffiffiffi

7p

þffiffiffiffiffi

21p ¼ 1:8

126. An S-wave incident on the free surface of a semi-infinite medium with s ¼ 0.25 is

given by (in units of 103 m2)

c i ¼ 10ffiffiffi

3p

; 2; 4

exp 5i1

4x1 þ

ffiffiffi

3p

4x2 þ

ffiffiffi

3p

2x3 4t

Calculate:

(a) The amplitude of the components of the reflected P-wave.

(b) The components of the reflected S-wave.

(a) From the direction cosines we calculate the incidence i and emergence f angles and

azimuth az of the incident S-wave:

n1 ¼1

4¼ sin i cos az

n2 ¼

ffiffiffi

3p

4¼ sin i sin az

n3 ¼

ffiffiffi

3p

2¼ cos i ) i ¼ 30 ) f ¼ 60

1

4¼

1

2cos az ) az ¼ 60

Since Poisson’s ratio is 0.25 then l ¼ m ) a ¼ffiffiffi

3p

b ¼ 4ffiffiffi

3p

, and

sin ic

a¼

1

b) sin ic ¼

1ffiffiffi

3p ) ic ¼ 35

Since i < ic we have a reflected P-wave. The reflection coefficient at a free surface for a

reflected P-wave from an incident S-wave is given by

VSP ¼4b 1þ 3a2ð Þ

4abþ 1þ 3a2ð Þ2ð126:1Þ


where a ¼ tan e and b ¼ tan f, and f is the emergence angles of the incident S-wave and e is

that of the reflected P-wave (Fig. 126). The relation between f and e according to Snell’s law is

cos f

b¼

cos e

a) cos e ¼

a

bcos f ¼

ffiffiffi

3p 1

2) e ¼ 30

Substituting in Equation (126.1):

VSP ¼4ffiffiffi

3p

1þ 1ð Þ

4þ 1þ 1ð Þ2¼

ffiffiffi

3p

We can write the potential of the incident S-wave referred to the incidence plane (x1, x3) by

means of the rotation matrix

cos az sin az 0

sin az cos az 0

0 0 1

0

@

1

A

and substituting

1

2

ffiffiffi

3p

20

ffiffiffi

3p

2

1

20

0 0 1

0

B

B

B

@

1

C

C

C

A

10ffiffiffi

3p

2

4

0

@

1

A ¼B1

B2

B3

0

@

1

A)B1 ¼ 4

ffiffiffi

3p

B2 ¼ 16 ¼ B0

B3 ¼ 4

8

<

:

the potential is

c0i ¼ B1;B2;B3ð Þ exp ikbðcos fx1 þ sin fx3 4tÞ

¼ 4ffiffiffiffi

3;p

16; 4

exp 5i1

2x1 þ

ffiffiffi

3p

2x3 4t

ð126:2Þ

The scalar potential of the SV component is

cSV ¼ B0 exp ikbðcos fx1 þ sin fx3 4tÞ

¼ 16 exp 5i1

2x1 þ

ffiffiffi

3p

2x3 4t

S S

P

f f e

X3

X1

Fig. 126

234 Seismology

The amplitude of the potential of the reflected P-wave is

A ¼ B0VSP ¼ 16ffiffiffi

3p

From

kb ¼o

b¼ 5 ¼

o

4we have o ¼ 20 s1

and

ka ¼o

a¼

20

4ffiffiffi

3p km1

The scalar potential of the reflected P-wave referred to the original system of axes is

given by

’ ¼ A exp ika njxj at

where the direction cosines are now

n1 ¼ sin i cos az ¼ cos e cos az ¼

ffiffiffi

3p

2

1

2¼

ffiffiffi

3p

4

n2 ¼ sin i sin az ¼ cos e sin az ¼

ffiffiffi

3p

2

ffiffiffi

3p

2¼

3

4

n3 ¼ cos i ¼ sin e ¼1

2

Substituting these values we obtain

’ ¼ 16ffiffiffi

3p

exp i5ffiffiffi

3p

ffiffiffi

3p

4x1 þ

3

4x2

1

2x3 4

ffiffiffi

3p

t

The components of the displacement in mm are

uP ¼ r’ )

uP1 ¼ 20ffiffiffi

3p

uP2 ¼ 60

uP3 ¼ 40

8

>

>

<

>

>

:

(b) For the reflected S-wave we have to separate the SV and SH components. The SV

component can be deduced from the scalar potential

cr

SV ¼ B exp ikbðcos f x1 sin f x3 4tÞ ¼ B exp 5i1

2x1

ffiffiffi

3p

2x3 4t

We obtain B by means of the reflection Vss:

VSS ¼4ab 1þ 3a2ð Þ

2

4abþ 1þ 3a2ð Þ2¼ 0 ¼

B

B0

) B ¼ 0

The reflected S-wave doesn’t have an SV component.


For the reflected SH component we use the displacement of the u2 instead of the

potential. The displacements of the SH component of the incident wave are obtained from

(126.2):

uiSH ¼ c01;3 c0

3;1 ¼ 40

Referred to the reference of the plane of incidence, the displacement is given by

uiSH ¼ 40 exp 5i1

2x1 þ

ffiffiffi

3p

2x3 4t

The amplitude of the reflected SH wave is equal to that of the incident SH wave. Referred

to the incidence plane system of reference,

urSH ¼ 40 exp 5i1

2x1

ffiffiffi

3p

2x3 4t

The displacement of the reflected S-wave referred to the original system of axes is

uri ¼ ðBr

1;Br

2;Br

3Þ exp i51

4x1 þ

ffiffiffi

3p

4x2

ffiffiffi

3p

2x3 4t

Since the SV component is zero, Br

3 ¼ 0;Br

1 and Br

2 are found using the equations

urj j ¼ urSH ) ðBr

1Þ2 þ ðBr

2Þ2 ¼ 1600

uri ni ¼ 0 )1

4Br

1 þ

ffiffiffi

3p

4Br

2 ¼ 0

resulting in

Br

1 ¼ 20ffiffiffi

3p

Br

2 ¼ 20

127. A wave represented by the potential

w ¼ 4 exp 0:25ix1ffiffiffi

6p þ

x2ffiffiffi

3p þ

x3ffiffiffi

2p 4t

is incident on the surface x3 ¼ 0 of separation between two liquids. If the speed of

propagation in the second medium is 2 km s1, the pressure exerted by the incident

wave on the surface of separation is 5 109 Pa, and the transmitted energy is four

times greater than the reflected energy, calculate:

(a) The energy transmitted to the second medium.

(b) The potentials of the transmitted and reflected waves referred to the same

coordinate system as the incident potential.

(a) The intensity or energy per unit surface area of the wavefront of an incident

P-wave is given in units of J m2 by

Iinc ¼ A20o

2k2aar ð127:1Þ

236 Seismology

From the given potential we have

A0 ¼ 4 103 m2

ka ¼ 0:25 km1

a ¼ 4 km s1

ka ¼o

a) o ¼ 1 s1

We need to know the value of the density r. Since the medium is liquid l ¼ K (bulk

modulus) and then l ¼ P/y, where P is the pressure and y the cubic dilatation (change of

volume per unit volume). For liquids the shear modulus m is zero and from the velocity of

P-waves we obtain,

a ¼


lþ 2m

r

s

¼

ffiffiffi

l

r

s

) l ¼ a2r

The cubic dilatation is obtained from the potential ’:

y ¼ r2’ ¼ k2aA ¼1

164 ¼

1

4

Then, we obtain

l ¼ a2r ¼P

y¼

5 109

1

4 16 106

Pa

m2 s2) r ¼

5

4g cm3

By substitution in (127.1) the incident energy is

Iinc ¼ 16 11

16 4

5

4¼ 5 Jm2

The energy transmitted into the second medium is

Itras ¼ A02o2k 02a a0r0 ¼ W 2A2

0k02a a

0r0 ð127:2Þ

and the energy reflected is

Irefl ¼ Ao2k 02a a0r0 ¼ V 2A2

0k02a a

0r0 ð127:3Þ

where W and V are the transmission and reflection coefficients, respectively:

W ¼A0

A0

¼2r tan e

r0 tan eþ r tan e0

V ¼A

A0

¼r0 tan e r tan e0

r0 tan eþ r tan e0

ð127:4Þ

The emergence angle e of the incident wave is

n3 ¼1ffiffiffi

2p ¼ cos i ¼ sin e ) i ¼ e ¼ 45


and from Snell’s law the emergence angle of the transmitted wave e0 is

cos e

a¼

cos e0

a0) cos e0 ¼

a0

acos e ¼

2

4

1ffiffiffi

2p ¼

1

2ffiffiffi

2p ) sin e0 ¼


11

8

r

¼

ffiffiffi

7p

2ffiffiffi

2p

Given that the transmitted energy is four times the reflected energy,

Itras

Iref¼ 4 ¼

W 2A20r

0o4

a0

V 2A20ro

4

a

)V 2

W 2¼

2r0a

5ra0

If we substitute in (127.4)

V ¼r0

5

4

ffiffiffi

7p

r0 þ5

4

ffiffiffi

7p

W ¼25

4

r0 þ5

4

ffiffiffi

7p

We have three equations for r’, V, and W. The solution for positive values of the variables

is

r0 ¼ 7:7 )W ¼ 0:23V ¼ 0:40

(b) The potential of the reflected P-wave is

’ref ¼ VA0 exp ika n1x1 þ n2x2 n3x3 atð Þ

¼ 1:6 exp i1

4

1ffiffiffi

6p x1 þ

1ffiffiffi

3p x2

1ffiffiffi

2p x3 4t

To determine the potential of the transmitted wave we have to calculate the direction

cosines of the transmitted ray. The azimuth is the same as that of the incident wave which

can be deduced from the direction cosines and the value of i:

n1 ¼ sin i cos az ¼1ffiffiffi

6p ) cos az ¼

1ffiffiffi

3p

n2 ¼ sin i sin az ¼1ffiffiffi

3p ) sin az ¼

ffiffiffi

2p

ffiffiffi

3p

The direction cosines of the transmitted ray in the medium M0 are

n01 ¼ sin i0 cos az ¼1

2ffiffiffi

2p

1ffiffiffi

3p ¼

1

2ffiffiffi

6p

n02 ¼ sin i0 sin az ¼1

2ffiffiffi

3p

n03 ¼ cos i0 ¼

ffiffiffi

7p

2ffiffiffi

2p

238 Seismology

and the potential of the transmitted wave is

’tras ¼ 0:92 exp i1

2

1

2ffiffiffi

6p x1 þ

1

2ffiffiffi

3p x2 þ

ffiffiffi

7p

2ffiffiffi

2p x3 2t

128. Two liquid media are separated at x3 ¼ 0, the first of volumetric coefficient

K ¼1

2 109 Pa and density 1 g cm3. The amplitudes of the components of an

incident wave of frequency 3 Hz are ui ¼ 18p 1; 1;ffiffiffi

6p

mm and those of the wave

transmitted to medium 2 are63

ffiffiffi

2p

p

7

ffiffiffiffi

2;p ffiffiffi

2p

;ffiffiffi

3p

mm. Given that the amplitude of

the transmitted potential is twice that of the reflected potential, find expressions for

the incident, reflected, and transmitted potentials.

In liquids only P-waves are propagated and their displacements can be deduced from the

scalar potential

’ ¼ A0 exp ika n1x1 þ n2x2 þ n3x3 atð Þ ) uP ¼ r’

Then in our case the components of the displacement in mm are

uP1 ¼@’

@x1¼ A0kan1 ¼ A0ka sin i cos az ¼ 18p

uP2 ¼@’

@x2¼ A0kan2 ¼ A0ka sin i sin az ¼ 18p

uP3 ¼@’

@x3¼ A0kan3 ¼ A0ka cos i ¼ 18p

ffiffiffi

6p

and we find az ¼ 45; i ¼ 30; e ¼ 60, A0 ¼ 6 103 m2.

The emergence angle of the refracted wave, e0, can be found from its displacements,

uPtras ¼63p

ffiffiffi

2p

ffiffiffi

7p

ffiffiffi

2p

ffiffiffi

7p ;

ffiffiffi

2p

ffiffiffi

7p ;

ffiffiffi

3p

ffiffiffi

7p

) n03 ¼ sin e0 ¼

ffiffiffi

3p

ffiffiffi

7p ) cos e0 ¼

2ffiffiffi

7p

The P-wave velocity in the medium of the incident wave is

a ¼


lþ 2m

r

s

¼

ffiffiffiffi

K

r

s

¼1ffiffiffi

2p km s1

Using Snell’s law we find the velocity of the medium of the refracted wave,

cos e

cos e0¼

a

a0) a0 ¼

2ffiffiffi

2p

ffiffiffi

7p

From the values of the velocities in the two media we calculate their densities:

a2

a02¼

7

16¼

Kr0

K 0r¼

1

2r0

K 0

a0 ¼ffiffiffiffi

K 0

r0

q

¼2ffiffiffi

2p

ffiffiffi

7p ) K 0 ¼

8

7r0

9

>

>

>

>

>

=

>

>

>

>

>

;

) r0 ¼3

2r ¼

3

2g cm3


The reflection Vand transmissionW coefficients are found using their expressions and from

them we get the relation between the amplitude A0 of the incident wave potential and those

of the reflected A and refracted A0 waves, and substituting the value for A0 ¼ 6, we obtain

A0 ¼ 6 ) A ¼ 3 103 m2

From these values we can write the potentials of the incident, reflected, and transmitted

waves:

’inc ¼ 6 exp i6pffiffiffi

2p 1

2ffiffiffi

2p x1 þ

1

2ffiffiffi

2p x2 þ

ffiffiffi

3p

2x3

1ffiffiffi

2p t

’ref ¼ 3 exp i6pffiffiffi

2p 1

2ffiffiffi

2p x1 þ

1

2ffiffiffi

2p x2

ffiffiffi

3p

2x3

1ffiffiffi

2p t

’tras ¼ 6 exp i3p

ffiffiffi

7p

ffiffiffi

2p

ffiffiffi

2p

ffiffiffi

7p x1 þ

ffiffiffi

2p

ffiffiffi

7p x2 þ

ffiffiffi

3p

ffiffiffi

7p x3 2

ffiffiffi

2p

ffiffiffi

7p t

129. Two liquids in contact have speeds of propagation of 4 and 6 km s1. The density

of the first is 2 g cm3 and is less than that of the second. For waves of normal

incidence, the reflected and transmitted energies are equal. A wave of v ¼ 1 s1 and

with a potential of amplitude A0 ¼ 2103 cm2 is incident from the first onto the

second at an angle of 30. Calculate:

(a) The transmitted and reflected energies.

(b) An expression for the transmitted potential.

(a) For normal incidence, the reflection and transmission coefficients in terms of the

refractive index m ¼ a/a0 and the density contrast m ¼ r0/r are given by

Vn ¼m n

mþ n

Wn ¼2

mþ n

If the reflected energy is equal to the transmitted energy, then

V 2n ¼ mnW 2

n )ðm nÞ2

ðmþ nÞ2¼

4mn

ðmþ nÞ2ð129:1Þ

Substituting n ¼ a/a0 ¼ 2/3, from (129.1) we obtain the value of m:

m2 4mþ4

9¼ 0 )

m ¼ 3:9 4

m ¼ 0:1

Trying both values we obtain for r0

m ¼r0

r¼ 0:1 ) r0 ¼ 0:2 g cm3

m ¼ 4 ) r0 ¼ 8 g cm3

240 Seismology

But the problem states that r ¼ 4 < r0, so r0 ¼ 8 g cm3.

For a wave with incidence angle 30 (e ¼ 60) the emergence angle of the transmitted

wave e0 is, according to Snell’s law (Fig. 129), given by

cos e

a¼

cos e0

a0) cos e0 ¼

3

4) sin e0 ¼


19

16

r

¼

ffiffiffi

7p

4

The partition of energy between the reflected and refracted waves is given by

mnsin e0

sin eW 2 þ V 2 ¼ 1

The reflection V and transmission W coefficients are

V ¼m sin e n sin e0

m sin eþ n sin e0¼

4

ffiffiffi

3p

22

3

ffiffiffi

7p

4

4

ffiffiffi

3p

2þ2

3

ffiffiffi

7p

4

¼ 0:8

W ¼2 sin e

m sin eþ n sin e0¼

ffiffiffi

3p

4

ffiffiffi

3p

2þ2

3

ffiffiffi

7p

4

¼ 0:46

The incident, reflected, and transmitted energies per unit time and surface area are (Problem127)

Einc ¼ro4

aA20 sin e ¼ 17:3 erg cm2 s

Eref ¼ro4

aA2 sin e ¼

ro4

aA20V

2 sin e ¼ 11:1 erg cm2 s

Etrans ¼r0o4

a0A02 sin e0 ¼

r0o4

a0A20W

2 sin e0 ¼ 7:5 erg cm2 s

M

M

e

e

i = 30°

r = 2 g cm–3

a = 4 km s–1

a = 6 km s–1

Fig. 129


(b) The potential of the transmitted wave is

’tras ¼ A0 exp ik x3 tan e0 þ x1 ctð Þ

where

A0 ¼ A0W ¼ 2 103 0:46 ¼ 920 cm2

k ¼ k 0a0 cos e0 ¼

o

a0cos e0 ¼

1

8km1

c ¼a0

cos e0¼ 8 km s1

’tras ¼ 920 exp i1

8

ffiffiffi

7p

3x3 þ x1 8t

130. An SV wave is incident on the free surface of an elastic medium of Poisson ratio

0.25. If the potential of the wave is (in units of 103m2)

ci ¼5ffiffiffi

2p ;

5ffiffiffi

2p ; 0

exp i1

2ffiffiffi

2p x1 þ

1

2ffiffiffi

2p x2 þ

ffiffiffi

3p

2x3 4t

find the components of the amplitude of the reflected P-wave referred to this set of axes.

From the direction cosines we find the incidence angle i, the emergence angle f, and the

azimuth az of the incident SV wave (Fig. 130):

n3 ¼ cos i ¼ sin f ¼

ffiffiffi

3p

2) i ¼ 30and f ¼ 60


2cos az ¼

1

2ffiffiffi

2p ) az ¼ 45

Bearing in mind that Poisson’s ratio is 0.25, from Snell’s law we find the emergence angle

e of the reflected P-wave:

s ¼ 0:25 ) l ¼ m ) a ¼ffiffiffi

3p

b ¼ 4ffiffiffi

3p

km s1

cos f

b¼

cos e

a) cos e ¼

ffiffiffi

3p

2) e ¼ 30

X3

SV S

ff

e

iP

Fig. 130

242 Seismology

The reflection coefficient for the reflected P-wave gives us the relation between the

amplitude of the potential, B0, of the incident SV wave and A, that of the reflected P wave:

VSP ¼A

B0

¼4 tan f 1þ 3 tan2 eð Þ

4 tan e tan f þ 1þ 3 tan2 eð Þ2¼

ffiffiffi

3p

To find B0 we write the potential of the incident SV wave referred to the (x1, x3) plane of

incidence using the rotation matrix

cos az sin az 0

sin az cos az 0

0 0 1

0

B

@

1

C

A

5ffiffiffi

2p

5ffiffiffi

2p

0

0

B

B

B

B

@

1

C

C

C

C

A

¼

0

5

0

0

B

@

1

C

A

B0 ¼ 5 103m2 ) A ¼ 5ffiffiffi

3p

103m2

Referred to this system of axes the potential of the reflected P-wave is given by

’ ¼ A exp ik x3 tan eþ x1 a tð Þ

k ¼kb

cos f¼

112

¼ 2 km1

’ ¼ 5ffiffiffi

3p

exp i2 x31ffiffiffi

3p þ x1 4

ffiffiffi

3p

t

The amplitudes of the displacements of the reflected P-wave referred to this set of axes are

u1 ¼@’

@x1¼ 10

ffiffiffi

3p

mm

u3 ¼@’

@x3¼ 10 mm

Referred to the original set of axes the horizontal components are

u10 ¼ u1 cos az ¼10

ffiffiffi

3p

ffiffiffi

2p mm

u20 ¼ u1 sin az ¼10

ffiffiffi

3p

ffiffiffi

2p mm

Ray theory. Constant and variable velocity

131. Assume that the Earth’s crust consists of a single layer of thickness H and a

constant speed of propagation of seismic waves of v1 on top of a mantle of velocity of

propagation 20% greater than the crust. Given that a focus on the surface produces a

reflected wave that takes 17.2 s to reach a distance of 99 km, and that this is the

243 Ray theory. Constant and variable velocity

critical distance, calculate the values of H, v1, and v2. Plot the travel-time curve (t, x)

for this specific case with numerical values.

The critical distance xc is the distance at which a ray that is reflected with the critical angle

at the top of the mantle arrives at the surface and is given by the equation (Fig. 131a)

xc ¼ 2H tan ic ¼ 99 km ð131:1Þ

where H is the thickness of the crust. Since we know the relation between the velocities in

the crust and the mantle, we can calculate the critical angle

v2 ¼ 1:2v1 )sin ic

v1¼

1

v2) sin ic ¼

1

1:2) ic ¼ 56:44

If we substitute in Equation (131.1) we obtain the thickness of the crust, H:

99 ¼ 2H tan 56:44 ) H ¼ 32:8 km

The travel time of the critically reflected ray is

t ¼ 2H

v1 cos ic¼ 2

H

v1 cos 56:44¼ 17:2

) v1 ¼ 232:84

17:2 cos 56:44¼ 6:9 km s1 ) v2 ¼ 8:3 km s1

To draw the travel-time curve for different distances of the direct, reflected, and critically

refracted waves we use the equations

t1 ¼x

v1

t2 ¼2

v1


x2

4þ H2

r

t3 ¼x

v2þ2H


v22 v21

p

v1v2

v2

v1

xc

icic

SF

H

Fig. 131a

244 Seismology

We obtain the following values

The travel-time curves are drawn in Fig. 131b.

x(km) t1 (s) t2 (s) t3 (s)

0 0 9.5 –

30 4.3 10.4 –

60 8.7 12.9 –

90 13.0 16.1 –

99 14.3 17.2 17.2

120 17.4 19.8 19.7

150 21.7 23.7 23.4

0

0

10

20

30

t (S

)

40

50 xc 150

x (km)

200

2

3

1

250 300

ti

Fig. 131b


132. In a seismogram recorded at a regional distance, the S-P time lag is 5.5 s, and the

focus is at a depth x/2, where x is the epicentral distance. The model Earth has a single

layer of Poisson ratio 0.25 and constant S-wave velocityffiffiffi

3p

km s1. Calculate:

(a) The depth of the focus.

(b) The epicentral distance.

(a) For a direct wave from point F to point P (Fig. 132) the difference of the arrival

times of the P- and S-waves (the S-P interval) is

tS-P ¼ 5:5 ¼

FP

bFP

a

The distance FP can be expressed in terms of x as (Fig. 132)

FP ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ h2p

¼


x2 þx

2

2r

¼ x

ffiffiffi

5p

2

The S-P interval is given by

5:5 ¼ x

ffiffiffi

5p

2

a b

ab

Since Poisson’s ratio is 0.25 and knowing the S-wave velocity we obtain

s ¼ 0:25 ) a ¼ bffiffiffi

3p

) 5:5 ¼ x

ffiffiffi

5p

2

ffiffiffi

3p

1ffiffiffi

3p ffiffiffi

3p

x ¼ 21 km

h ¼x

2¼ 10:5 km

133. The Earth consists of a layer of thickness 20 km and seismic wave velocity

6 km s1 on top of a medium of speed of propagation 8 km s1. A seismic focus is

P

x

h

F

Fig. 132

246 Seismology

located at a depth of 10 km. Calculate the difference in travel times between the

reflected and the critical refracted waves observed on the surface at a distance of

150 km from the epicentre.

This problem is similar to Problem 131, but now the focus is at depth h ¼10 km.

The critical distance in this case is given by (Fig. 133)

xc ¼ 2H hð Þ tan ic

sin ic ¼v1

v2) ic ¼ sin1 6

8

¼ 48:6

) xc ¼ 2 20 10ð Þ tan 48:6ð Þ ¼ 34:0 km

Since the distance 150 km is greater than the critical distance there arrive critically

refracted rays. The travel times of the reflected (t2), and critically refracted (t3) rays at that

distance are

t2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 þ 2H hð Þ2q

v1¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1502 þ 2 20 10ð Þ2q

6¼ 25:5 s

t3 ¼x

v2þ

2H hð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v22 v21

p

v1v2¼

150

8þ

2 20 10ð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

82 62p

8 6¼ 22:1 s

The time difference between the travel times of the two rays is

t3 t2 ¼ 22:06 24:49 ¼ 3:4 s

134. Consider a crust of thicknessH and constant speed of propagation v1 on a mantle

of constant speed of propagation v2. A seismic focus is located at depth H/2, the

critical distance is 51.09 km, the delay time is 4.96 s, and the critical angle is 48.59.

Calculate the values of H, v1 and v2, and the depth of the focus.

For a focus at depth h ¼ H/2, the travel times of the critically refracted (t3) rays and the

critical distance are given by the expressions (Fig. 133).

H

h

x S

v1

v2

ic

icic

i

F

Fig. 133


t3 ¼x

v2þ


v22 v21

p

v1v2

xc ¼ 2H hð Þ tan ic ) 51:09 ¼ 2H H

2

tan 48:59ð Þ

) H ¼ 30 km; h ¼ 15km

Knowing the depth and thickness of the crust, using Snell’s law, the value of the critical

angle, and the delay time ti, we find the velocities v1 and v2:

sin ic

v1¼

1

v2) v1 ¼ 0:75v2

ti ¼2H hð Þ


v22 v21

p

v1v2) 4:96 ¼

2 30 15ð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v22 0:75v2ð Þ2q

0:75v22

) v2 ¼ 8 km s1

v1 ¼ 6 km s1

135. In a seismogram, the S-P time difference is equal to 5.31 s, and corresponds to a

regional earthquake that occurred at a depth h ¼ 2H, where H is the thickness

of the crust. Given that the crust is formed by a layer of constant P-wave velocity

of 3 km s1, that below it there is a semi-infinite mantle of double that speed of

propagation, and that Poisson’s ratio is 0.25, determine:

(a) An expression for the travel-time of the P- and S-waves.

(b) The epicentral distance for an emerging P-wave with a take-off angle of 30 at the

focus.

(a) The travel time corresponding to the ray given in Fig. 135 is given by

t ¼FA

2vþAS

v

F

F

S

ih

i0i0H

2H

2v

xS

v

A

Fig. 135

248 Seismology

where v ¼ a for P-waves and v ¼ b for S-waves. Using Snell’s law we find the relation

between the incidence angle at the focus ih and at the station i0:

sin ih

2v¼

sin i0

v) sin i0 ¼

1

2sin ih

From Fig. 135 we obtain

cos ih ¼H

FA) FA ¼

H

cos ih

cos i0 ¼H

AS) AS ¼

H

cos i0

From these equations we deduce the expression for the travel time:

t ¼H

2v cos ihþ

H

v cos i0ð135:1Þ

For the epicentral distance we obtain

x ¼ F0Aþ AS0 ¼ H tan ih þ H tan i0 ð135:2Þ

Using Equations (135.1) and (135.2) and putting v ¼ a we obtain the travel time and

epicentral distance for P-waves and putting v ¼ b for S-waves.

(b) For a P-wave with take-off angle at the focus (ih) of 30, we first find the value of

the angle at the station i0,

sin ih

2a¼

sin i0

a) i0 ¼ 14:47

Substituting in (135.1) we obtain

tP ¼

H

2a

ffiffiffi

3p

2

þH

a0:97¼ 1:61

H

a

The travel time of the S-wave with take-off angle jh ¼30 can also be calculated using

(135.1). Since Poisson’s ratio is 0.25, the velocity of the S-wave is

s ¼ 0:25 ) a ¼ffiffiffi

3p

b ) b ¼affiffiffi

3p ¼

ffiffiffi

3p

km s1

The incidence angle at the station, j0, using Snell’s law, is given by

sin jh

2b¼

sin j0

b) j0 ¼ 14:47

and the travel time is

tS ¼

H

2b cos jhþ

H

b cos j0¼ 2:79

H

a

Since we know the S-P time interval we can obtain the value of h:

tS-P ¼ 5:31 ¼ 2:79

H

3 1:61

H

3) h ¼ 13:61 km


The epicentral distance is found using (135.2):

x ¼ 13:61 tan 30 þ 13:61 tan 14:47 ¼ 11:41 km

136. Consider a crust composed of two layers of thickness 12 and 18 km, and constant

P-wave speeds of propagation of 7 and 6 km s1, respectively, on top of a semi-infinite

mantle of constant speed of propagation 8 km s1. There is a seismic focus at a depth

of 6 km below the surface. For a station located at 100 km epicentral distance,

calculate the travel time of the direct, reflected, and critical refracted waves (neglect-

ing waves with more than a single reflection or critical refraction).

The travel times of the direct ray t1 and the ray reflected on the bottom of the first layer

t2 are given by (Fig.136)

t1 ¼


h2 þ x2p

v1¼


62 þ 1002p

7¼ 14:3 s

t2 ¼


2H1 hð Þ2þx2q

v1¼


2 12 6ð Þ2þ1002q

7¼ 14:5 s

As the velocity of the second layer is less than that of the first layer there is no critical

refraction at that boundary. There is critical refraction at the boundary between the second

layer and the mantle where the velocity is greater. Using Snell’s law, we can calculate the

H2

H1

h

E Sx

F

ic

i1i1

ic

2 (

H1–

h)

F ∗

n2

n3

n1

Fig. 136

250 Seismology

critical angle ic and from this value the incidence angle at the focus i1 for the critically

refracted ray:

sin i1

v1¼

sin ic

v2¼

1

v3) sin ic ¼

6

8) ic ¼ 48:6

sin i1 ¼7

8) i1 ¼ 61:0

The travel time t3 of the critically refracted ray at the bottom of the second layer is given by

t3 ¼FA

v1þAB

v2þBC

v3þCD

v2þDS

v1ð136:1Þ

If the epicentral distance x is 100 km, the different segments of (136.1) are

cos i1 ¼H1 h

FA¼

H1

DS) FA ¼ 12:4 km; DS ¼ 24:8 km

cos ic ¼H2

AB) AB ¼ CD ¼ 27:2 km

BC ¼ x 2H2 tan ic H1 hð Þ tan i1 H1 tan i1 ¼ 26:8 km

Finally, by substitution in (136.1) we obtain,

t3 ¼ 17:7 s

137. Consider a two-layered structure of thickness H and speed of propagation v and

3v on top of a half-space medium of speed of propagation 2v. At a depth 3H below the

surface there is a seismic focus. Write the expressions (as functions of H, v, and ih) for

the travel times of waves that reach the surface without being reflected. Give the

range of values of ih.

In this problem the focus is located at the half-space medium at depth h¼3H under its

boundary. Applying Snell’s law we can find the relation between the velocities, the

incidence angles at the focus and at the bottom of each layer, and the critical angle at the

boundary between the second layer and the half-space (Fig. 137):

sin ih

2v¼

sin i2

3v¼

sin i1

vsin ic

2v¼

1

3v) ic ¼ 41:8

ð137:1Þ

The rays which leave the focus and arrive at the surface at a distance x are only those with

angles less than the critical angle (Fig. 137). The travel time for these rays is

t ¼FA

2vþAB

3vþBS

v¼

H

2v cos ihþ

H

3v cos i2þ

H

v cos i1ð137:2Þ

According to Equation (137.1) we have the relation between the incidence angles:

sin i2 ¼3

2sin ih ) cos i2 ¼


1 sin2 ih

q

¼1

2


4 9 sin2 ih

q

sin i1 ¼1

2sin ih ) cos i1 ¼


1 sin2 ih

q

¼1

2


4 sin2 ih

q


Substituting in (137.2) we write the travel time as function of the take-off angle ih:

t ¼H

v

1

2 cos ihþ

2

3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 9 sin2 ihp þ

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 sin2 ihp

!

We find a similar expression for the epicentral distance x:

x ¼ H tan ih þ H tan i2 þ H tan i1

x ¼ H tan ih þsin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 sin2 ihp þ

3 sin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 9 sin2 ihp

!

The range of values of the take-off angle for rays which arrive at the surface is

0< ih <42.

138. A semi-infinite medium consists of two media of velocities v and 3v separated by a

vertical surface. In the first medium there is a focus of seismic waves at a depth a

below the free surface and at the same distance a from the surface separating the two

media. Write the expressions for the direct, reflected, and transmitted waves arriving

at the free surface, and plot the travel time curve (t, x) in units of a/v and a (neglecting

waves with more than a single reflection).

In this situation we have the following rays arriving at the surface: direct in the first

medium, reflected at the boundary, and critically refracted and refracted to the

second medium. We consider two cases for rays arriving at distances 0 < x < a and

distances x > a.

E

B

A

x S

v

3v

F

H

H

H

2v

i2

i1

i0

ic

ih

Fig. 137

252 Seismology

(a) For 0 < x <a, the travel time t1 of the direct wave is (Fig. 138a)

t1 ¼FS

v¼


x2 þ a2p

v

The travel time t2 of the reflected ray is

t2 ¼FPþ PS

v¼

APþ PS

v¼

AS

v¼


2a xð Þ2þa2q

v

The reflected rays exist also for negative distances, but we will not consider them.

The travel time t3 of the critically refracted ray (Fig. 138b) is

t3 ¼FA

vþAB

3vþAS

vð138:1Þ

E

AF

S

P

x

i

ia

a a

i

v 3v

Fig. 138a

E

B

A

3v

F

a

a

v

S

ic

ic

ic

x

Fig. 138b


Using Snell’s law the critical angle is given by

sin ic

v¼

1

3v) ic ¼ 19:47

and

cos ic ¼a

FA¼

a x

AS

The distance AB is given by

AB ¼ a a tan ic a xð Þ tan ic ¼ a tan ic x 2að Þ


t3 ¼ 2:22a

v0:94

vx

The critical distance (distance of the ray reflected with the critical angle) is given by

(Fig. 138c)

xc ¼ a SB ð138:2Þ

From Fig. 138c we obtain

tan ic ¼BP

SB¼

a a tan ic

SB) SB ¼ 1:83a

and substituting in (138.2)

xc ¼ 0:83a

Critically refracted rays exists for distances 0.83a < x < a. Here we consider only those

for 0 < x < a.

ES

F

xc

ic

a

a

v 3v

ic

ic

ic

Fig. 138c

254 Seismology

(b) For distances x > a, we have the rays refracted at the boundary between the two

media when the incidence angle is less than the critical angle (Fig. 138d), that is,

i <19.47:

a

a

e

E

F

A

SX

∗

i

v 3v

Fig. 138d

0.00.0

0.0

1.0

t (a

/v)

1.5

3

1

2

2.0

0.2 0.4 0.6

x (a)

0.8 1.0

Fig. 138e


t4 ¼FA

vþAS

3v

FA ¼a

cos i

AS ¼x a

cos e

Using Snell`law

sin i

v¼

sin e

3v) sin e ¼ 3 sin i

The travel-time is given by

t4 ¼a

v cos iþ

x a

3vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 9 sin2 ip

The travel-time curves for direct (1), reflected (2), critically refracted (3) and transmitted

(4) waves are given in Fig. 138e.

139. Given the structure in the diagram, calculate the arrival times of the direct and

(non-reflected) transmitted waves for x 0, where x ¼ 0 is a point on the free surface

in the vertical above the focus.

At x ¼ 0 )ih ¼ 0, the travel-time of the vertical ray is (Fig. 139a)

t ¼a

2vþa

v¼

3a

2v

For rays arriving at x > 0 and leaving the focus with take-off angles 0 < ih < 45, the

travel-times are given by

t ¼FA

2vþAS

v¼

a

2v cos ihþ

a

v cos r

cos ih ¼a

FA

cos r ¼a

AS

ð139:1Þ

2a

Focus

a

a2v

ν

Fig. 139

256 Seismology

Applying Snell’s law,

sin ih

2v¼

sin r

v) sin r ¼

1

2sin ih ð139:2Þ

Substituting in (139.1):

t ¼a

v

1

2 cos ihþ

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 sin2 ihp

!

ð139:3Þ

The relation between the epicentral distance x and the incidence angle ih is

x ¼ F0Aþ A0S

F0A ¼ a tan ih

A0S ¼ a tan r

x ¼ a tan iþ tan rð Þ ¼ a tan iþsin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 sin2 ip

h

!

From Fig. 139a we deduce:

tan ih ¼F0A

a

tan r ¼x F0A

a¼

x

a tan ih

Using Equation (139.2),

sin r

cos r¼

sin ih2

cos r¼

x

a tan ih )

1

cos r¼

2x

asin ih

2

cos ih

A

r

2a

v

a

a

F

F

X

ih

2v

A SE

Fig. 139a


By substitution in (139.3),

t ¼2x

v sin ih

3a

2v cos ih

For ih ¼ 45 the epicentral distance is

x ¼ a 1þ1ffiffiffi

7p

This is the limit of the epicentral distance at which these rays arrive. The corresponding

time limit is

t ¼a

v

1ffiffiffi

2p þ

4ffiffiffiffiffi

14p

For angles ih > 45 (Fig. 139b), the travel time and epicentral distance, as a function of the

take-off angle ih, are

t ¼FA

2vþAS

v¼

a

2v sin ihþ

x a

v sin r¼

a

2v sin ihþ

x a

vsin ih

2

x ¼ aþ A0S

tan r ¼A0S

2a F0A

F0A ¼a

tan ih

x ¼ aþ 2aa

tan ih

sin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 sin2 ihp

E A

x

S

n

A2n

F

a

a

F

ih

ih

r

2a

Fig. 139b

258 Seismology

Using

cos ih

2v¼

cos r

v) cos r ¼

1

2cos ih

sin r ¼1

2


4 cos2 ihp

tan r ¼


4 cos2 ihp

cos ih

we obtain for x and t,

x ¼ aþ 2aa

tan ih


4 cos2 ihp

cos ih

t ¼a

2v sin ihþ

2ðx aÞ

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4 cos2 ihp

For i ¼ 90, as expected the ray doesn’t arrive at the free surface.

140. For the structure in Fig. 140a, write the equations of the travel times of the

direct, reflected, and transmitted waves (neglecting waves with more than a single

reflection) as a function of the epicentral distance. Determine the times of intersection,

and the minimum and maximum distances in each case in terms of a/v and v. Plot the

travel-time curves.

The travel-time of the direct wave for distance 0 < x < 1 is (Fig. 140a)

t ¼


4x2 þ a2p

2v

For the ray reflected on the horizontal surface at depth a the travel time is

t ¼


x2 þ3a

2

2s

v¼


4x2 þ 9a2p

2v

ν

2ν

a

a

xS

a /2

F∗

Fig. 140a


The range of distances for this ray is

xmin ¼ 0

xmax )aa

2

¼xmax

3a

2

) xmax ¼ 3a

For the reflected ray on the surface at depth 2a the travel time is (Fig. 140b)

t ¼


x2 þ7a

2

2s

v¼


4x2 þ 49a2p

2v

The range of distances is

3a

3

2a¼

xmin

7

2a

) xmin ¼ 7a ) tmin ¼7ffiffiffi

5p

2

a

v

xmax ¼ 1

The critically refracted ray on the surface at depth a, using the general expression, is given by

t ¼x

v2þ


v22 v21

p

v1v2¼

x

2vþa

v

3ffiffiffi

3p

4

The minimum distance for this ray corresponds to the critical distance:

sin ic

v¼

1

2v) ic ¼ 30

xc ¼a

2tan ic þ a tan ic ¼

ffiffiffi

3p

2a

and the maximum distance is

xmax ¼ aþaffiffiffi

3p ¼

a 3þffiffiffi

3p

3

2v

v

a

a

S

Fa/2

x

Fig. 140b

260 Seismology

On the surface at depth 2a there is no critically refracted ray, since the minimum take-off

angle ih at that surface is

tan ih ¼aa

2

¼ 2 ) ih ¼ 63:4

greater than the critical angle 30.

The travel-time curves are drawn after rewriting the equations in units of a/v and a, and

are represented in Fig. 140c

1. Direct ray:tv

a

2

x

a

2

¼ 14; 0

x

a< 1 (a hyperbola)

2. Reflected ray on the surface at depth a:tv

a

2

x

a

2

¼9

4; 0

x

a 3 (a hyperbola)

3. Reflected ray on the surface at depth 2a:tv

a

2

x

a

2

¼49

4; 7

x

a 1 (a hyperbola)

4. Critically refracted ray on surface at depth a:tv

a¼

1

2

x

aþ3ffiffiffi

3p

4; 0:87

x

a 1:58 (due

to the short range of distances this is not noticeable in the figure)

0

1

3

2

42

4

6

t (a

/v)

8

10

2 4 6

X (a)

8 10 12

Fig. 140c


141. For the structure in the diagram, assume a seismic focus at the surface, and

calculate the travel time of the direct, reflected, and critical refracted waves for

epicentral distances between 0 and a. Calculate the critical distance, and the expres-

sion for the transmitted wave.

Since the focus is at the free surface, the travel time of the direct ray is simply given by

(Fig. 141a)

t ¼x

v0 < x < a

a

a3ν

45°

ν

F

*

Fig. 141

a

a

a√

45°

45°

2

P

ν3ν

S

x

F

F

Fig. 141a

262 Seismology

The travel time of the reflected wave is (Fig. 141a)

t ¼FP

vþPS

v¼

F0P

vþPS

v¼

1

v


x2 þ 2a2 2axp

The critical angle is given by

sin ic

v¼

1

3v) sin ic ¼

1

3) cos ic ¼

2ffiffiffi

2p

3) ic ¼ 19:47

The travel-time of the critically refracted ray is (Fig. 141b)

t ¼FA

vþAB

3vþBS

v

FA ¼

affiffiffi

2p

cos ic¼

3

4a

The distance BS can be found using the sine law in triangle SBD:

sin 90 icð Þ

a x¼

sin 45

BS) BS ¼ a xð Þ

3

4

and the distance AB (calling d ¼ BD) is

AB ¼ affiffiffi

2p

a cos 45 a

ffiffiffi

2p

2tan ic d ¼ a

ffiffiffi

2p

2 a

ffiffiffi

2p

2tan ic d

a a

a

45°

45°

45°

ν

A

2

2

B

DSF

x

ic

ic

√

*β

Fig. 141b


The distance d, using the sine law, is given by

sin 45þ icð Þ

d¼

sin 90 icð Þ

a x) d ¼ a xð Þ

sin 45þ icð Þ

sin 90 icð Þ

AB ¼ a

ffiffiffi

2p

21 tan icð Þ a xð Þ

sin 45þ icð Þ

sin 90 icð Þ¼ 0:96x 0:50a

The travel time is given by

t ¼ 0:43x

vþ 1:33

a

v

Finally we determine the critical distance xc from the triangle AS0D (Fig. 141c):

bþ 90 icð Þ þ 45 ¼ 180 ) b ¼ 64:47

sin 90 icð Þ

a xc¼

sin baffiffiffi

2p

affiffiffi

2p tan ic

) xc ¼ 0:52a

142. A medium consists of a flat crust of thicknessH and constant speed of propagation

v1 on a semi-infinite mantle of constant speed of propagation v2. For a focus at the

surface, at a distance x the direct wave arrives at a time t1 ¼ x/a, the critical distance is

xc ¼2affiffiffi

3p , and the direct and critical refracted waves intersect at the distance x ¼ 2a

ffiffiffi

3p

.

(a) Calculate the crust’s thickness, its speed of propagation, the mantle’s speed of

propagation, and the critical angle.

a

a

45°

45°

45°

ν

a2

2

ic

ic

β

A

DF

xc

s

Fig. 141c

264 Seismology

(b) Assume now that this is a layer which dips downwards at 45 with the parameters

of the model being those determined in the previous part. Calculate the travel

times of the reflected and critical refracted waves at x ¼ a, 3a, and 5a.

(a) We determine the velocity of the crust from the travel time of the direct ray:

t1 ¼x

v1¼

x

a) v1 ¼ a

The critical distance is given by

xc ¼ 2H tan ic ¼ 2Hv1ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v22 v21

p ¼ 2Haffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v22 a2p ¼ 2

affiffiffi

3p ð142:1Þ

Equating the travel times of the direct and critically refracted ray for the value of the

distancex ¼ 2affiffiffi

3p

we obtain

t1 ¼ t3 )x

v1¼

x

v2þ2H


v22 v21

p

v1v2ð142:2Þ

From Equations (142.1) and (142.2) we obtain the values of H and v2:

H ¼ a

v2 ¼ 2a

The critical angle may be estimated from

x0c ¼ 2H tan ic )2affiffiffi

3p ¼ 2a tan ic ) ic ¼ 30

(b) We now consider the case of a dipping layer with dip angle y ¼ 45. The critical

distance is now given by the equation (Fig. 142)

S

xc

xc sinθ

H

F

ic

θ

θ

*

Fig. 142


xc cos y ¼ H tan ic þ H þ xc sin yð Þ tan ic ð142:3Þ

so

xc ¼2H tan ic

cos y sin y tan ic

The critical distance along the horizontal free surface xc is found by substituting H, y, and icin (142.3)

xc ¼ 3:86a

The travel times of the reflected and critically refracted rays for a dipping layer are given by

the equations

t2 ¼


4H2 þ x2 þ 4Hx sin yp

v1

t3 ¼x cos y

v2þx sin yþ 2H

v1v2


v22 v21

q

By substitution of the values of H, v1, v2, and y, we obtain

t2 ¼


4a2 þ x2 þ 2ffiffiffi

2p

axp

a

t3 ¼xffiffiffi

2p

4aþx

ffiffiffi

2p

2þ 2a

2a

ffiffiffi

3p

¼xffiffiffi

2p

4a1þ

ffiffiffi

3p

þffiffiffi

3p

For the required values of x, we obtain the following values of the travel time in units

of a/v:

Since the critical distance is 3.86a, it only exists for x ¼5a.

143. In a flat medium, the velocity increases linearly with depth according to the

expression v ¼ v0 þ kz, where v0 is the velocity at the surface, k is a constant, and z is

the depth. For a focus at depth h, calculate an expression for the take-off angle at the

focus ih in terms of the epicentral distance x, and v0, h, and k.

At the maximum depth of penetration of the ray r the incidence angle of the ray with the

vertical is equal to 90 (Fig. 143). Using Snell’s law we can relate the angles at the focus ih,

at the bottom of the ray 90, and i0, the incidence angle at the station on the surface:

sin i0

v0¼

1

v0 þ kr¼

sin ih

v0 þ kh) sin ih ¼

v0 þ kh

v0 þ krð143:1Þ

x t2 (a/v) t3 (a/v)

a 2.80 –

3a 4.64 –

5a 6.57 6.56

266 Seismology

The problem is solved if we can express r as a function of x, v0, h, and k. The epicentral

distance x is the sum of a and b (Fig. 143):

b ¼


R2 v0

k

2r

But we know that for a distribution of velocities which increases linearly with depth the

rays are circular and their radius R is

R ¼v0

kþ r ) b ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v0

kþ r

2

v0

k

2r

and

a ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v0

kþ r

2

v0

kþ h

2r

Therefore,

b ¼ x a )

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

v0

kþ r

2

v0

k

2r

¼ x


v0

kþ r

2

v0

kþ h

2r

Solving for r we obtain

r ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

x2 h2 2v0

kh

2x

0

@

1

A

2

þv0

kþ h

2

v

u

u

u

t v0

k

By substitution of this expression for r in (143.1) we obtain the required expression

for ih.

F

E

R

ba

x

S

hr

V0

i0

ih

ν0/ k

∗

Fig. 143


144. Consider a semi-infinite medium in which the velocity increases linearly with depth

according to the expression v ¼ 4 þ 0.1 z. There is a seismic focus at a depth of 10 km.

Calculate the epicentral distance reached by a wave leaving the focus at an angle of 30.

The velocity at the focus is found directly by putting in the equation for the distribution of

velocity, z ¼ h:

vh ¼ 4þ 0:1 10 ¼ 5 km s1

According to Snell’s law we find the velocity at the point of greatest depth penetration

(i ¼ 90) of the ray (Fig. 144):

sin ih

vh¼

1

vm) vm ¼

5

sin 30¼ 10 km s1

From this value we find the depth to that point:

vm ¼ 10 ¼ 4þ 0:1 r ) r ¼ 60 km

Knowing that the rays are circular of radius R,

R ¼ r þv0

k¼ 60þ 40 ¼ 100 km

As in the previous problem, the epicentral distance (from point E to S) is (Fig. 144):

x ¼ aþ b

where

a ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R2 v0

kþ h

2r

¼ 86:60 km

b ¼


R2 v0

k

2r

¼ 91:65 km

so

x ¼ 178:25 km

x

ν0/k

a b

R

r

ih

F

E

h

V0

S

i0

Fig. 144

268 Seismology

145. A flat medium consists of a layer of thickness H and constant speed of propaga-

tion v on top of a medium of variable speed of propagation v ¼ v0 þ k(z H) where z

is the depth and k is a constant. If there is a focus at the surface:

(a) Write expressions for the epicentral distance x and the travel time t as functions of

the angle of incidence i0 at the surface.

(b) If H ¼ 10 km, k ¼ 0.1 s1, and v0 ¼ 6 km s1, calculate the angle of incidence of a

wave that reaches an epicentral distance of 140 km.

(a) As we saw in Problem 143, for a distribution with a linear increase of velocity with

depth, now in themediumunder the layer, v¼ v0þ k(zH), the rays are circular with

radius R ¼v0

kþ r where r is the maximum depth of penetration of the ray (Fig. 145).

The travel-time of the ray that crosses the layer and penetrates the medium is given by

t ¼ 2FP

v0þ2

ksinh1 kx

0

2v0ð145:1Þ

In the layer of constant velocity the path is a straight line and in the medium it is circular.

The epicentral distance x (from F to S) is (Fig. 145)

x ¼ x0 þ 2H tan i0

The length of the straight ray in the layer is

FP ¼H

cos i0


t ¼2H

v0 cos i0þ2

ksinh1 kx0

2v0

ð145:2Þ

Since the layer has constant velocity the angle i0 is the same at the focus as at the bottom of

the layer at the boundary with the medium. According to Snell’s law

x

x

i0

i0

r

Q

S

P

F

H

i0ν0

ν0/k

Fig. 145


sin i0

v0¼

1

v0 þ kr) vm ¼ v0 þ kr ¼

v0

sin i0

where r is the maximum depth reached by the ray in the medium and vm the velocity at that

depth. According to Fig. 145,

x0

2

2

þv0

k

2

¼v0

kþ r

2

¼v2mk2

¼v20

k2 sin2 i0) x0 ¼ 2

v0

kcot i0 ð145:3Þ

The epicentral distance x is given by

x ¼ x0 þ 2H tan i0 ¼2v0

kcot i0 þ 2H tan i0 ð145:4Þ

Substituting in (145.2) the expression for x 0 in terms of i0 (143.3) we obtain

t ¼2H

v0 cos i0þ2

ksinh1 cot i0ð Þ

(b) By substituting the given values in (145.4), we obtain

140 ¼2 6

0:01cot i0 þ 2 10 tan i0 )

i0 ¼ 45

i0 ¼ 80:5

146. Beneath a layer of thickness H of velocity distribution v ¼ v0 þ kz there is a semi-

infinite medium of speed of propagation v1 ¼ 2(v0 þ kH).

(a) Determine expressions (as functions of the above parameters) for the critical

distance, the time of intersection of the reflected wave, and the maximum distance

of the direct wave.

(b) For H ¼ 10 km, v0 ¼ 1 km s1, and k ¼ 0.1 s1, calculate these parameters and

plot the travel-time curves.

(a) In a layer of thickness H with variable velocity the epicentral distance x for a

reflected ray is given by

x ¼ 2

ðH

0

tan i dz

Using the ray parameter p ¼ sin i/v we can write

sin i ¼ vp ) tan i ¼pvffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 p2v2p

The epicentral distance for a ray reaching the bottom the layer is given by

x ¼ 2

ðH

0

vpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 p2v2p dz ¼

2

k

ðH

0

kpðv0 þ kzÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 p2ðv0 þ kzÞ2q dz ð146:1Þ

For a ray incident at the bottom of the layer at the critical angle, we have

p ¼sin ic

v0 þ kH¼

1

2ðv0 þ kHÞ

270 Seismology

Substituting this expression in (146.1) and evaluating the integral, making the change of

variable u ¼ v0 þ kz, we obtain the critical distance

xc ¼ 4 v0 þ kHð Þ

k

ffiffiffi

3p

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1v20

4 v0 þ kHð Þ2

s" #

ð146:2Þ

The intercept time for x ¼ 0, corresponding to the time of the reflected vertical ray (p ¼ 0),

is given by

ti ¼ 2

ðH

0

dz

v0 þ kz¼

2

klnv0 þ kH

v0ð146:3Þ

The maximum distance xmax corresponds to the last ray propagated inside the layer

without penetrating into the medium and has a circular path of radius R ¼ H þv0

k(Fig. 146a):

xmax

2

2

þv0

k

2

¼ H þv0

k

2

) xmax ¼ 2H


1þ2v0

kH

r

ð146:4Þ

(b) For the particular case with the values, H ¼10 km, v0 ¼ 1 km s1, and k ¼ 0.1 s1,

the velocity at the bottom of the layer H is

vH ¼ v0 þ kH ¼ 1þ 10 0:1 ¼ 2 km s1

The velocity of the medium is

v1 ¼ 2 v0 þ kHð Þ ¼ 2 1þ 10 0:1ð Þ ¼ 4 km s1

The critical distance, using Equation (146.2), is

xc ¼ 4 2

0:1

ffiffiffi

3p

2


11

4 2ð Þ2

s" #

¼ 20ffiffiffi

3p

2ffiffiffi

5p

¼ 8:2 km

S

xmax

xc

ic

F

H

n0/k

n1

Fig. 146a


The intercept time (146.3) of the reflected ray is

ti ¼2

klnv0 þ kH

v0¼

2

0:1ln1þ 0:1 10

1¼ 13:9 s

and the maximum distance for the ray in the layer (146.4) is

xmax ¼ 2 10

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ2 1

0:1 10

r

¼ 20ffiffiffi

3p

¼ 34:6 km

The travel-time curve for rays inside the layer is calculated using the expression

t ¼2

ksinh1 kx

2v0

¼ 20 sinh1ð0:05xÞ

and is represented in Fig. 146b.

30

25

20

t (s

)

15

10

5

0 10 20

x (km)

30 40

Fig. 146b

272 Seismology

147. A medium has a distribution of velocity with depth of the form v ¼ v0 eaz, with

0 < a <1. Write as functions of the epicentral distance x the expressions for the ray

parameter, travel-time, and maximum depth reached.

If r is the maximum depth reached for a ray with ray parameter p (Fig. 147), the epicentral

distance x is given by

x ¼

ð

r

0

pvdzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 p2v2p ¼

ð

r

0

pv0eazdz


1 p2v20e2az

p

¼2

asin1ðpv0e

azÞ

r

0

¼2

a½sin1ðpv0e

arÞ sin1ðpv0Þ

and, as p ¼1

v0ear, we have

x ¼2

a

p

2 sin1ðpv0Þ

h i

From this expression we obtain

p ¼1

v0cos

ax

2

The travel-time is given by

t ¼

ðx

0

pdx ¼1

v0

ðx

0

cosax

2dx ¼

2

v0asin

ax

2

To find the maximum depth of penetration r of a ray arriving at distance x, we write

p ¼1

v0ear¼

1

v0cos

ax

2) r ¼

1

aln cos

ax

2

148. In a semi-infinite medium of speed of propagation v ¼ 6 expz

2

, the P-wave

emerges with an angle of incidence of 30. Calculate the difference in arrival times at

a given station of the P-wave and the PP-wave (the wave reflected once at the free

surface). At what angle of incidence does the PP-wave emerge?

For a velocity distribution increasing with depth of the type v ¼ v0ea z (in our case with

v0 ¼ 6, a¼ 1/2) rays follow a curved path. For a focus on the free surface the ray parameter

p and the travel times t are given by (Problem 147)

p ¼1

v0cos

ax

2

t ¼2

av0sin

ax

2

ð148:1Þ

x

r

SF

Fig. 147


For a ray with incidence angle at the surface i0 ¼ 30, the ray parameter of the direct

P-wave is given by

p ¼sin i0

v0¼

1

2 6¼

1

12

Substituting this value in (148.1) we obtain, for the epicentral distance x,

1

12¼

1

6cos

x

2 2

) x ¼4p

3km

The corresponding travel time is

tP ¼

2

1

26

sin1

2

4p

3

1

2

¼ 0:58 s

The travel time of the reflected PP-wave (Fig. 148) is double that of the direct P-wave

arriving at the distance x/2:

tPP ¼ 2

2

1

26

sin1

2

4p

6

1

2

¼ 0:67 s:

The difference between the two times is

tPP t

P ¼ 0:67 0:58 ¼ 0:09 s:

To calculate the incidence angle of the PP-wave, we determine first the ray parameter

p corresponding to the distance x/2:

x

2¼

4p

3 2¼

4p

6

so

p ¼1

6cos

1

2

4p

6

1

2

¼

ffiffiffi

3p

12

x

SF

Fig. 148

274 Seismology

From the value of p, using Snell’s law, we obtain i0:

p ¼sin i0

v0) sin i0 ¼

ffiffiffi

3p

126 ) i0 ¼ 60

149. A layer of thickness H has a velocity distribution v ¼ v0 exp(az) where a <1.

Beneath it there is a semi-infinite medium of speed of propagation v1 ¼ 2v0 exp(aH).

Determine in terms of v0, v1, a, and H:

(a) The distance and critical angle.

(b) The time of intersection of the reflected wave.

(c) The maximum distance of the direct wave.

(d) Calculate the values of these parameters if v ¼ 1 km s1, H ¼ 10 km, and

a ¼ 0.1 km1.

(a) The distance for a ray reaching a depth H is given by (Fig. 149)

x ¼ 2

ðH

0

tan idz ð149:1Þ

and using the definition of the ray parameter p,

p ¼sin i

v) sin i ¼ vp

cos i ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 v2p2p

tan i ¼vpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 v2p2p


x ¼ 2

ðH

0


1 v2p2p dz ¼ 2

ðH

0

pv0eaz


1 p2v20e2az

p dz ð149:2Þ

FS

ic

xc

xmax

H

n1

Fig. 149


Introducing the change of variable

u ¼ v0eaz

du ¼ v0eazadz

we find

x ¼2

asin1pv0e

aH sin1pv0

For the critical angle ic at the bottom of the layer we have

p ¼sin i0

v0¼

sin ic

vH¼

1

v1) ic ¼ sin1 vH

v1¼ sin1 v0e

aH

v1

By substitution in (149.2) we find, for the critical distance xc,

xc ¼2

asin1 v0e

aH

v1 sin1 v0

v1

¼2

asin1 vH

v1

sin1 v0

v1

¼2

aic i0ð Þ

(b) The intercept timeof the reflected ray corresponding to the vertical ray (x¼ 0 andp¼ 0) is

ti ¼ 2

ðH

0

dz

v¼ 2

ðH

0

1

v0eazdz ¼

2

av01 eaH

(c) The maximum distance of a ray contained in the layer is given by (149.2)

xmax ¼ 2

ðH

0


1 p2v2p dz ¼

2

asin1pv0e

aH sin1pv0

ð149:3Þ

At the point of greatest depth penetration the incidence angle is 90 and, according to

Snell’s law,

p ¼sin 90

vH¼

eaH

v0¼

1

vH¼ p

By substitution in (149.3),

xmax ¼2

a

p

2 sin1eaH

h i

¼2

acos1eaH

(d) Substituting the data of the problem we obtain

ic ¼ sin1 1 e0:110

5:62¼ 30

ti ¼2

0:1 11 e0:110

¼ 12:6 s

xmax ¼2

0:1cos1e0:110 ¼ 23:9 km

xc ¼2

0:1sin1 1

2 sin1 1

5:62

¼ 6:8 km

276 Seismology

Ray theory. Spherical media

150. Assume that the Earth consists of two concentric regions of constant velocity: the

core of radius R/2 and the mantle. The speed of propagation in the core is twice that of

the mantle. Calculate:

(a) The maximum angular distance of the direct ray in the mantle.

(b) The critical angular distance of the refracted ray in the core.

(c) Plot the paths of the waves that propagate through the Earth’s interior, and the

travel-time curves of these waves in units of R/v, where v is the speed of propaga-

tion in the mantle.

(a) The travel time of the direct ray in the mantle in terms of the angular distance is

given by

t1 ¼ 2R

vsin

2

where 0 D Dmax and Dmax is the maximum distance for a ray contained in the mantle.

According to Fig. 150a the last ray which propagates in the mantle without entering the

core corresponds to angular distance Dmax which in our case is

cosmax

2¼

R

2R) max ¼ 120

(b) The critical angle for a ray incident at the core is

sin ic

v¼

1

2v) ic ¼ 30

To calculate the critical distance Dc we consider the relation (Fig. 150b)

Δmax

R/2

F

FR

R

S∗

∗

n

2n

Δ/2

Fig. 150a

277 Ray theory. Spherical media

yþ bþ a ¼ 180

sin a

R

2

¼sin b

R

ic þ b ¼ 180 ) b ¼ 150 ) a ¼ 14:5 ) y ¼ 15:5

c ¼ 2y ¼ 31

(c) The travel time of a ray reflected at the mantle–core boundary is (Fig. 150c)

t2 ¼FP

vþPS

v¼ 2

FP

vð150:1Þ

SF

R

P

ic

n

2n

ΔcR/2

β

α

q

Fig. 150b

F

Δ/22ν

nP

S

R

R/2

Fig. 150c

278 Seismology

According to the cosine law,

FP2¼

R

2

2

þ R2 2R

2R cos

2


t2 ¼R

v


5 4 cos

2

r

The travel times for the minimum and maximum angular distances are

min ¼ 0 ) t ¼R

v

max ¼ 120 ) t ¼ffiffiffi

3p R

v

The travel time for a ray which enters the core, that is, with i1 < ic, can be determined

according to Fig. 150d using Snell’s law:

sin i1

v¼

sin i2

2v

sin i2 ¼ 2 sin i1

ð150:2Þ

Adding the times of the paths through the mantle and the core:

t3 ¼FP

vþPQ

2vþQS

v¼

2FP

vþPQ

2v

FP2¼ R2 þ

R2

4 2R

R

2cos

a

2

PQ ¼ 2R

2sin

a

2

Because a ¼ 180 – 2i2, we obtain

t3 ¼2R

v


5

4 sin

2þ i2

s

þR

2vcos i2 ð150:3Þ

for values of the incidence angle 0 < i < ic, corresponding to distances Dc < D < 180.

The relation between the incidence angle i1 and angular distance D is given by

i2i1

i1

i2

nF

R/2 R

S

P

Q

α

Δ

2n

Fig. 150d


sin i1

R¼

sin 90 þ i1 i2

2

R=2

sin i1 ¼ 2 cos i1 i2

2

) ¼ 2 i1 i2 cos1 sin i1

2

ð150:4Þ

Using Equations (150.2), (150.3), and (150.4) we can calculate the travel times of the direct

ray in the mantle, the reflected ray, and the transmitted ray through the core. Some values

for the transmitted rays in the core are given in the table.

The travel-time curves for direct rays (1), reflected rays (2), and rays refracted in the core

(3) are shown in Fig. 150e.

Δº

00.0

0.5

t (R

/v)

1.0

1.5

2.0

2

3

1

50 100 150

Fig. 150e

i1 () i2 (

) D () t3 (R/v)

0 0 180.0 1.50

10 20.3 169.9 1.53

20 43.2 153.4 1.60

30 90 89.0 1.46

280 Seismology

151. Assume that the Earth consists of two concentric regions of constant velocity: the

core of radius R/2 and the mantle. The speed of propagation in the core is half that

of the mantle. Plot the travel-time curves of the waves that propagate in the interior of

the Earth in units of R/v where v is the speed of propagation in the mantle.

This problem is similar to Problem 150, but now the velocity of the core is less than that of

the mantle. In the mantle we have direct and reflected rays. As in Problem 150 the

maximum angular distance for the direct wave is 120. The travel times for the direct

(t1) and reflected (t2) rays are

t1 ¼ 2R

vsin

2ð151:1Þ

t2 ¼R

v


5 4 cos

2

r

ð151:2Þ

Since the velocity of the core is less than that of the mantle there is no critical angle. All

rays incident at the core are refracted into it. According to Snell’s law the refracted angle i2is less than the incident angle i1 (Fig. 151a):

sin i1

v¼

sin i2v

2

sin i2 ¼1

2sin i1

i2

i1

i2

i1

F ∗

n n/2

P

S

R/2

R

QΔ

a

Fig. 151a


The travel-time for a ray crossing the mantle and the core is (Fig. 151a)

t3 ¼FP

vþPQv

2

þQS

v¼

2FP

vþ2PQ

v

where

FP2¼ R2 þ

R2

4 2R

R

2cos a ¼ R2 þ

R2

4 R2 cos 90þ i2

2

FP2¼

5R2

4þ R2 sin i2

2

PQ ¼ 2R

2cos i2 ¼ Rcos i2

so

t3 ¼2R

v


5

4þ sin i2

2

s

þ2R

vcos i2 ð151:3Þ

Δ (º)

t (R

/n

)

2

1

0

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

25 50 75 100 125 150 175

3

Fig. 151b

282 Seismology

The relation between the incidence angle at the mantle–core boundary, i1, and the angular

distance, D, of a ray which crosses the core is

sinð180 i1Þ

R¼

sin i1 i2 þ

2 90

R=2

sin i1 ¼ 2 cos i1 i2 þ

2

) ¼ 2 cos1 sin i1

2

i1 þ sin1 1

2sin i1

ð151:4Þ

The range of distance for this ray is 120 < D <180.

From Equations (151.1), (151.2), and (151.3) we can calculate the values for the travel-

times of the direct, reflected, and refracted rays. Some values for t3 are given in the table.

The travel-time curves for rays that are direct (1), reflected (2), and refracted in the core (3)

are shown in Fig. 151b.

152. Consider a spherical Earth of radius R formed by two hemispherical media of

constant velocities of propagation v and 2v. For a focus on the surface of the

hemisphere of velocity v at the point of intersection of the diameter perpendicular

to the plane that separates the two media, calculate the travel times and travel-time

curves of the direct, reflected, and critical refracted waves at the surface of separation

of the two media, in units of R/v. Calculate the expression for the travel time of waves

that propagate through the medium of speed of propagation 2v.

The travel time for angular distances D 90 are given by (Fig. 152a)

t1 ¼FS

v¼ 2

R

vsin

2ð152:1Þ

The travel time of the ray reflected at the plane boundary between the two hemispheres is

(Fig. 152a)

t2 ¼FP

vþPS

v¼

F0S

v

According to Fig. 152a (triangle OSF0) the relation between the angles a and D is

ð180 Þ þ 2a ¼ 180 ) a ¼

2

sin a ¼SS

0

F0S) F0S ¼

SS0

sin a

SS0 ¼ R sin ) F0S ¼R sin

sin a¼ 2R cos

2

i1 () i2 (

) D () t3 (R/v)

0 0 180.0 1.50

10 20.3 149.4 1.48

20 43.2 114.0 1.40

30 90 31.0 1.07


Then for D < 90

t2 ¼2R

vcos

2ð152:2Þ

The critical angle, according to Snell’s law, is given by

sin ic

v¼

1

2v) ic ¼ 30

The travel-time of the critically refracted ray is (Fig. 152b)

t3 ¼FA

vþAB

2vþBS

vð152:3Þ

According to Fig. 152b

FA ¼R

cos ic

BS ¼SS0

cos ic¼

R cos

cos ic

AB ¼ OS0 OA BS

0

OS0¼ R sin

OA ¼ R tan ic

BS0¼ BS sin ic

F

S

PO

R

2ν

ν

α

α

F

S

∆

Fig. 152a

284 Seismology

Substituting ic ¼30:

FA ¼2Rffiffiffi

3p ; BS ¼

2R cosffiffiffi

3p ; AB ¼ R sin

Rffiffiffi

3p

R cosffiffiffi

3p

and substituting in (152.3) we obtain, for Dc D 90,

t3 ¼R

2vsinþ

ffiffiffi

3p

1þ cosð Þh i

ð152:4Þ

The critical distance can be calculated from ic¼ 30 using Fig. 152a (triangleOSP) anda¼D/2

90 c þ aþ ic þ 90 ¼ 180 ) c ¼ ic þ a ) c ¼ 60

The travel time of the rays that cross the boundary and penetrate in to the medium of

velocity 2v is given by (Fig. 152c)

t4 ¼FP

vþPS

2vð152:5Þ

According to Snell’s law,

sin i

v¼

sin i0

2v) sin i0 ¼ 2 sin i

and we have that

FP ¼R

cos i

PS ¼SP0

sin i0

SP0¼ R sin 180 ð Þ R tan i

ic

ic ic

ic

F

S

S

F

P

2n

n

O

R

A B

Δ

∗

Fig. 152b


F

F

O

2n

n

S

P

R

P

i

i

i

Δ

Fig. 152c

Δ (°)0

0

0.5

1.0

1.5

2.0

1

2

3

20 40 60 80

t (R

/n

)

Fig. 152d

286 Seismology

Substituting in Equation (152.5) we obtain, for D > 90,

t4 ¼3R

4v cos iþR sin

4v sin ið152:6Þ

Travel-time curves for t1, t2, and t3 are shown in Fig. 152d.

153. Assume an Earth formed by a mantle of thickness R and a core of radius R/2,

with velocities v and 2v. There is a seismic focus at depth R/4 below the surface.

Calculate the travel-time curves of the direct and reflected waves.

The travel time of the direct ray is (Fig. 153a)

t1 ¼FP

v

The distance FP can be expressed in terms of R and D using the cosine law in triangle FOP:

FP ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3R

4

2

þ R2 2R3R

4cos

s

Then, we obtain

t1 ¼R

v

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

25

163

2cos

r

ð153:1Þ

The maximum distance for the direct ray is

max ¼ 1 þ2

cos1 ¼

R

23R

4

) 1 ¼ 48:2

cos2 ¼

R

2R) 2 ¼ 60:0

max ¼ 48:2þ 60:0 ¼ 108:2

Δ1

Δ2

P

O

S

F

RR/2R /4∗

Δ

ΔMAX2n n

Fig. 153a


The travel time for the reflected ray (Fig. 153b) is

t2 ¼FP

vþPS

v

The distances FP and PS are expressed in terms of R, D1, and D2 using the cosine law in

triangles FOP and SOP (Fig. 153b):

FP ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3

4R

2

þR

2

2

2R

2

3R

4cos1

s

PS ¼


R2 þR

2

2

2R

2cos2

s

Then, we obtain

t2 ¼R


13

163

4cos1

r

vþR


5

4 cos2

r

vð153:2Þ

Now we need to express D1 and D2 in terms of the take-off angle i at the focus (F). Using

Snell’s law for a spherical medium, we relate i and i0, the incidence angle at the station (S):

3R

4sin i

v¼

R

2sin r

v¼

R sin i0

v) sin i0 ¼

3

4sin i ð153:3Þ

According to Fig. 153b for triangle FOP we have

iþ aþ1 ¼ 180

sin a

3R

4

¼sin i

R

2

and we obtain

2 sin 1 þ ið Þ ¼ 3 sin i ð153:4Þ

S

F∗

R /4 R/2 O R

2n n

Δ2

Δ1

P

b

a

r

i

r

i

Fig. 153b

288 Seismology

and for triangle POS

b ¼ 180 2 i0

sin i0

R

2

¼sin b

R

and

2 sin i0 ¼ sin 2 þ i0ð Þ ð153:5Þ

Equations (153.3), (153.4), and (153.5) allow us to calculate D1 and D2 from the take-off

angle i at the focus. The travel-times are given in the following table.

For angular distance D greater than 108.2 there are no reflected rays. The travel-time

curves corresponding to the direct (1) and reflected (2) rays are shown in Fig. 153c.

0

0.4

0.6

0.8

0.8

0.8

1.4

20

1

2

40 60

Δ ( )

80 100

t (R

/n

)

Fig. 153c

i () i0 () D1 () D2 (

) D () t2 (R/v)

0 0 0 0 0 0.75

10 7.5 5.1 7.6 12.7 0.76

30 22.0 18.6 26.6 45.2 0.92

40 28.8 34.6 45.8 80.4 1.19

41.8 30.0 47.1 58.9 106.0 1.41


154. Consider a spherical Earth of radius R ¼ 3000 km and constant P-wave speed of

propagation of 4 km s1. Within it there is a core of radius R/2 and constant velocity

v1. At a station at epicentral distance D from an earthquake with focus at the surface,

the observed time interval is tS-P ¼ 547.0 s. Given that Poisson's ratio is 1/6, and that

the arrival of the P-wave is at 12 h 23 m 20.4 s, calculate:

(a) The epicentral distance.

(b) The time of the earthquake.

(a) For a spherical Earth of constant velocity the travel time of the direct ray is given by

t ¼ 2FO0

v¼ 2

R

vsin

2ð154:1Þ

Taking into account the presence of the core themaximumdistance for thedirect ray is (Fig. 154)

cosmax

2¼

R

2R) max ¼ 120

Since Poisson’s ratio is 1/6 we have

s ¼1

6¼

l

2 lþ mð Þ) m ¼ 2l

and we can calculate the relation between the velocities of the P-wave (a) and the S-wave (b),

a ¼


lþ 2m

r

r

b ¼

ffiffiffi

m

r

r

9

>

>

=

>

>

;

) a ¼

ffiffiffi

5

2

r

b ) b ¼ 2:53 km s1

Using (154.1) and assuming the same path for P- and S-waves, from the time interval S-P

we obtain the distance:

tS-P ¼ 2R sin

2

1

b1

a

) ¼ 77:7

Δmax

F

O

Δ

n1 n

ROR/2∗

S

Fig. 154

290 Seismology

Notice that D < Dmax.

(b) To calculate the time of origin we subtract from the arrival time of the P-wave the

value of the travel time for that distance:

tP ¼

2R

asin

2¼ 2

3000

4sin

77:7

2¼ 940:9 s

The time of origin is then given by

t0 ¼ 12 h 23 m 20:4 s 940:9 s ¼ 12 h 07m 39:5 s

155. Consider the Earth formed by a sphere of radius R ¼ 4000 km, Poisson's ratio

1/8, and constant S-wave speed of propagation 3 km s1. Within it, there is a liquid

core of radius R/2. There occurs an earthquake with a focus in the interior of the

Earth. At a station of epicentral distance D the observed time interval is tS-P ¼ 600 s.

This focus may be at a depth of either R/10 or 2R/5. Calculate the correct depth of the

focus, and the epicentral distance.

First we determine the maximum distance for direct rays which don’t penetrate into the

core, which according to Fig. 155 corresponds to Dmax ¼ D1 þ D2:

cos1 ¼

R

2R h

cos2 ¼

R

2R) 2 ¼ 60

If the depth of the focus is R/10 then

h ¼R

10) cos1 ¼

5

9) 1 ¼ 56 ) max ¼ 116

and if it is 2R/5, then

h ¼2

5R ) cos1 ¼

5

6) 1 ¼ 33:56 ) max ¼ 94

P

F

h R/2 R

S

O

Δ2

Δ1Δ

ν2 ν1

Fig. 155


For a point on the surface at distance D, the S-P time interval implies, assuming the same

path for P- and S-waves,

tS-P ¼

FP

bFP

a¼

FP

aba bð Þ ð155:1Þ

From the value of Poisson’s ratio the relation between the P and S velocities is given by

s ¼1

8¼

l

2 lþ mð Þ) l ¼

m

3) a ¼


lþ 2m

r

s

¼

ffiffiffiffiffiffi

7m

3r

s

¼

ffiffiffi

7

3

r

b

Substituting in (155.1) the S-P interval equal to 600 s we obtain the length of the ray:

FP ¼ 600ab

a bð Þ¼ 5212 km

Using the cosine law for triangle FOP

FP2¼ R hð Þ2 þ R2 2R R hð Þ cos

cos ¼FP

2 ðR hÞ2 R2

2RðR hÞ

If h ¼2

5R then D ¼ 106, but this result is not possible because the maximum distance of

the direct ray for that depth is 94. If h ¼R

10then D ¼ 86, this result is possible because

this distance is less than the maximum distance. The depth is, then, 400 km.

156. Consider the Earth of radius R and constant velocity v with a core of radius 6R/10

and constant speed of propagation 2v. An earthquake occurs with focus at 8R/10 from

the centre of the Earth. A wave emerges from that focus with a take-off angle of 15.

(a) Will it pass through the core?

(b) What epicentral distance will it reach?

(c) What will be the travel time of the wave (in units of R/v)?

(a) First we calculate the maximum epicentral distance for a ray which doesn’t

penetrate the core. According to Fig. 156a the maximum distance is

max ¼ 1 þ2

cos1 ¼

6

10R

8

10R

) 1 ¼ 41:4

cos2 ¼

6

10R

R) 2 ¼ 53:1

max ¼ 41:4þ 53:1 ¼ 94:5

From this value we calculate the take-off angle ih for this ray:

1 þ ih ¼ 90 ) ih ¼ 48:6

292 Seismology

For take-off angles less than 48.6 the rays travel through the core.

Since the velocity in the core is greater than in the mantle, to find out which rays

penetrate into the core, we also need to know the critical angle. Rays with incidence angle

at the core–mantle boundary with i > ic are totally reflected and don’t penetrate into the

core. According to Snell’s law the critical angle is given by

sin ic

v¼

1

2v) ic ¼ 30:0

We calculate, using Snell’s law, the angle of incidence i corresponding to the take-off angle

of 15 (Fig. 156b):

8

10R sin ih

v¼

6

10R sin i

v) i ¼ 20:2

Since the incidence angle i (20.2) is less than the critical angle (30) and less than the

angle corresponding to the maximum distance (48.6), the ray with take-off angle of 15

penetrates into the core.

P

O

2nF

6R/10 R

8R/10

ih

S

Δ2

Δ1 nΔ

Fig. 156a

A

B

R

S

2n

6R/10

8R/10

O

F

iih

i2

i2

i

i0

Δ1Δ2

Δ3

n

Fig. 156b


(b) Applying Snell’s lawwe find the angle of the transmitted ray in the core i2 (Fig. 156b):

8

10R sin ih

v¼

6

10R sin i2

2v) i2 ¼ 43:7

By consideration of triangles FOA and AOB, we determine D1 and D2 (Fig. 156b):

ih þ1 þ 180 i ¼ 180 ) 1 ¼ 5:2

i2 þ2 þ i2 ¼ 180 ) 2 ¼ 92:6

and using Snell’s law we determine i0 the incidence angle at the surface and D3:

6

10R sin i2

2v¼

6

10R sin i

v¼

R sin io

v) io ¼ 11:9

3 ¼ i io ¼ 8:3

The epicentral distance of the ray is

¼ 1 þ2 þ3 ¼ 106

(c) The travel time is

t ¼FA

vþAB

2vþBS

v

where

FA ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

8R

10

2

þ6R

10

2

28

10R

6

10R cos1

s

¼ 0:21R

AB ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

6

10R

2

þ6

10R

2

26

10

6

10R2 cos2

s

¼ 0:87R

BS ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

6

10R

2

þ R2 26

10RR cos3

s

¼ 0:42R

so

t ¼ 1:07R

v

157. Assume a spherical Earth of radius R ¼ 6000 km and constant S-wave speed of

propagation 4.17 km s1. Poisson's ratio is 1/4. At a station at epicentral distance 60

an earthquake is recorded with a time interval tS-P ¼ 554 s. Calculate the depth of the

earthquake.

Given that Poisson’s ratio is 0.25, the P-wave velocity is

s ¼1

4¼

l

2 lþ mð Þ) l ¼ m ) a ¼

ffiffiffi

3p

b ¼ 7:22 km s1

294 Seismology

From the time interval S-P we can calculate the length of the ray FS (Fig. 157):

tS-P ¼

FS

bFS

a¼ FS

a b

ab

FS ¼tS-Pab

a b¼ 5469 km

ð157:1Þ

The distance along the ray in terms of the angular epicentral distance D, using the cosine

law, is

FS ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R2 þ R hð Þ2 2R R hð Þ cos

q

Substituting FS from (157.1):

5370 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R2 þ R hð Þ2 2R R hð Þ cos

q

ð157:2Þ

We substitute in (157.2) the values R ¼ 6000 km and D ¼ 60 and solve for h, finding two

possible solutions:

h1 ¼ 4706 km

h2 ¼ 1294 km

158. Assume a spherical Earth of radius R and P-wave velocity which can be

expressed by the equation v(r) ¼ a br2. The speed of propagation at the surface

of the Earth is v0 and at the centre of the Earth it is 2v0. What angular distance D does

a wave reach which penetrates to a depth equal to half the Earth’s radius?

If the velocity distribution inside the Earth is v(r) ¼ a br2, the ray paths are circular with

radius given by (Fig. 158)

F

Δ

R – h

S

R

Fig. 157


r ¼r

pdv

dr

ð158:1Þ

From the conditions of the problem

r ¼ R ) v ¼ v0 ¼ a bR2

r ¼ 0 ) v ¼ 2v0 ¼ a

and

b ¼v0

R2

a ¼ 2v0

)

) v ¼ v0 2r2

R2

The radius of curvature of the ray which penetrates to r ¼ R/2 is that corresponding to the

ray parameter

p0 ¼r0

v0¼

R

2v0ð158:2Þ

The velocity at depth R/2 is

v0 ¼ v0 2

R

2R

0

B

@

1

C

A

20

B

B

@

1

C

C

A

¼ v07

4

Substituting the ray parameter in (158.2):

p0 ¼R

v07

2

¼2R

7v0

7/4 R

7/4 R

R/2 O R

SF

Δ/2Δ/2

Fig. 158

296 Seismology

The derivative of the velocity is

dv

dr¼

d

dra br2

¼ 2br ¼ 2v0

R2r

Substituting in (158.1) we obtain, for the radius of curvature (Fig. 158),

r ¼r

p 2brð Þ¼

7R

4

The epicentral (angular) distance corresponding to this ray is found by applying the cosine

law to the triangle POS:

7

4R

2

¼9

4R

2

þ R2 29R

4R cos

2) ¼ 96:4

159. The Earth consists of a mantle of radius R and speed of propagation v ¼ a=ffiffi

rp

,

and a core of radius R/2 and speed of propagation 4v0, where v0 is the speed of

propagation at the Earth’s surface. Calculate:

(a) The maximum epicentral distance corresponding to a wave that travels only

through the mantle.

(b) The critical angle of the wave reflected at the core, and the angle at which it leaves

the surface.

(c) The epicentral distance Dc corresponding to the critical angle.

(d) Plot the travel-time curve, specifying Dc and Dm.

(a) Thevalue ofa in the velocity distribution is found from the value of velocity at the surface:

r ¼ R ! v ¼ v0 ¼ aR12 ) a ¼ v0R

12

The velocity distribution is

v ¼ v0R

r

a

¼ v0R

r

12

For this general type of distribution of velocity with depth a < 1, the angular distance for a

surface focus (Fig. 159a) is given by

¼2

1þ acos1 p

0

ð159:1Þ

where

¼r

v) 0 ¼

R

v0

The maximum distance for a ray which travels only through the mantle, that is that reaches

depth R/2, can be calculated from the velocity at that depth, vm:

vm ¼ v0R

R

2

0

B

@

1

C

A

12

¼ v0ffiffiffi

2p

) p ¼

R

2vm

¼

R

2

v0ffiffiffi

2p ¼

1

2ffiffiffi

2p

R

v0¼ p


By substitution in (159.1) we obtain, for the maximum distance,

m ¼2

1þ1

2

cos1

1

2ffiffiffi

2p

R

v0R

v0

0

B

B

@

1

C

C

A

¼ 92:4

(b) The critical angle of a reflected ray at the mantle-core boundary applying Snell’s law is

sin ic

v0ffiffiffi

2p ¼

1

4v0) ic ¼ 20:7

The take-off angle at the surface i0 for this ray is found by again applying Snell’s law:

R sin i0

v0¼

R

2sin 20:7ð Þ

v0ffiffiffi

2p ) i0 ¼ 7:2

(c) To find the critical distance we use the expression

c ¼ 2

ð

r0

rp

p

r


2 p2p ð159:2Þ

where

rp ¼R

2

r0 ¼ R

p ¼r0 sin i0

v0¼

R

8v0

¼r

v¼

r

v0Rr

12

¼r3=2

v0ffiffiffi

Rp

Substituting the values of the problem in (159.2):

c ¼1

4

ðR

R=2

dr

r

R3=2


r3

R3

64

r

i0ic

Δc

R/2 RO

Δ/2Δ/2

SF ∗

Fig. 159a

298 Seismology

This integral is of the type

ð

dx

xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

xn anp ¼

2

nffiffiffiffiffi

anp cos1

ffiffiffiffiffi

an

xn

r

so we can write the solution

c ¼8

6cos1 1

8

cos1 1ffiffiffi

8p

¼ 18

(d) The travel time of the rays in the mantle is given by

t ¼201þ a

sin 1þ að Þ

2

¼R

v0

4

3sin

3

4for 0 92:4

By substitution of values of D we find the travel time curve given in Fig. 159b.

160. A spherical medium of radius R has a constant speed of propagation v0 from the

surface down to R/2, and from R/2 to the centre a core of variable speed of propaga-

tion v ¼ v0R

r

1=2

.

(a) What value should i0 have for the waves to penetrate into the core?

(b) Calculate the epicentral distance reached by a wave leaving a focus at the surface

at angle i0.

(a) The velocity at the top of the core (r ¼ R/2) is

1.2

1.0

0.8

0.6

t (R

/n)

0.4

0.2

0

0 20 40

Δ ()

Δ mΔ c

60 80

Fig. 159b


v1 ¼ v0R

R

2

0

B

@

1

C

A

12

¼ v0ffiffiffi

2p

Applying Snell’s law we find the critical angle ic for incident rays at the core (Fig. 160):

R sin ic

v0¼

R

2ffiffiffi

2p

v0) ic ¼ 45

The take-off angle i0, for a focus at the surface corresponding to the critical angle, using

Snell’s law, is

R sin i0

v0¼

R

2sin ic

v0) i0 ¼ 20:7

The rays that penetrate into the core must leave the focus with take-off angles less than

20.7.

(b) For a ray with take-off angle i0 which penetrates the core the epicentral distance is the

sum of that corresponding to the part that has travelled through the mantle, D1, plus the

part that has travelled through the core, D2:

¼ 21 þ2

Since in the core the velocity varies with depth with the law given in the problem, the

epicentral distance is given by

2 ¼2

1þ acos1 p

1

where a is the exponent of the velocity distribution

v ¼ v0ra ) a ¼

1

2

i0

Δ1 Δ2

Δ1

n0

∗F

R/2 R

P

S

l

O

Fig. 160

300 Seismology

and p is the ray parameter, which can be obtained using Snell’s law:

p ¼R sin i0

v0¼

R

2sin i1ffiffiffi

2p

v0

and

¼r

v) 1 ¼

R

2ffiffiffi

2p

v0

Then, we find

2 ¼2

1þ 12

cos1

R sin i0

v0R

2ffiffiffi

2p

v0

0

B

B

@

1

C

C

A

¼4

3cos1 2

ffiffiffi

2p

sin i0

The distance D1 can be determined using the sine and cosine laws for the triangle FOP

(Fig. 160):

R

2sin i0

¼l

sin1

) 1 ¼ sin1 2l sin i0

R

l2 þ R2 2Rl cos i0 ¼R2

4) l ¼ R cos i0

1

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4R2cos2i0 3R2p

and we find the expression in terms of i0:

¼ 2sin1 2 cos i0 1

2


4cos2i0 3p

sin i0

þ4

3cos1 2

ffiffiffi

2p

sin i0

161. Consider a spherical Earth of radius 6000 km and surface velocity of 6 km s1, with

a velocity distribution of the type v rð Þ ¼ a=ffiffi

rp

. At the distance reached by a wave

emerging at a take-off angle of 45 from a focus on the surface, calculate the interval

between the arrival times of the direct P- and reflectedPP-waves (the PP-wave is one that

is reflected at the surface at themidpoint between the focus and the point of observation).

The velocity distribution is of the type v ¼ v0R

r

a

where

r ¼ R ) v ¼ v0 ¼affiffiffi

Rp ) a ¼ v0

ffiffiffi

Rp

v ¼ v0R

r

12

and the ray parameter p for a ray with take-off angle of 45 is

p ¼R sin i0

v0¼

60001ffiffiffi

2p

6¼ 707 s


For this type of velocity distribution the relation between the ray parameter and the

epicentral distance is

p ¼ 0 cos1þ a

2

ð161:1Þ

In this problem the value of 0 is

¼r

v) 0 ¼

R

v0¼ 1000 s

Substituting the values in (161.1) we obtain the distance for the ray with take-off angle

of 45:

707 ¼ 1000 cos3

4

) ¼ 60

The PP-wave which arrives at D¼ 60 travels twice the distance which a P-wave does for a

distance of 30. For this type of velocity distribution the travel time for a distance D is

given by

t ¼201þ a

sin1þ að Þ

2

In our case for the P- and PP-waves at distance 60 we substitute the values of the problem

and find

tPð60Þ ¼ 943 s

tPPð60Þ ¼ 2tPð30

Þ ¼ 1021 s

The time interval between the PP- and P-waves 60 is

tPP tP ¼ 1021 943 ¼ 78s

162. In an elastic spherical medium of radius r0, the velocity increases with depth

according to v ¼ arb. If v0 ¼ 6 km s1, r0 ¼ 6000 km, and, at a point at distance

D ¼ 90, the slope of the travel-time curve is 500 s, determine:

(a) The value of b.

(b) The value of rp and of vp of the wave reaching an epicentral distance of 90.

(a) For this type of velocity distribution the travel time in terms of the epicentral

distance is given by

t ¼201þ b

sin 1þ bð Þ

2

ð162:1Þ

As we know the velocity at the surface,

¼r

v) 0 ¼

r0

v0¼ 1000 s

302 Seismology

The ray parameter p is known, because it is equal to the slope of the travel-time curve

which for D ¼ 90 is given as 500 s. Using the relation between p and D for this type of

velocity distribution,

p ¼dt

d¼ 0 cos 1þ bð Þ

2

500 ¼ 1000 cos 1þ bð Þp

4

h i

) cos 1þ bð Þp

4

h i

¼1

2) 1þ bð Þ

p

4¼

p

3) b ¼

1

3

(b) At the point of greatest penetration rp for D ¼ 90, we have the relation

p ¼rp sin 90

vp¼

rp

vp) rp ¼ pvp

and also

vp ¼ v0r0

rp

13

From these two equations we obtain rp and vp:

rp ¼ v0pr130r

13

p ) rp ¼ v340r

140p

34 ¼ 3564 km

vp ¼rp

p¼

3564

500¼ 7:1 km s1

163. Consider a spherical Earth of radius R, the northern hemisphere with a constant

speed of propagation v0, and the southern hemisphere with a speed of propagation of

v ¼ v0R

r

12

.

(a) Calculate the travel time of seismic waves for a focus on the equator and stations

on the same meridian.

(b) In which hemisphere does the wave at a distance of 60 arrive first?

(a) In the northern hemisphere the velocity is constant and the rays have straight paths

and their travel time is (Fig. 163)

tN ¼

2R

v0sin

20 < < 90 ð163:1Þ

In the southern hemisphere the velocity increases with depth and the rays have curved

paths. Their travel time is given by

tS ¼

201þ b

sin 1þ bð Þ

2ð163:2Þ

where

0 ¼R

v0and v ¼ v0

R

r

b

) b ¼1

2

Substituting in (163.2) we obtain for the travel time in the southern hemisphere


tS ¼

4

3

R

v0sin

3

40 < < 90 ð163:3Þ

(b) The travel times for waves in the northern and southern hemisphere are given by

Equations (163.1) and (163.3). By substitution of D ¼ 60 we obtain

tN ¼

R

v0

tS ¼

R

v0

2ffiffiffi

2p

6¼ 0:47

R

v0

The waves arrive first in the southern hemisphere.

164. Consider a spherical medium of radius R consisting of two concentric regions

(mantle and core), the core of radius R/2. The speeds of propagation are v ¼ ar1=2

for the mantle and v ¼ aR1=6r1=3 for the core. The surface velocity is v0. For a wave

leaving a focus with angle of incidence 14.5, calculate the angular distance D at

which it reaches the surface.

We calculate a by applying the boundary conditions

r ¼ R ) v ¼ v0 ) v0 ¼ aR12 ) a ¼ v0R

12

Mantle : v ¼ v0R

r

12

Core : v ¼ v0R

r

13

We determine the ray parameter corresponding to the ray with take-off angle i0 ¼ 14.5,

using Snell’s law (Fig. 164a):

p ¼r sin i

v¼

R sin 14:5

v0¼

R

4v0

S2

R

n0

S1

Δ

F ∗

Fig. 163

304 Seismology

At the bottom of the mantle at depth R/2 the velocity is

v1 ¼ v0R

R

2

0

B

@

1

C

A

12

¼ v0212

The incident angle i of this ray on the mantle–core boundary, applying Snell’s law, is given by

R

2sin i

v0212

¼R sin i0

v0) i ¼ 45

On the top of the core the velocity is

v2 ¼ v0R

R

2

0

B

@

1

C

A

13

¼ v0213

which is less than at the bottom of the mantle and there is no critical angle. Applying

Snell’s law again we obtain the angle i2 of the refracted ray in the core:

sin i

v1¼

sin i2

v2) i2 ¼ 39

The take-off angle of the last ray which travels only in the mantle is given by

R sin i0

v0¼

R

2

v0212

) i0 ¼ 20:7

In our case the angle 14.5 is less and the ray penetrates into the core.

The epicentral distance is the sum of the distances corresponding to the paths in the

mantle and in the core:

¼ 21 þ2

Δ1 Δ2Δ1

i0i2

i0

R/2 O R

F∗

S

i

Fig. 164a


The distance corresponding to the path in the core is given by

2 ¼2

1þ bcos1 p

0

where

p ¼R

4v0

0 ¼

R

2

v0213

and where p is the ray parameter and ¼ r/v. Substituting the values we obtain

2 ¼ 76:4

To calculate D1 we suppose that there is no core and a ray with take-off angle i0 ¼ 14.5

would arrive at distance D3 which is related with D1 by (Fig. 164b)

21 ¼ 3 4

The distances D3 and D4 can be determined using the equation

¼2

1þ bcos1 p

0

where for D3

0 Rð Þ ¼R

v0

R

2

¼

R

2

v0R

R

2

0

B

@

1

C

A

12

¼R

v02

32

Δ1 Δ4

Δ3

Δ1

F∗

O R

S

Fig. 164b

306 Seismology

and we obtain

3 ¼2

1þ 12

cos1

R

4v0R

v

0

B

B

@

1

C

C

A

¼ 100:6

and by similar substitutions for D4

4 ¼2

1þ1

2

cos1

R

4v0

R

v023

2

0

B

B

B

B

@

1

C

C

C

C

A

¼ 60:0

Then, 2D1 ¼ 100.6 – 60 ¼ 40.6 and the epicentral distance is

¼ 40:6þ 76:4 ¼ 117:0

Surface waves

165. A Rayleigh wave in a semi-infinite medium has a 20 s period. If the P-wave

velocity is 6 km s1 and Poisson’s ratio is 0.25, calculate the depth at which u1 ¼ 0,

and at which depth the particle movement becomes prograde.

Since Poisson’s ratio is 0.25, we find the relation between P- and S-waves:

s ¼1

4¼

l

2 lþ mð Þ) l ¼ m ) a ¼


lþ 2m

r

s

¼

ffiffiffiffiffiffi

3m

r

s

¼ffiffiffi

3p

b

a ¼ 6 ) b ¼6ffiffiffi

3p ¼ 3:4 km s1

For a half-space the velocity of Rayleigh waves is

cR ¼ 0:919b ¼ 3:2 km s1

The displacement u1 is given by

u1 ¼@’

@x1

@c

@x3

where the potentials are given by

’ ¼ A exp ikrx3 þ ik x1 cRtð Þð Þ

c ¼ B exp iksx3 þ ik x1 cRtð Þð Þ

r ¼ i 1c2Ra2

1=2

¼ 0:85i

s ¼ i 1c2R

b2

1=2

¼ 0:39i

307 Surface waves

Then,

u1 ¼ 0 ) ikA exp 0:85kx3ð Þ 0:39ikB exp 0:39kx3ð Þ ¼ 0 ð165:1Þ

so

k ¼2p

l¼

2p

TcRffi 0:1 km1

We can write B in terms of A using the boundary condition of zero stress at the free surface:

t31 ¼ 0jx3¼0 ) 2rA 1 s2

B ¼ 0

so

B ¼ 1:47iA

Substituting in (165.1) we obtain the value of x3:

exp 0:85kx3ð Þ ¼ 0:39 1:47 exp 0:39kx3ð Þ

x3 ¼ 12 km

At 12 km depth u1 is null and for greater values of depth the particle motion is prograde

while for lesser values of depth it is retrograde.

166. Given a layer of thickness H and shear modulus m¼ 0 on top of a half-space or

semi-infinite medium in which l¼ 0, study (without expanding the determinant)

whether there exist surface waves that propagate in the x1-direction. Are they

dispersive waves?

In the liquid layer (m ¼ 0) the P- and S-velocities are

b0 ¼ 0 ¼

ffiffiffiffi

m0

r

s

) a0 ¼


l0 þ 2m0

r

s

¼

ffiffiffiffi

l0

r

s

and in the solid half-space

l ¼ 0 ) a ¼

ffiffiffiffiffiffi

2m

r

s

¼ffiffiffi

2p

b

The relation between the stress and strain is

tij ¼ lydij þ 2meij

where eij ¼12ui; j þ uj;i

.

In the layer: m0 ¼ 0 )t0ii ¼ l0 e011 þ e022 þ e033ð Þt0ij ¼ 0

In the half-space: l ¼ 0 ) tij ¼ 2 meijIf there are surface waves propagating in the x1-direction, their displacements in terms of

the potentials are given by (Fig. 166)

u1 ¼ ’;1 c;3

u2 ¼ u2

u3 ¼ ’;3 þ c;1

308 Seismology

The boundary conditions at the free surface are null normal stresses:

x3 ¼ H )t033 ¼ 0

t031 ¼ 0

and at the boundary between the liquid layer and the solid half-space continuity of the

normal component of the displacement and stress and zero tangential stresses,

x3 ¼ 0 )

u3 ¼ u03

t33 ¼ t033

t32 ¼ t032 ¼ 0

t031 ¼ 0

8

>

>

>

<

>

>

>

:

In the liquid layer there is only the P-wave potential ’. Taking (x1, x3) as the incidence

plane

’0 ¼ A exp ikr0x3 þ ik x1 ctð Þð Þ þ B exp ikr0x3 þ ik x1 ctð Þð Þ

r0 ¼


c2

a02 1

r

where c is the velocity of wave propagation in the x1-direction

In the half-space

c ¼ C exp iksx3 þ ik x1 ctð Þð Þ

u2 ¼ E exp iksx3 þ ik x1 ctð Þð Þ

’ ¼ D exp ikrx3 þ ik x1 ctð Þð Þ

r ¼


c2

a2 1

r

s ¼


c2

b2 1

s

In the layer we have only guided P-waves and r0 is real, while in the half-space for surface

waves, r and s must be imaginary. Then a > b > c > a0 must be satisfied.

x3

x1H

0

μ = 0

λ = 0

Fig. 166

309 Surface waves

From the boundary conditions we obtain the following equations:

x3 ¼ H

t033 ¼ 0 ) A 1þ r02

eikr0H þ B 1þ r

02

eikr0H ¼ 0

x3 ¼ 0

t31 ¼ 0 ) 2Dr C þ Cs2 ¼ 0

u03 ¼ u3 ) Ar0 Br0 ¼ Dr þ C

t033 ¼ t33 ) l0 1þ r02

Aþ Bð Þ ¼ 2m Dr2 þ Cs

For a solution of the system the determinant must be zero:

eikr0H eikr0H 0 0

0 0 s2 1 2r

r0 r

0 1 r

l0 1þ r02ð Þ l0 1þ r

02ð Þ 2ms 2mr2

¼ 0

Expanding the determinant and working r0, r, and s in terms of the variable c, we obtain c

(k), the velocity of waves in the x1-direction. They have the form of guided waves in the

liquid layer and surface waves in the half-space. Since the velocity c(k) is a function of the

wavenumber the waves are dispersive.

167. There is a liquid layer of density r and speed of propagation a on top of a rigid

medium (half-space). Derive the dispersion equation of waves in the layer by bound-

ary conditions and by constructive interference in terms of v. Plot the dispersion

curve for the different modes.

Given that the layer is liquid the only potential is ’:

’ ¼ A exp ikrx3 þ B exp ikrx3ð Þð Þ exp ik x1 ctð Þ ð167:1Þ

The boundary conditions at the free surface are zero normal stress and at the boundary

between the liquid layer and the rigid half-space zero normal component of displacement

(Fig. 167a):

x3 ¼ H ) t33 ¼ 0

x3 ¼ 0 ) u3 ¼ 0

where

t33 ¼ ly ¼ r2’ ¼ ro2’ ¼ 0

u3 ¼ ’;3

r ¼


c2

a2 1

r

By substitution of (167.1) we obtain

AeikrH þ BeikrH ¼ 0

A B ¼ 0 ) A ¼ B

310 Seismology

Then, 2A cos krH ¼ 0. For waves propagating in the layer, r must be real and c > a. The

solution is given by

krH ¼ nþ1

2

p; n ¼ 0; 1; 2; . . . ð167:2Þ

The solution can also be found by the method of constructive interference. The condition

of constructive interference implies that waves coinciding at a given wavefront (AB) are in

phase, that is, the distance along the ray must be an integer multiple of the wavelength,

taking into account possible phase shifts (Fig. 167b). In our case on the free surface,

x3 ¼ H, there is a phase shift of p (l/2) and we write the condition as (Fig. 167b)

A Pþ PQþ QBla

2¼ nla

or

2p

laðAPþ PQþ QBÞ p ¼ 2pn

x1

x3

H

0

liquid layer

rigid half-space

Fig. 167a

x3

x1

H

0P

A

B

O

Fig. 167b

311 Surface waves

Substituting

APþ PQþ QB ¼ 2H cos i

we obtain

2p

la2H cos i p ¼ 2p nþ 1ð Þ ð167:3Þ

According to Snell’s law,

sin i ¼a

c) cos i ¼


1a2

c2

r

¼a

c


c2

a2 1

r

¼a

cr

anda

cka ¼ k.

Substituting in (167.3), we obtain the same solution obtained in (167.2):

kaa

cHr ¼ nþ

1

2

p

This expression can also be written as

krH ¼ nþ1

2

p)oH

c

ffiffiffiffiffiffiffiffiffiffiffiffi

c2

a21

r

¼ nþ1

2

p) c¼1

a2 nþ

1

2

2p2

H2o2

" #12

ð167:4Þ

The fundamental mode (FM) corresponds to n ¼ 0, and n 1 to the higher modes (HM).

In the FM and the higher modes, the frequency oc corresponding to the zero in the

denominator in (167.4) is called the cut-off frequency, as there are no values of c for o <

oc. For a mode of order n the cut-off frequency is given by

oc ¼

p nþ1

2

a

H

The dispersion curve is shown in Fig. 167c.

c

FM 1 HM 2 HM

ω

α

πα

2H3πα

2H5πα

2H

Fig. 167c

312 Seismology

168. Consider an elastic layer of coefficients l and m, thickness H, and density r on a

rigid semi-infinite medium. Derive the dispersion equation c(v) for P-SVand SH-type

channelled waves for the fundamental mode (FM) and the first higher mode (1HM).

Plot the dispersion curve for the SH motion.

For a SH-wave which propagates in the x1-direction its displacement is given by

u2 ¼ E exp iksx3ð Þ þ F exp iksx3ð Þð Þ exp ik x1 ctð Þ

The P- and SV-waves are given by their scalar potentials ’ and c:

’ ¼ A exp ikrx3ð Þ þ B exp ikrx3ð Þð Þ exp ik x1 ctð Þ

c ¼ C exp iksx3ð Þ þ D exp iksx3ð Þð Þ exp ik x1 ctð Þ

where r and s were defined in Problem 166.

The boundary conditions for SH-waves are null stress at the free surface and null

displacement at the boundary with the rigid medium (Fig 168a):

x3 ¼ H ) t32 ¼ 0 ¼ m@u2@x3

x3 ¼ 0 ) u2 ¼ 0

By substitution we have

EeiksH FeiksH ¼ 0

E þ F ¼ 0 ) F ¼ E

EðeiksH þ eiksH Þ ¼ 0 ) cosðksHÞ ¼ 0 ) ksh ¼ nþ1

2

p

Substituting s and putting k ¼ o/c:

Ho

c


c2

b2 1

s

¼ nþ1

2

p ) c ¼1

b2 nþ

1

2

2p2

H2o2

!12

ð168:1Þ

This equation give us, for the SH component, the frequency dependence of the velocity c(o).

The boundary conditions for P and SV are similarly

α, β, µ

rigid medium

0

H x1

x3

Fig. 168a

313 Surface waves

x3 ¼ H ) t31 ¼ 0; t33 ¼ 0

x3 ¼ 0 ) u1 ¼ 0; u3 ¼ 0

where

t33 ¼ lðe11 þ e33Þ þ 2me33

t31 ¼ 2me31

u1 ¼ ’;1 c;3

u3 ¼ ’;3 þ c;1

Substituting the expression for the potentials we obtain

ðlþ 2mÞðr2AeikrH þ r2BeikrH Þ þ lðAeikrH þ BeiksH Þ þ 2msðCeiksH DeiksH Þ ¼ 0

2rðAeikrH BeikrH Þ þ ð1 s2ÞðCeiksH þ DeiksH Þ ¼ 0

ðAþ BÞ sðC DÞ ¼ 0

rðA BÞ þ C þ D ¼ 0

For a solution we put the determinant of the system of equations equal to zero:

1 1 s s

r r 1 1

2reikrH 2reikrH ð1 s2ÞeiksH ð1 s2ÞeiksH

½lþ r2ðlþ2mÞeikrH ½lþ r

2ðlþ2mÞeikrH 2mseiksH 2mseiksH

¼ 0 ð168:2Þ

Expanding the determinant and putting it equal to zero, we obtain the dependence with

frequency of the velocity c(o) which gives us the dispersion curve.

For the wave with SH component the dispersion curve is given in Fig. 168b:

c ¼1

b2 nþ

1

2

2p2

H2o2

!12

c

FM 1 HM 2 HM

ω

β

πβ

2H3πβ

2H5πβ

2H

Fig. 168b

314 Seismology

For n ¼ 0 the curves correspond to the fundamental mode and for 1 n to the higher

modes. For all modes, including the fundamental mode, there is a cut-off frequency oc ¼

(nþ1)pb/2H, with n ¼ 0 for the fundamental mode and n 1, for higher-order modes.

169. For a liquid layer of thickness H with a rigid medium above and below, derive

the dispersion equation c(v) of the fundamental and higher modes. For the FM, at

what height above the layer is the motion circular?

Given that the medium is a liquid, motion is represented only by the scalar potential f:

’ ¼ A exp ikrx3 þ B exp ikrx3ð Þð Þ exp ik x1 ctð Þ ð169:1Þ

where r ¼


c2

a2 1

r

.

The boundary condition at the two boundaries between the liquid and rigid solid is that

the normal component of the displacement is null (Fig. 169):

x3 ¼ 0 ) u3 ¼ 0

x3 ¼ H ) u3 ¼ 0

Substituting u3 ¼ ’;3 we have

A B ¼ 0

AeikrH BeikrH ¼ 0ð169:2Þ

which leads to the equation

A eikrH eikrH

¼ 0 ð169:3Þ

Consider first that r is real, that is, c >a. Then, from (169.1)

2iA sin krH ¼ 0 ) krH ¼ np; n ¼ 0; 1; 2; :::

with n ¼ 0, fundamental mode (FM), and n 1 for higher modes.

For the FM, n ¼ 0 and r ¼ 0, and then

Hk


c2

a2 1

r

¼ 0 ) c ¼ a

The displacements from (169.1) and (169.2) are

x3

x1

H

0

rigid medium

liquid layer

rigid medium

Fig. 169

315 Surface waves

u3 ¼@’

@x3¼ Aikr exp ikrx3 expikrx3ð Þ exp ikðx1 ctÞ

u1 ¼@’

@x1¼ Aik exp ikrx3 þ expikrx3ð Þ exp ikðx1 ctÞ

For the FM r ¼ 0, then u3 ¼ 0 and this is a P-wave, with only a u1 component, which

propagates in the x1-direction. For all HM the displacements have both components

For the first higher mode (1HM), n ¼ 1:

Ho

c


c2

a2 1

r

¼ p

c2 ¼1

1

a2

p2

o2H2

If

1

a2

p2

o2H2¼ 0 then o ¼

ap

H) c ! 1

The cut-off frequency is oc > ap/H. For each higher mode there is a cut-off frequency onc

> nap/H.

If r ¼ ir_is imaginary, then c < a and this implies that 2 sinhðkr

_HÞ ¼ 0 which is

impossible (1< sinhx <1).

The particle motion inside the layer is circular when

u1 ¼ u3 ð169:4Þ

so

u1 ¼ u3 ) ð1 rÞ exp ikrx3 þ ð1þ rÞ expikrx3 ¼ 0

Taking only the amplitudes of the displacements,

ð1 rÞ cos krx3 þ ð1þ rÞ cos krx3 ¼ 0 ) cos krx3 ¼ 0

krx3 ¼ nþ1

2

p ) x3 ¼ nþ1

2

p

kr

For the FM, we have seen that u3 ¼ 0, so there is no circular motion. For the 1HM, the

height in the layer at which the motion is circular is

x3 ¼3p

2kr

The height inside the layer at which the motion is circular depends in each higher mode of

the frequency.

170. In the hypothetical case of a layer of thickness H and speed of propagation b0 on

top of a semi-infinite medium of speed of propagation b, the phase shifts at the free

surface and the contact plane are

p

4and tan1


c2

b0 1

s


1c2

b2

s

2

6

6

6

6

6

4

3

7

7

7

7

7

5

:

316 Seismology

Determine:

(a) The dispersion equation using constructive interference.

(b) The cut-off frequency of the fundamental mode and first higher mode.

(c) Plot the dispersion curve of the FM and 1HM using units of c/b and H/l for

b ¼ 2b0.

(a) The distance from A to B along the ray path is (Fig. 170a)

AB ¼ 2H cos i

According to Snell’s law

sin i ¼b0

c

cos i ¼b0

c


c2

b02 1

s

and the wavenumber k associated with velocity c is

kb0 sin i ¼ kb0b0

c¼ k

As explained in Problem 167, the condition for constructive interference is that the distance

AB along the ray path be an integer multiple of the wavelength, taking into account the

phase shift at the free surface and the boundary surface between the two media:

2kB0H cos iþp

4 tan1 s0

s

¼ 2pn ð170:1Þ

where

s0 ¼


c2

b02 1

s

s ¼


1c2

b2

s

By substitution in (170.1) we obtain the dispersion equation

HA

B

β

β

i

Fig. 170a

317 Surface waves

tan 2kHs0 2n1

4

p

¼


c2

b02 1

1c2

b2

v

u

u

u

u

u

u

t

ð170:2Þ

(b) For the fundamental mode (FM), n ¼ 0:

tan 2kH


c2

b02 1

s

þp

4

" #

¼


c2

b02 1

s


1c2

b2

s > 0 where b0 < c < b

Given that tan B > 0 )p

4 B

p

2so that b’< c < b, then

If k ¼ 0 then

tanp

4

¼ 1 ¼


c2

b02 1

s


1c2

b2

s ) c ¼

ffiffiffi

2p

bb0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b2 þ b02p

so

tan 2kHs0 þp

4

¼ tan Bð Þ ¼ 1 ) B ¼p

2

If c ¼ b then

2kH


c2

b02 1

s

þp

4¼

p

2) k ¼

p

8Hs0) s0 ¼


b2

b02 1

s

so

tan 2kHs0 þp

4

¼ 0

If c ¼ b0 then

s0 ¼


c2

b02 1

s

¼ 0 ) tan kHs0 þp

4

¼ 0 ) k ¼ 1 ) kHs0 ¼ p

4

But kHs0 must be positive, so this last solution is not possible.

For the first higher mode (1HM), n ¼ 1:

tan 2kHs0 7

4p

¼s0

s

The tangent function is positive for the range [0, p/2]:

0 2kHs0 7p

4

p

2

For c ¼ b0

318 Seismology

tan 2kHs0 7p

4

¼ 0 ) kHs0 ¼7p

8) k ¼ 1

For c ¼ b

tan 2kHs0 7p

4

¼ 1 ) 2kHs0 7p

4¼

p

2) k ¼

9p

8Hs0

(c) Taking b’¼ b/2, the dispersion equation for the FM is

tan 2kH


4c2

b2 1

s

þp

4

" #

¼


4c2

b2 1

s


1c2

b2

s )c

b

2

¼2

5 3 sin 8pH

l


4c

b

2

1

s

" #

But we have seen that

2kHs0 þp

4

p

2)

H

l

1

16s0

c ¼ b ) s0 ¼ffiffiffi

3p

H

l

1

16ffiffiffi

3p

Giving values to H/l we can calculate the corresponding values of c/b by means of a

numerical method. For example, we obtain,

For the first higher mode (1HM), we arrive at the same equation, given that tan(a þ p/4)

¼ tan(a þ 7p/4), but vary the intervals of H/l and kHs0:

7p

8 kHs0

9p

8

7

16s0

H

l

9

16s0

ForH

l¼

7

16s0at one limit we have the value s0¼ 0 which corresponds to c ¼ b0 and c/b ¼

0.5.

ForH

l¼

9

16s0we have s ¼ 0 and c ¼ b, and consequently

H

l¼

9

16ffiffiffi

3p .

The dispersion curves for the fundamental mode and the first higher mode are shown in

Fig.170b.

H/l c/b

0.0 0.63

0.01 0.68

0.02 0.78

0.036 1

319 Surface waves

171. In a structure with a layer of thickness H and speed of propagation b0 on top of a

medium of speed of propagation b, the phase shifts at the free surface and the contact

plane are:

d1 ¼ p

2and d2 ¼ sin1


1c2

b2

s


c2

b02 1

s

8

>

>

>

>

>

<

>

>

>

>

>

:

9

>

>

>

>

>

=

>

>

>

>

>

;

Calculate:

(a) The dispersion equation of the Love wave.

(b) For the fundamental and first higher mode, and the minimum and maximum

frequencies as functions of H, b, and b0.

(c) For this mode, given b0 ¼ b/2, the maximum and minimum frequencies, and the

corresponding values of c.

(a) As in Problem 170, the condition of constructive interference with the phase shifts

given in this problem results in (Fig. 171)

4p

lH cos i

p

2 sin1 s

s0

¼ 2pn ð171:1Þ

where

c

1.0

0.5

1 HMFM

β

Hλ0.10 0.20 0.301

316 316

9

Fig. 170b

320 Seismology

s ¼


1c2

b2

s

s0 ¼


c2

b02 1

s

According to Snell’s law

sin i ¼b0

c

cos i ¼b0

c


c2

b02 1

s

Substituting in (171.1) we find the dispersion equation for Love waves:

2Hks0 p

2 2pn ¼ sin1 s

s0

) cos 2Hk


c2

b02 1

s !

¼


1c2

b2

s


c2

b02 1

s

(b) The fundamental mode (FM) corresponds to the values

0 2Hks0 p

2) 1


1c2

b2

s


c2

b02 1

s 0

The velocity at the limit of lowest frequencies, k ¼ 0, is given by

k ¼ 0 )s

s0¼


1c2

b2

s


c2

b02 1

s ¼ 1 ) c2 ¼2b02b2

b02 þ b2

k ¼p

4Hs0)

s

s0¼ 0 ) s ¼ 0 ) c ¼ b ) s0 ¼

1

b0


b2 b02q

H

A

B

β

β

i

Fig. 171

321 Surface waves

For the first higher mode (1HM)

3p

2 2Hks0 2p ) 0


1c2

b2

s


c2

b02 1

s 1

In the lowest limit

k1 ¼3p

4Hs0)

s

s0¼ 0 ) s ¼ 0 ) c ¼ b ) s0 ¼

1

b0


b2 b02q

The minimum value of the frequency is

k1 ¼3pb0

4Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b2 b02p

In the highest frequency limit

k2 ¼p

Hs0)

s

s0¼ 1 ) s ¼ s0 ) c2 ¼

2b02b2

b2 þ b02) s0 ¼


b2 b02

b2 þ b02

s

The maximum value of the frequency is given by

k2 ¼pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b2 þ b02p

Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

b2 b02p

(c) If we put b0 ¼ b=2 in the 1MS,

k1 ¼3p

b

2

4H


b2 b2

4

r ¼pffiffiffi

3p

H4¼

2p

l)

H

l¼

ffiffiffi

3p

8; c ¼ b

k2 ¼p


b2 þb2

4

r

4H


b2 b2

4

r ¼pffiffiffi

5p

Hffiffiffi

3p ¼

2p

l)

H

l¼

ffiffiffi

5p

2ffiffiffi

3p ; c ¼ b

ffiffiffi

2

5

r

172. Consider a layer of thickness H and parameters m0 and r0 on top of a semi-

infinite medium of parameters m ¼ 4m0 and r ¼ r0. Ifc

b¼ a and

H

l¼ b:

(a) Write the dispersion equation of the Love wave in terms of a and b.

(b) Calculate the values of b corresponding to a ¼1

2;3

4and 1 for the fundamental

mode and first higher mode.

(c) For which values of b is the node of the amplitude of the first higher mode at a

depth of H/2?

(a) The dispersion equation for Love waves is

322 Seismology

tan kH


c2

b02 1

s( )

¼mffiffiffiffiffiffiffiffiffiffiffiffi

1 c2

b2

q

m0ffiffiffiffiffiffiffiffiffiffiffiffiffi

c2

b02 1

q ð172:1Þ

In this problem,

m ¼ 4m0

r ¼ r0

then,

b ¼

ffiffiffi

m

r

r

! b0 ¼

ffiffiffiffiffiffi

m

4r

r

¼b

2

We now introduce a and b:

a ¼c

b)

c

b0¼ 2a

b ¼H

l) k ¼

2p

l¼

2pb

H


tan 2pbffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4a2 1p

¼ 4


1 a2p


4a2 1p

(b) For the FM

a ¼ 1; b ¼ 0 and c ¼ b

a ¼1

2; b ! 1 and c ¼ b0

a ¼3

4) tan 2pb


49

16 1

r

" #

¼ 4


19

16

r


49

16 1

r ) b ¼ 0:17

For the 1HM,

a ¼1

2; b ! 1

a ¼ 1 ) tan 2pbffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4a2 1p

¼ 0 ) 2pbffiffiffi

3p

¼ p ) b ¼1

2ffiffiffi

3p ¼ 0:29

a ¼3

4) tan 2pb


43

4

2

1

s0

@

1

A ¼ 4

ffiffiffiffiffi

7

20

r

) pbffiffiffi

5p

¼ tan1 4

ffiffiffiffiffi

7

20

r

!

þ p ) b ¼ 0:61

(c) Inside the layer the amplitude of the displacements of the Love wave are given by

u02 ¼ 2A0 cos ks0H 1x3

H

h i

cos k s0H þ x1 ctð Þ

323 Surface waves

The nodes are the points where the amplitude is zero. For the 1HM the node is located at

the value of x3 which satisfies the relation

ks0H 1x3

H

¼3p

2

If we want a node located at x3 ¼ H/2 then

k ¼3p

s0H) b ¼

3

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4a2 1p

If we substitute the values of a, ½, 1, and ¾ we obtain for b infinity, 0.86, and 1.34. The

infinite value of b corresponds to l¼ 0.

Focal parameters

173. Consider three stations with coordinates:

St1 ¼ 36.2 N, 4.8 E; St2 ¼37.0 N, 2.4 E; St3 ¼38.6 N, 4.0 E

An earthquake is recorded at the three stations with the following respective S-P

time intervals:

tSP1 ¼ 26:7 s

tSP2 ¼ 27:0 s

tSP3 ¼ 22:5 s

Given that the focus is at the surface, the P-wave velocity is constant and equal to 6

km s1, and Poisson’s ratio is 1/3, calculate the coordinates of the epicentre.

Given that Poisson’s ratio is 1/3,

s ¼1

3¼

l

2 lþ mð Þ) l ¼ 2m ) a ¼


lþ 2m

r

s

¼ 2b

where a and b are the velocities of the P- and S-waves, respectively.

From the S-P time intervals, calling x the epicentral distance (Fig. 173):

tSP ¼

x

b

x

a¼

x

2b) x ¼ 2tSPb

Given that a ¼ 6 km s1 then b ¼ 3 km s1.

The epicentral distances corresponding to each station in kilometres and degrees are

x1 ¼ 26:7 6 ¼ 160 km ¼160

111:11¼ 1:44

x2 ¼ 27 6 ¼ 162 km ¼162

111:11¼ 1:46

x3 ¼ 22:5 6 ¼ 135 km ¼135

111:11¼ 1:22

324 Seismology

If xe and ye are the geographical coordinates of the epicentre,

36:2 xeð Þ2 þ 4:8 yeð Þ2 ¼ 1:442

37:0 xeð Þ2 þ 2:4 yeð Þ2 ¼ 1:462

38:6 xeð Þ2 þ 4:0 yeð Þ2 ¼ 1:222

Solving the system we find two possible solutions:

xe; yeð Þ ¼37:40 N; 3:83 Eð Þ37:63 N; 4:71 Eð Þ

174. A focus is at a depth h, in a medium with constant P- and S-wave speeds of

propagation a and b. Calculate an expression for the depth h in terms of the time

interval between the P and sP phases.

If L is the distance along the ray, the travel-time of the P-wave from the focus to the station

is given by (Fig. 174)

tP ¼

L

a

36°N

35°N

37°N

38°N

39°N

40°N

1°E

St3

St2

St1

E1

E2

3°E2°E 4°E 5°E 6°E

Fig. 173

325 Focal parameters

An sP-wave leaves the focus upward as an S-wave is reflected at the Earth’s surface and

converted into a P-wave that travels to the station. Its travel time is (Fig. 174)

tsP ¼

FS

bþSF

0

aþL

a

If we consider the length along the ray L from F and F0 to the station to be the same for both

waves, then the sP-P time interval is (Fig. 174)

tsP t

P ¼ t0 ¼

FS

bþSF

0

að174:1Þ

At the focus at depth h, the take–off angle of the direct P-wave is ih and the take-off angle

of the sP-wave is jh. We can write

FS ¼h

cos jh

cosðjh þ ihÞ ¼SF

0

FS) SF

0¼

h

cos jhðsin jh sin ih cos jh cos ihÞ

sin ih ¼a sin jh

b


t0 ¼ h

cos jh

bþcos ih

a

175. The displacement vector l of an earthquake is (0, 1, 0) and the vector normal to

the plane of displacement n is (0, 1, 0). Determine:

(a) The components of the P-wave displacement at the point of azimuth 45 and angle

of incidence 30.

S

h

AF

ih

ih

jh F

A

Fig. 174

326 Seismology

(b) The kind of mechanism it represents.

(a) The elastic displacements due to a dislocation Du in the direction of li on a plane of

normal ni (Fig. 175) are given by

uPk ¼ u lnk lkdij þ m linj þ nilj

Gki;j

where Green’s function corresponding to the P-wave in the far field for an infinite medium

is given by

Gki ¼1

4pra2rgigkd t

r

a

and its derivative is

Gki;j ¼1

4pra3rgigkgj

_d t r

a

where gi are the direction cosines of the line from the focus to the observation point.

The amplitude of the displacement is then

uPk ¼u

4pra3rlnslsdij þ m linj þ ljni

gigkgj ð175:1Þ

In our problem the direction cosines of the ray of the waves arriving at the point are

g1 ¼ sin i cos az ¼

ffiffiffi

2p

4

g2 ¼ sin i sin az ¼

ffiffiffi

2p

4

g3 ¼ cos i ¼

ffiffiffi

3p

2

X3

X1

X2

l i= n i e

az

ir

P

Fig. 175


The orientation of the source is given by li ¼ (0, 1, 0) and ni ¼ (0, 1, 0), and substituting in

(175.1) gives

uP1 ¼ A lþ 2mg22

g1

uP2 ¼ A lþ 2mg22

g2

uP3 ¼ A lþ 2mg22

g3

A ¼u

4pa3rr

Substituting the direction cosines of the ray we obtain,

uP1 ¼ A lþ1

4m

ffiffiffi

2p

4

uP2 ¼ A lþ1

4m

ffiffiffi

2p

4

uP3 ¼ A lþ1

4m

ffiffiffi

3p

2

(b) The mechanism corresponds to a fault on the (x1, x3) plane which opens in the

direction of its normal, x2, under tensional forces in that direction.

176. The focal mechanism of an earthquake can be represented by a double-couple

(DC) model. The orientation of the fault plane is azimuth 30, dip 90, and slip angle

0. Calculate:

(a) What kind of fault it is. Sketch it, indicating the direction of motion.

(b) The auxiliary plane.

(c) The azimuth of the stress axis.

(d) A wave incident at a station has azimuth 180 and angle of incidence at the focus

of 90. Calculate the amplitude of the components of the P-wave at that station.

(a) Given that the dip of the plane is 90, the fault plane is vertical, and since the slip

angle is 0, the motion is horizontal. Thus, it corresponds to a right lateral strike-

slip fault (Fig. 176).

Δu

ϕA

δA

n

l

N

Fig. 176

328 Seismology

(b) From the azimuth (’ ¼ 30), dip (d ¼ 90), and slip (l ¼ 0) we can calculate the

unit vectors li and ni which give the direction of the fracture and of the normal to

the fault plane:

n1 ¼ sin d sin’ ¼ 1

2¼ sinYn cosFn

n2 ¼ sin d cos’ ¼

ffiffiffi

3p

2¼ sinYn cosFn

n3 ¼ cos d ¼ 0 ¼ cosYn

l1 ¼ cos l cos’þ cos d sin l sin’ ¼

ffiffiffi

3p

2¼ sinYl cosFl

l2 ¼ cos l sin’ cos d sin l cos’ ¼1

2¼ sinYl cosFl

l3 ¼ sin l sin d ¼ 0 ¼ cosYl

’A ¼ Fn þ 90 ) Fn ¼ 120

dA ¼ Yn ¼ 90

lA ¼ sin1 cosYl

sinYn

¼ 0 ) Yi ¼ 90

whereY and F are the spherical coordinates for the vectors n and l (r ¼ 1, unitary vectors)

and for the auxiliary plane,

’B ¼ 30þ 90 ¼ 120

dB ¼ 90

lB ¼ sin1 cosYn

sinYl

¼ 0

(c) The T-axis is on the same plane as ni and li at 45 between them, and the direction

cosines are

T1T2T3

0

@

1

A ¼n1 n2 n3l1 l2 l3Z1 Z2 Z3

0

@

1

A

1ffiffi

2p

1ffiffi

2p

0

0

@

1

A ð176:1Þ

where Zi is the axis normal to ni and li, that is, Zi ¼ ni li which results in Zi ¼ (0, 0, 1).

Substituting in (176.1) we obtain Ti ¼

ffiffiffi

3p

1

2ffiffiffi

2p ;

ffiffiffi

3p

þ 1

2ffiffiffi

2p ; 0

. The azimuth of the T-axis is

FT ¼ tan1 T2

T1

¼ tan1

ffiffiffi

3p

þ 1ffiffiffi

3p

1

¼ 75

(d) The direction cosines of the direction from the focus to the station are

g1 ¼ sin ih cos az ¼ 1

g2 ¼ sin ih sin az ¼ 0

g3 ¼ cos ih ¼ 0


The amplitude of the displacements for a shear fracture or double-couple (DC) source in an

infinite medium is given by

R ih; azð Þ ) uPj ¼ A nilk þ nk lið Þgigkgj

uP1 ¼ 0:84A

uP2 ¼ 0

uP3 ¼ 0

A ¼M0

4pa3rr

177. An earthquake is caused by a shear fracture. The vectors n and l (normal and

direction of travel) are, in terms of the angles F and Y,

n ¼ ð57:13; 66:44Þ

l ¼ ð305:96; 50:09Þ

Calculate the orientation of the fault plane, the auxiliary plane, and the tension (T )

and pressure (P) stress axes.

The orientation of the fault plane and auxiliary plane in terms of the angles’, d, and l (azimuth,

dip, and slip) are found directly from the given values (Fig. 177), using the following relations:

’A ¼ Fn þ 90 ¼ 147:13

dA ¼ Yn ¼ 66:64

lA ¼ sin1 cosYl

sinYn

¼ 135:68

’B ¼ Fl þ 90 ¼ 35:96

dB ¼ Yl ¼ 50:09

lB ¼ sin1 cosYn

sinYl

¼ 31:12

Z

X

P

N

n

Y

T

l

fn

qp

qn

Fig. 177

330 Seismology

To calculate the T and P axes, we calculate first the direction cosines of the l and n axes

from the given angles:

x1 ¼ sinY cosF

x2 ¼ sinY sinF

x3 ¼ cosY

)

n1 ¼ 0:50; l1 ¼ 0:45

n2 ¼ 0:77; l2 ¼ 0:62

n3 ¼ 0:40; l3 ¼ 0:64

Given that the T and P axes are on the same plane as n and l at 45 between them, we can

write, as in Problem 176,

T1T2T3

0

@

1

A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3

0

@

1

A

1ffiffiffi

2p

1ffiffiffi

2p

0

0

B

B

@

1

C

C

A

ð177:1Þ

where the Z-axis is normal to n and l and is found by Z ¼ n l. Its direction cosines

are (Z1, Z2, Z3) ¼ (0.72, 0.14, 0.66). By substitution of ni, li, and Zi in (177.1) we

obtain

T1 ¼ 0:67 ¼ sinYT cosFT

T2 ¼ 0:11 ¼ sinYT sinFT

T3 ¼ 0:74 ¼ cosYT

9

>

=

>

;

) TðYT ¼ 42:27;FT ¼ 9:32Þ

In the same way for the axis P

P1

P2

P3

0

@

1

A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3

0

@

1

A

1ffiffiffi

2p

1ffiffiffi

2p

0

0

B

B

@

1

C

C

A

! PðYP ¼ 80:02;FP ¼ 268:03Þ

178. The seismic moment tensor relative to the geographical axes (X1, X2, X3) (north,

east, nadir) is

Mij ¼2 1 1

1 0 1

1 1 2

0

@

1

A

Find the values of the principal stresses, and the orientation of the tension and

pressure stress axes.

First we calculate the eigenvalues ofMij. SinceMij is a symmetric tensor its eigenvalues are

real and the corresponding eigenvectors mutually orthogonal (Problem 111):

2 s 1 1

1 0 s 1

1 1 2 s

¼ 0 ) s ¼3

2

1

0

@

1

A

Ordered by magnitude, the three eigenvalues are

s1 ¼ 3; s2 ¼ 2; s3 ¼ 1


The diagonalized matrix is

Mij ¼3 0 0

0 2 0

0 0 1

0

@

1

A

In this formMij is referred to the coordinate system formed by the eigenvectors or principal

axes. Given that the sum of the elements of the principal diagonal is not zero, the source

has net volume changes. Then, we can separate Mij into two parts: an isotropic part with

volume changes (ISO) and a deviatoric part without volume changes. The second part can

be separated into two parts: a part corresponding to a double-couple or shear fracture (DC)

and a part corresponding to a non-double-couple source usually expressed as a compen-

sated linear vector dipole (CLVD). Thus the moment tensor is separated into three parts,

namely

M ¼ M ISO þMDC þMCLVD

The isotropic part is given by

M ISO ¼ s0 ¼1

3s1 þ s2 þ s3ð Þ ¼

4

3

The deviatoric part (DCþCLVD) is given by

M 0ij ¼ Mij dijso

and in our case

M 0ij ¼

s1 0 0

0 s2 0

0 0 s3

0

@

1

A ¼

5

30 0

02

30

0 0 7

3

0

B

B

B

B

@

1

C

C

C

C

A

Now we separate this part into two parts, DC and CLVD:

M 0ij ¼ MDC þMCLVD

M 0ij ¼

1

2s1 s3ð Þ 0 0

0 0 0

0 0 1

2s1 s3ð Þ

0

B

B

B

@

1

C

C

C

A

þ

s2

20 0

0 s2 0

0 0 s2

2

0

B

B

B

@

1

C

C

C

A

M 0ij ¼

2 0 0

0 0 0

0 0 2

0

B

@

1

C

Aþ

1

30 0

02

30

0 0 1

3

0

B

B

B

B

B

@

1

C

C

C

C

C

A

332 Seismology

The orientation of the P and T axes is calculated from the double-couple part MDC:

MDCij ¼ 2

1 0 0

0 0 0

0 0 1

0

@

1

A

We can find the n and l axes:

MDCij ¼ M0ðlinj ljniÞ

~n :1ffiffiffi

2p ; 0;

1ffiffiffi

2p

) nðYn ¼ 45;Fn ¼ 0Þ

~l :1ffiffiffi

2p ; 0;

1ffiffiffi

2p

) lðYl ¼ 45;Fl ¼ 0Þ

In the same way as in Problems 176 and 177, we determine T and P from n and l, finding

first Z ¼ n l:

T1T2T3

0

@

1

A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3

0

@

1

A

1ffiffiffi

2p

1ffiffiffi

2p

0

0

B

B

B

@

1

C

C

C

A

T1 ¼ 1 ¼ sinYT cosFT

T2 ¼ 0 ¼ sinYT sinFT

T3 ¼ 0 ¼ cosYT

! TðYT ¼ 90;FT ¼ 0Þ

For the P-axis,

P1

P2

P3

0

@

1

A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3

0

@

1

A

1ffiffiffi

2p

1ffiffiffi

2p

0

0

B

B

B

@

1

C

C

C

A

PðYP ¼ 0;FP ¼ 0Þ

179. The magnitude Ms of an earthquake as calculated for surface waves of period

20 s is 6.13.

(a) Calculate the amplitude of these waves at a station 3000 km away. If the instru-

ment’s amplification is 1500, what will be the amplitude of the seismogram’s

waves and the seismic energy?

(b) IfMs ¼Mw, and the area of the fault is 12 km 8 km with m¼ 4.4104 MPa, find

the fault slip Du.

(a) The surface wave magnitude Ms is given by

Ms ¼ logA

Tþ 1:66 logþ 3:3


where A is the ground motion amplitude, T is the period of the wave, and D is epicentral

distance in degrees. Knowing the magnitude and period of the waves we can calculate the

wave amplitude:

6:13 ¼ logA

Tþ 1:66 log

3000

111:11þ 3:3

) logA

T¼ 0:454 ) A ¼ 2:84 20 1500 ¼ 8:5 cm

We have reduced the ground motion to the amplitude of the seismogram using the

amplification of the instrument (1500).

Knowing the magnitude we can calculate the seismic energy:

logEs ¼ 11:8þ 1:5Ms ) Es ¼ 1021ergs ¼ 1014 J

(b) Mw ¼ 6.13 ¼ 2/3 log M0 – 6.1; M0 ¼ 2.191018 Nm

If M0 ¼ mDu S, with S ¼ 12 8 ¼ 9.6107 m2, then

u ¼M0

mS¼

2:19 1018 Nm

4:4 1010 Nm2 9:6 107 m2¼ 0:52m

334 Seismology

5 Heat flow and geochronology

Heat flow

180. Assume that the temperature variation within the Earth is caused by gravita-

tional forces under adiabatic conditions. Knowing that the coefficient of thermal

expansion at constant pressure is aP ¼ 2105 K1 and the specific heat at constant

pressure is cP ¼ 1.3 kJ kg1 K1, determine an expression for the gradient of the

temperature with depth. Compare it with the value observed at the surface which is

30 K km1, knowing that, at 200 km depth, T ¼ 1600 K.

Under adiabatic conditions, there is no heat flow and the variation of pressure with depth z

is a function of gravity g and density r:

dP ¼ rgdz ð180:1Þ

Using the first and second laws of thermodynamics

dU ¼ dQ PdV

dQ ¼ TdS

where Q is the heat, U is the internal energy, S is the entropy, T is the absolute temperature,

P is the pressure, and V is the volume.

If we use the specific variables (variables divided by mass) we can write

du ¼ dq pdv

dq ¼ Tds

Considering that

dS ¼@S

@T

P

dT þ@S

@P

T

dP

we can write

dq ¼ Tds ¼ T@s

@T

P

dT þ T@s

@P

T

dP

According to the definition of specific heat at constant pressure, cP and the increase in heat

dq are given by

335

cP ¼ T@S

@T

P

dq ¼ cPdT þ T@s

@P

T

dP

ð180:2Þ

The Gibbs function G is defined as

G ¼ u Tsþ pv

Taking the differential in this expression, and taking into account the second law of

thermodynamics

du ¼ Tds pdv

we obtain

dG ¼ vdp sdT

If we compare this expression to the differential of the Gibbs function

dG ¼@G

@p

T

dpþ@G

@T

p

dT

we differentiate again and using the Schwartz theorem we obtain

@v

@T

P

¼ @s

@P

T

But the coefficient of thermal expansion is defined as

ap ¼1

v

@v

@T

p

In consequence we can write Equation (180.2) as

dq ¼ cpdT Tvapdp

In our case the process is adiabatic and in consequence using this equation and Equation

(180.2) we obtain

dT

dz¼

Tapg

cpð180:3Þ

where we have taken into account that the variables are by unit mass so rv ¼ 1, and

substituting the values we obtain:

dT

dz¼

1600 2 105 10KK1 ms2

1:3 103 J kg1 K1¼ 0:25K km1

We observe that this result is two orders of magnitude lower than the observed values. This

shows that observations correspond to heat flow at the lithosphere and are not satisfied by

purely adiabatic conditions.


181. If the Earth’s temperature gradient is 1 C/30 m, calculate the heat loss per

second due to conduction from its core. Compare this with the average power

received from the Sun.

Data:

Thermal conductivity K ¼ 4 Wm1 C1.

Earth’s radius R ¼ 6370 km.

Solar constant: 1.35 kWm2.

The heat flow is given by

_q ¼ KAdT

dr

where A is the area of the Earth’s surface

A ¼ 4pR2 ¼ 5:10 1014 m2

and

dT

dr¼

1C

30m


_q ¼ 6:80 1013 J s1 ¼ 1:63 1013 cal s1

the average power received on the Earth’s surface by the radiation from the Sun is

1:35 103W

m25:10 1014 m2 ¼ 6:89 1017 J s1 ¼ 1:65 1017 cal s1

In consequence, from these values we can see that on the Earth’s surface the average solar

power is much larger than that due to the heat flow from inside the Earth.

182. At the Earth’s surface, the heat flow is 60 mWm2 and T0 ¼ 0 C. If all the heat is

generated by the crust at whose base the thermal conductivity is K¼ 4Wm1 C1, and

T is 1000 C, determine the thickness of the crust and the heat production per unit volume

If we assume that all the heat is generated at the crust and there is no heat flow from the

mantle at the crust base, then we can write

_qjz¼H ¼ 0

Using the temperature equation for a flat Earth for one-dimensional heat-flow and the

stationary case we can write

T ¼ e

2Kz2 þ

_q0Kzþ T0

e ¼_q0H

For z ¼ H:

TH ¼_q0H

2Kþ T0 ) H ¼

ðTH T0Þ2K

_q0

337 Heat flow

Substituting the values given in the problem we obtain

H ¼ 133:3 km

and the heat production by unit of volume is

e ¼_q0

H¼ 4:5 104 mWm3

183. Consider the crust to be H ¼ 30 km thick and the heat flow at the surface to be

60 mWm2.

(a) If all the heat is generated in the crust, what is the value of the heat generated per

unit volume? (Take K ¼ 3 W m1 K1)

(b) If all the heat is generated in the mantle with a distribution Aez=H mWm3, what

is the value of A? What is the temperature at 100 km depth?

(a) We solve the heat equation for a stationary one-dimensional case, assuming a flat

Earth with one-dimensional flow in the z-direction (vertical) positive downward.

In this case the solution of the heat equation is given by

T ¼ e

2Kz2 þ

_q0Kzþ T0

where e is the heat generated by unit volume and time, K is the thermal conductivity, q0 and

T0 are the heat flow and temperature at the surface of the Earth, respectively

If all the heat is generated at the crust we can write (Fig. 183)

z ¼ H ! _q0 ¼ 0

The heat generated by unit volume is

_qjz¼H ¼ 0 ¼ KdT

dz

z¼H

¼ K e2z

2Kþ

_q0K

z¼H ¼ 0

e ¼_q0H

¼60 103

30 103¼ 2 106 Wm3

(b) If all heat is generated in the mantle with distribution

e ¼ Aez=H

the heat equation is

d2T

dz2¼

e

K¼

A

Kez=H

and the solution is given by

T ¼ A

KH2ez=H þ Czþ D ð183:1Þ

where C and D are constants of integration. They may be estimated from the boundary

conditions at the surface


z ¼ 0 ! T ¼ T0 ¼ A

KH2 þ D ! D ¼ T0 þ H2 A

K

z ¼ 0 ! _q ¼ _q0 ¼ 0 ! C ¼ A

KH


T ¼ A

KH2ez=H

AH

Kzþ T0 þ H2 A

K¼

AH

KHez=H zþ H

þ T0 ð183:2Þ

If the heat has its origin in the mantle, the flow at the base of the crust is

_qjz¼H ¼ _qo ¼ KdT

dz

z¼H

) A ¼_q0

H e1 1ð Þ

The temperature at z = 100 km may be estimated from (183.2) assuming that T0 = 0:

TðzÞ ¼_q0

K e1 1ð ÞHez=H zþ H

Tðz ¼ 100Þ ¼60 103

3 e1 1ð Þ30 103e100=30 100 103 þ 30 103

¼ 2249K

184. Calculate the thickness of the continental lithosphere if its boundary coincides

with the 1350 C geothermal, knowing that the surface temperature is 15 C,

the heat flow at the surface is _q0 ¼ 46mWm2, the lithospheric mantle’s thermal

conductivity is K = 3.35 Wm1 K1, and the radiogenic heat production is

P = 0.01 103

mWm3

The geothermal equation at depth z is given by:

Tz ¼ T0 þ_q0Kðz z0Þ

P0

2Kðz z0Þ

2 ð184:1Þ

T0

TH

H

q0

•

qH

•

Fig. 183

339 Heat flow

whereK is the thermal conductivity, T0 is the temperature at the surface of the Earth (in K), _q0

is the heat flow at the surface, and P0 is the radiogenic heat production at the Earth’s surface.

At the Earth’s surface z0 = 0, and Equation (184.1) becomes:

Tz ¼ T0 þ_q0Kz

P0

2Kz2

P0

2Kz2

_q0Kzþ ðTz T0Þ ¼ 0

So

z ¼

_q0K

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

_q0K

2

4P0

2KðTz T0Þ

s

2P0

2K

¼_q0


_q20 2P0KðTz T0Þp

P0

Substituting the data given in the problem:

K ¼ 3.35 W m1 K1

_q0 ¼ 46 103 Wm2

P0 ¼ 0.01106 mW m3

Tz ¼ 1623 K

T0 ¼ 288 K

we obtain two solutions, but only z = 98.27 km is realistic (the second one gives a depth

larger than the Earth’s radius).

185. On the surface of an Earth of radius 6000 km, the temperature is 300 K, the heat

flow is 6.7mWm2, and the thermal conductivity is 3Wm1K1. If the heat produc-

tion per unit volume inside the Earth is homogeneously distributed, what is the

temperature at the centre of the planet?

We begin solving the problem of heat conduction inside a sphere with constant internal

heat generation per unit volume e and conductivity K. The differential equation for heat

conduction with spherical symmetry is

1

r2

d

drr2 dT

dr

¼ e

Kð185:1Þ

Integrating twice and using the boundary conditions:

Surface: r = R ! T = T0Center: r = 0 ! T finite

we obtain the solution

T ¼ T0 þe

6KR2 r

2

ð185:2Þ


_q ¼ KdT

dr) _q0ðr ¼ RÞ ¼ 6:7mWm2

¼ Kd

drT0 þ

e

6KR2 r

2

¼e r

3


Solving for the heat production e

e ¼ 3:35 109 Wm3

From Equation (185.1) the temperature at the Earth’s centre is

Tr¼0 ¼ 300þ3:35 109

6 3 62 1012 ¼ 7000K

186. Consider a spherical Earth of radius R = 6000 km and a core at R/2, in which

there is a uniform and stationary distribution of ε heat sources per unit volume. The

heat flow at the surface is 5 mW m2, the thermal conductivity is 3 W m1 K1, and

the temperature at the core–mantle boundary is 4000 K. Calculate the temperature at

the Earth’s surface.

We consider the problem as one of heat conduction inside a sphere with conductivity K and

constant heat generation per unit volume e inside the core (radius R/2). We begin with

Equation (185.1)

1

r2

d

drr2 dT

dr

¼ e

K

The boundary conditions at the core–mantle boundary and its centre are

r ¼ R=2 ! T ¼ TN

r ¼ 0 ! T finite

where TN is the temperature at the core–mantle boundary

Integrating twice we obtain

T rð Þ ¼ TN þe

6K

R

2

2

r2

!


_q0 r ¼ Rð Þ ¼ KdT

dr

r¼R

¼eR

3) e ¼

_q0 3

R¼ 2:5 109 Wm3

Then, the temperature at the Earth’s surface is

T r ¼ Rð Þ ¼ 4000þ2:5 109

6 3

6000 103

2

2

60002 106

!

¼ 250K

187. Consider the Earth of radius R0 = 6000 km formed by a spherical crust

with its base at 500 km and constant thermal conductivity K. If the temperature

at the base of the crust is T1 and at the surface of the Earth is T0 = 0 C,

determine:

(a) An expression for the heat flow through the crust.

341 Heat flow

(b) An expression for the temperature distribution within the Earth.

(c) The temperature at the base of the crust if, at that depth, _q ¼ 5:5 1013 W and

K = 4 W m1 C1.

(a) We assume a spherical Earth where the temperature varies only in the radial

direction. Then we can solve the problem as one of spherical unidirectional flow.

For the stationary case, when the conductivity and heat generation are constant, the

Fourier law may be written as

_q ¼ KAdT

drð187:1Þ

where A ¼ 4pr2 is the area in the normal direction to the heat flow. Integrating this

equation:

_q

4p

ðR0

r1

dr

r2¼ K

ðT0

T1

dT

where the conditions at the Earth’s surface are, r = R0 !T = T0 and at the base of the crust,

r = r1 !T = T1.

Solving Equation (187.1), assuming that K is constant, we obtain

_q ¼ 4pK1r1 1

R0

ðT0 T1Þ ¼4pKr1R0

R0 r1

ðT0 T1Þ ð187:2Þ

(b) The temperature distribution inside of the Earth may be obtained by integration of

Equation (187.1):

_q

4p

ð

r

r1

dr

r2¼ K

ðT

T1

dT

T rð Þ ¼ T1 þR0ðr r1Þ

ðR0 r1ÞrðT0 T1Þ

(c) The radial distance to the base of the crust is r2 = 5500 km, so, using expression

(187.2), we obtain

T1 ¼ T0 þ_qðR0 r1Þ

4pKr1R0

¼ 0þ5:5 1013 500 103

4p 4 5500 103 6000 103¼ 16579 C

This result implies a constant increase of temperature from the Earth’s surface

of 1 ºC each 33.2 m similar to the observed gradient in the real Earth of 1 ºC

per 30 m

188. Assume that the heat flow inside the Earth is due to solar heating of the

Earth’s surface. Calculate the maximum penetration of this flow in the diurnal and

annual cycles. Take as typical values for the Earth K = 3 Wm1 K1, r = 5.5 g cm3,

Cv = 1 kJ kg1

K1.


We assume the heat propagation inside the Earth coming from the solar radiation on its

surface as unidirectional flow thermal diffusion (inside the Earth) with periodic variation of

surface temperature. The diffusivity equation is

k@2T

@z2¼

@T

@tð188:1Þ

where the thermal diffusivity is k ¼K

rCv

, K is the thermal conductivity, r is the density,

and Cv is the specific heat at constant volume. We solve Equation (188.1) using the

separation of variables

T z; tð Þ ¼ Z zð Þy tð Þ

Substituting in (188.1) we obtain the solution

Z zð Þ ¼ Aeaz þ Beaz

y tð Þ ¼ Ceka2t

where a is the constant of separation of variables. Using the boundary condition of periodic

flow and the temperature T0 at the Earth’s surface,

z ¼ 0 ) T ¼ T0eiot

and as Z(z) exists only inside the Earth, B = 0. At the surface, z = 0, so

ACeka2t ¼ T0e

iot

Then

AC ¼ T0

ka2 ¼ io

But putting, i ¼ 121þ ið Þ2, we have

a ¼ 1þ ið Þ

ffiffiffiffiffiffi

o

2k

r

Then, we can write the temperature variation inside of the Earth as:

T z; tð Þ ¼ T0 exp

ffiffiffiffiffiffi

o

2k

r

zþ i


o

2kz

r

þ ot

This equation corresponds to a periodic wave, with angular frequency o propagating for

positive z values (to the Earth’s interior) and with the amplitude decreasing with depth. The

propagation velocity and wavelength are given by

v ¼

ffiffiffiffiffiffi

2k

o

r

l ¼ 2pv ¼ p

ffiffiffiffiffiffi

8k

o

r

343 Heat flow

The values of l corresponding to the daily and annual cycles give their maximum penetration:

Daily cycle:

o ¼2p

24 60 60¼ 7:2 105 s1

k ¼K

rCv

¼3Wm1 K1

5:5 103 K gm3 103 J Kg1 K1¼ 0:5 106 m2 s1

Then

l ¼ p


8 0:5 106

7:2 105

r

¼ 0:74m

Annual cycle:

o ¼2p

365 24 60 60¼ 2 107 s1

l ¼ 14m

The penetration of the solar radiation as periodic heat conduction inside the Earth is very

shallow due to the poor heat conduction.

189. Consider a lithospheric plate of 100 km thickness created from asthenospheric

material originating from a ridge in the asthenosphere with constant temperature Ta and

in which no heat is generated. Given that k = 106m² s1, that the temperature at the base

of the lithosphere is 1100 C, and in the asthenosphere is 1300 C, calculate the age of the

plate, and, if the velocity of drift is 2 cm yr1, how far it has moved away from the ridge.

The heat propagation inside the plate is given by:

K

rcv

@2T

@x2þ@2T

@z2

¼ u@T

@xð189:1Þ

where T is the temperature, r is the density, cv is the specific heat at constant volume, and u

is the horizontal velocity of the plate in the x-direction (normal to the plate front). If we

assume that the horizontal conduction of heat is insignificant in comparison with the

horizontal advection and vertical conduction, we can write, using the following change

of variable t = x/u,

K

rcv

@2T

@z2¼

@T

@t

Integrating this equation and using the boundary conditions at the ridge and surface:

x ¼ 0 ! T ¼ Ta

z ¼ 0 ! T ¼ 0

we obtain for the temperature distribution

T z; tð Þ ¼ Taerfz

2ffiffiffiffiffi

ktp


where

k ¼K

rcv

erf xð Þ ¼2ffiffiffi

pp

ðx

0

ey2dy

Substituting the data of the problem,

1100 ¼ 1300erfL

2ffiffiffiffiffi

ktp

) erfL

2ffiffiffiffiffi

ktp

¼ 0:846

Values of the error function, erf(x), may be obtained from tables. If erf(x) = 0.846, x = 1.008, then

L

2ffiffiffiffiffi

ktp

¼ 1:008 ) t ¼L2

4k 1:0082¼

1010

4 106 1:0082

¼ 2:5 1015 s ¼ 79Myr

If the displacement velocity is 2 cm yr1, the plate has moved 1580 km.

190. If the concentrations of 235U and 235Th in granite are 4 ppm and 17 ppm,

respectively, and the respective values of heat production are 5.7 104 W kg1

and 2.7 105 W kg1, respectively, calculate the heat flow at the base of a granite

column of 1 m² cross-section and 30 km height (the density of granite is 2.65 g cm3).

We estimate first the mass of the granite column:

M ¼ rV ¼ 2:65 103 1 30 103 ¼ 7:95 107 kg

If the concentration of 235U in the granite is 4 ppm, its quantity in the column is

235U :4 7:95 107

106¼ 318 kg

Then the heat flow due to the 235U is

_q ¼ 5:7 104 318 ¼ 181:26mWm2

For 235Th, the heat flow is

235Th : 7:95 107 17 106 ¼ 1351:5 kg

_q ¼ 1351:5 2:7 105 ¼ 36:49mWm2

Geochronology

191. The mass of 1 millicurie of 214Pb is 3 1014 kg. Calculate the value of the decay

constant of 214Pb.

The mass of the sample is

345 Geochronology

3 1014 kg ¼ 3 1017 g ¼N

N0

M

where N is the number of atoms in the sample, N0 is Avogadro’s number ¼ 6.02 1023,

and M is atomic number ¼ 214. Solving for N we obtain

N ¼3 1017 6:02 1023

214¼ 8:44 1038 atoms

The correspondence of a curie is

1 curie ¼dN

dt¼ lN ¼ 3:7 1010 disintegrations s1

where l is the decay constant and t is the time. Then

l ¼3:7 107

8:44 1038¼ 0:44 1031 s1

192. The isotope 40K decays by emission of b particle with a half-life of 1.83 109

years. How many b decays occur per second in one gram of pure 40K?

The average life t of a radioactive material is a function of the decay constant l:

t ¼1

l! l ¼

1

1:83 109¼ 0:55 109 yr1 ¼ 1:73 102 s1

The number of atoms N contained in 1 g of 40K may be estimated from Avogadro’s number

N0 and the atomic number M:

1 ¼N

N0

M ! N ¼1 6:02 1023

40¼ 0:15 1023 atoms

The rate of disintegration is given by

dN

dt¼ lN ¼ 0:26 1021 Bq

193. The half-life of 238U is 4468 106 yr and of 235U is 704 106 yr. The ratio235U/238U in a sample is 0.007257. Given that the ratio was 0.4 at the time of

formation, calculate the sample’s age.

From the half-life T1/2 we can obtain the decay constant l:

T1=2 ¼0:693

l

l238 ¼0:693

4468 106¼ 1:5510 1010 yr1

l235 ¼0:693

704 106¼ 9:8434 1010 yr1

The number N of disintegrating atoms at time t is given by

N ¼ N0 elt ð193:1Þ


where N0 is the number of atoms at time t ¼ 0. Then

N235 ¼ N 2350 el235t

N238 ¼ N 2380 el238t

If we divide these equations:

N235

N238

¼N 2350

N 2380

eðl238l235Þt

Substituting the values given in the problem we obtain

4:0095 ¼ 8:2924 1010t ! t ¼ 4:8 109 yr

194. Date a meteorite which contains potassium knowing that its content of 40K is

1.19 1014 atoms g1, of 40Ar is 4.14 1017 atoms g1, and that the half-life of40K!40Ar is 1.19 109 years.

We obtain the decay constant of the 40K from its half-life T1/2:

l ¼0:693

T1=2

l ¼0:693

1:19 109¼ 0:58 109 yr1

We can solve the problem considering it as a case of radioactive ‘parent’ atom disinte-

grating to a ‘daughter’ stable atom. At time t ¼ 0 we have n0 ‘parent’ atoms at the sample,

and at time t there remain NR radioactive atoms in the sample and NE daughter atoms,

from the disintegration of the n0 parent atoms:

n0 ¼ NRþ NE

But

n0

nt¼

NRþ NE

NR¼ 1þ

NE

NR

From Equation (193.1) we obtain the age of the sample:

nt ¼ n0 elt !

n0

nt¼ elt ¼ 1þ

NR

NRð194:1Þ

so

t ¼1

lln 1þ

NE

NR

NE, the number of atoms of 40K in the sample, can be estimated from the number of atoms

contained in 1 g of potassium:

1 ¼N

6:023 102340 ! N ¼ 1:506 1022 atoms g1

347 Geochronology

so

NE ¼ 1:506 1022 1:19 1014 ¼ 1:792 1036 atoms g1

Then the age of the meteorite from (194.1) is

t ¼1

lln 1þ

1:792 1036

4:41 1017

¼ 7:4 1010 yr

195. At an archaeological site, human remains were found and assigned an age of

2000 years. One wants to confirm this with 14C dating whose half-life is 5730 yr. If the

proportion of 14C/12C in the remains is 6 1013, calculate their age. (Assume that at

the initial time the 14C/12C ratio was 1.2 1012.)

The decay constant l may be obtained from the half-life:

T1=2 ¼0:693

l! l ¼

0:693

5730¼ 1:2094 104 yr1

The activity in a sample is given by

R ¼ R0 elt

where R0 ¼ (14C/12C)t¼ 0 and R ¼ (14C/12C).

Then the age of the remains is

t ¼1

lln

R0

R

¼1

1:2094 104ln

1:2 1012

9 1013

¼ 2379 yr

196. Mass spectrometry of the different minerals in an igneous rock yielded the

following table of values for the concentrations of 87Sr originating from the radio-

active decay of 87Rb and of 87Rb, with the concentration expressed relative to the

concentrations of 86Sr of non-radioactive origin.

Express on a 87Sr/86Sr–87Rb/86Sr diagram the isochron corresponding to the forma-

tion of the rock, and calculate the age of the rock. Take l ¼ 1.42 1011 yr1.

For the decay of 87Rb

87Sr

now¼ 87Sr

0þ 87Rb

nowelt 1

ð196:1Þ

Mineral 87Sr /86Sr 87Rb /86Sr

A 0.709 0.125

B 0.715 0.418

C 0.732 1.216

D 0.755 2.000

E 0.756 2.115

F 0.762 2.247


where [87Sr]now and [87Rb]now are the number of atoms of each isotope at time t, [87Sr]0 is the

amount of original number of atoms of the isotope 87Sr [87Sr]now, and l is the decay constant.

Equation (196.1) may be written as

87Sr86Sr

now

¼87Sr86Sr

0

þ87Rb86Sr

now

elt 1

ð196:2Þ

This equation corresponds to a line (y ¼ a þ bx) with intercept87Sr86Sr

0

and slope (elt 1),

which is called an isochron.

If we plot the values given in the problem (Fig. 196) we can obtain the equation of the

line by least-squares fitting:

y ¼ 0:025xþ 0:705

The age of the sample can be obtained from the slope b ¼ 0.025:

el t 1 ¼ b

t ¼lnð1þ bÞ

l

Substituting the values of b and l:

t ¼ 1:72 109 yr:

0.0 0.5 1.0 1.5 2.0 2.50.70

0.72

0.74

0.76

0.80

87S

r8

6S

r

0.78

87Rb86Sr

Fig. 196

349 Geochronology

197. Magma with a material proportion of 87Sr/86Sr equal to 0.709 crystallizes

producing a series of rocks with different concentrations of 87Rb with respect to the

content of 86Sr:

(a) Calculate the proportions of 87Sr/86Sr and 87Rb/86Sr that these rocks will have

after 500 Myr. Take l ¼ 1.421011 yr1.

(b) Express in a 87Sr/86Sr–87Rb/86Sr diagram the isochrons corresponding to t ¼ 0

and t ¼ 500 Myr.

(a) Using the same method as in the previous problem, we can write

87Sr86Sr

¼87Sr86Sr

0

þ87Rb86Sr

elt 1

¼ 0:709þ87Rb86Sr

e1:4210115108 1

Sample 87Rb/86Sr

A 1.195

B 2.638

C 4.892

D 5.671

0 1 2 3 4 5 6 70.70

0.72

0.76

0.74

0.78

t = 500 Ma

t = 0

87Rb86Sr

87S

r86S

r

Fig. 197


The results for each rock are given in the following table

(b) For t ¼ 500 Myr, we carry out a least-squares fitting to obtain the isochron,

which results in

y = 0.007x þ 0.709

In Fig. 197 the isochrones corresponding to t ¼ 0 and t ¼ 500 Myr are shown.

Sample 87Sr/86Sr 87Rb/86Sr

A 0.717 1.187

B 0.728 2.619

C 0.744 4.857

D 0.749 5.631

351 Geochronology

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