Gravitational Potential of Earth

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The Earth’s Gravitational Field

The study of the gravity field of the Earth implies the calculation, or

measurement, of the value of g at all points on, or above the earth’s surface.

Such a study reveals information about:

• The shape of the earth

• The mass distribution inside the earth

• The dynamics of earth’s rotation

• The presence of mass anomalies below the surface for commercial

use eg. oil, minerals, archaeology. This aspect will be covered in

Applied Geophysics.

This is possible because the value of g at a point outside the earth depends

on all the small mass components making up the earth, and their position in

the earth, including surface topography, (it therefore also depends on the

ellipticity of the earth), and on the earth’s rotation.

Measurements of g are made by very sensitive instruments called

gravimeters, and g is usually measured in mgal (1 mgal = 10-6 g ). The gal

(10-3 g ) is named after Galileo. Measurements of g are normally given withrespect to the reference ellipsoid.



The gravitational field.

If M is the mass of the Earth, the gravitational field g at distance r is

defined as the force of attraction acting on unit mass placed at the point.

Thus

rg ˆ2r

GM −=

the negative sign indicating that force acts in the opposite direction to r .

The gravitational potential

By virtue of its position in the gravitational field g of the earth, any mass m

has gravitational potential energy. The gravitational potential V is the

potential energy in the field due to M per unit mass, and is defined as the

work done to move the unit mass from infinity (where the potential is zero

by definition) to the point (with vector r from the earth’s centre). Thus

∫ ∫ ∫ ∞ ∞∞

−==•−=•=r r r

r

GM dr

r GM d

r

GM d V

22

1ˆ rrrg



The potential is the integral of the gravity field. Vice-versa, the

gravity field is the derivative, or the gradient, of the potential

V −∇=g

which, in Cartesian coordinates means that

∂∂

∂∂

∂∂

−= z

V

y

V

x

V g ,,

Gravitational potential outside a non-spherical body

The gravitational potential at point P is calculated by summing up the

potential due to the individual mass elements dM (Remember that potential

is a scalar, so this simplifies the addition).

∫ =Vol

dV P V )(

Now [ ]dM

r

s

r

sr

G

rs sr

GdM

q

GdM dV

2/12

2/122

cos21cos2

−

+

−=−+

−=−=

θ θ

We can use the binomial theorem to expand the denominator into a power

series in (s/r). Check that we get



)termsorder higher ()1cos3(2

1cos1cos21

2

22/1

2

+−

+

+=

−

+

−

θ θ θ r

s

r

s

r

s

r

s

and hence, using 1sincos 22 −= θ θ and ∫ =V

dU P U )( , we get (neglecting

h.o.t.)

∫ ∫ ∫ ∫ +−−−= dM sr

GdM s

r

GdM s

r

GdM

r

G P V θ θ

22

3

2

32sin

2

3cos)( (1)

Each of these terms has a physical significance and you should have a

“feeling” of what they represent:

• The first term is –GM/r and is the potential of a point mass, or of a

perfect sphere (equivalent to a point mass M at its centre). It thus

represents the potential of the Earth if it were a perfect sphere. It isalso the dominant term as r increases, ie as we go further away from

Earth – in fact, from a very large distance the Earth does look like a

spherical point mass.

• The second term represents the sum of all the moments of the dM’s

about O, but since O is the centre of mass, then this term must be

equal to zero.

• The third term is related to the moments of inertia of the body about

the x, y and z axes. In fact it turns out to be equal to ½ (Ix+Iy+Iz) or ½(A+B+C)

• The fourth term represents the moment of inertia of the object about

OP, which we call I.

Thus the equation for V may be written as

)3(2

)(3

I C B Ar

G

r

GM P V −++−−=

This is known as MacCullagh’s formula. Note that if the

earth were a perfect sphere, then A=B=C=I (perfect

symmetry), and the second term would be zero. Thus the



second term is the contribution to the potential from the

ellipticity of the Earth.

It can also be shown that I = Asin2θ + Ccos2θ, and

therefore MacCullagh’s formula becomes

−−+−= 2

1

cos2

3

)(2

3 θ AC r

G

r

Gm

V (2)

(Derive this formula, remembering to put A=B)

Representation of the gravitational potential in

spherical harmonics

There is another way of representing the earth’s

gravitational potential field, and we shall show that the 2ways are identical.

At a given point outside the Earth, the gravitational

potential V obeys Laplace’s Equation (this applies also for

the electrostatic potential in a source-free region)

z

y

x

θ

PI



02

2

2

2

2

2

=∂∂+

∂∂+

∂∂

z

V

y

V

x

V

It is more natural to express this equation in spherical polar coordinates:

0sin

1sin

sin

112

2

222

2

2=

∂∂+

∂∂

∂∂+

∂∂

∂∂

φ θ θ θ

θ θ

V

r

V

r r

V r

r r

where r is the distance from the earth’s centre,

θ is the co-latitude

φ is the longitudeAssuming Earth to have rotational symmetry about its axis

of rotation, V is independent of longitude and the last term

drops. In this case the solution becomes:

+

−−−= .........)(cos)(cos)(cos),(

2

2

21100θ θ θ θ P

r

a J P

r

a J P J

r

GM r V

(3)

where a is the equatorial radius

M is the mass of the Earth

J 0 , J 1 , J 2, are empirically determined constants

P0, P1, P2,….. are the Legendre Polynomials –



P0 = 1

P1 = cosθ

P2 = 1)θ(3cos2

1 2 −

P3 = )3cos θθ(5cos21 3 − etc

The Legendre Polynomials

This turns out to have exactly the same form as equation (1)

The coefficients Jn determine the relative importance of the nthterm and are related to the distribution of mass within the earth. It

is important to understand the significance of these terms and to

have a feeling of their relative importance (you are not expected to

know such equations by heart!).

For example,

• If the earth were a perfect sphere, then we expect V= -GM/r,

so in this case J0 = 1, J1 = J2 = J3 =….=0

• At very large distances from the earth, we expect the earth tolook like a small uniform sphere, and its gravitational field to

be as such, therefore we require J0 to be always 1. (The rest

of the terms will go to zero since a<<<r).



• J1 must be equal to zero as it would represent a centre of

mass offset from the earth’s centre (same reason as for

previous expansion)

• J2 is the most important coefficient, since the term is related

to the oblate spheroid.• For our purpose we may neglect higher order terms. For

example, the term in J3 would represent deviations from the

oblate spheroid.

Thus the potential V may be written as

−+−=

2

1cos

2

3 2

23

2

θ J r

GMa

r

GM V

This is exactly similar to equation (2) derived from MacCullagh’s

formula, and by comparison we can write

22Ma

AC J

−=

The values of the constants Jn are found empirically by fitting the

equation to the observed potential field as it influences satelliteorbits. For very precise studies of the earth’s shape, higher order

values of Jn are found. The coefficient J2 is found to be J2 =

1082.6 x 10-6.

The Geopotential

To fully describe the gravitational field we need to take into

account the earth’s rotation. We do this by introducing therotational potential as follows:

θ ω 222 sin

2

1r V V

g −=



(Note that on differentiating the last term we get θ θ ω sin)sin(2r ,

which is the radial component of the centripetal acceleration.)

Converting to latitude λ in place of co-latitude θ, sinθ = cosλ, and

we get

(4)

This equation describes the contribution to the gravitational

potential of (i) the central mass

(ii) the oblateness of the earth (flattening), considered a

perfect ellipsoid

(iii) the rotation of the earth

The flattening f

The free surface of the ellipsoid is a surface of equipotential. Thisfact is used to give us the flattening factor. By equating V at the

poles (λ=900) with V at the equator (λ = 0), show that the

flattening f is given by

GM

ca

a

c

c

a

Ma

AC

a

ca f

22

2

2

2 2

1

2

ω +

+

−=

−=

Now use the approximation ca ≈ on the RHS and the definition

GM

a

g

am

e

322ω ω

== to show that

22

32

m J f +=

Note that the constant m is the ratio of the centripetal to the

gravitational acceleration at the equator.

λ ω λ 2222

3cos

2

1

2

1sin

2

3)( r AC

r

G

r

GM V −

−−+−=



The International Gravity Formula

The value of the gravitational acceleration g on the rotating

reference ellipsoid is found by differentiating the geopotential (4)V −∇=g

In spherical polar coordinates with only radial and latitude

dependence, this is

∂∂

∂∂−=

θ

V

r r

V 1,g

These two gradients give the radial and tangential components of

gravity.

The resultant gravity vector is the resultant of the two, and comes

out to:

−+

−−+= λ λ 2sin

8

5

8sin

14

17

2

51

2

2

2 mf f

mf f m g g e (5)

where ge = 9.780327 ms-2 and is the value of g at the equator. The

direction of this resultant vector defines the true vertical.

The constants m and f are exactly known and the above equation

may be written in the following form:

g = ge [1 + β1 sin2λ + β2 sin22λ]

where β1 = 5.3024 x 10

-3

and β2 = -5.87 x 10

-6

This is known as the International Gravity Formula (IGF) and

gives the gravity on the surface of the rotating reference

ellipsoid. Note that the formula may be expanded to several orders

of accuracy.



Note also that β2 is about 1000 times smaller than β1.

Clairaut’s Theorem

If we go back to equation (5), and drop the terms in mf (which are

about 300 times smaller than m and f) and β2, we can get a simple

expression for the gravity at the poles in terms of that at the

equator by putting λ = 90o:

g poles = ge[1 + 5m/2 –f]

g p – ge = ge[5m/2 – f]

This is known as Clairaut’s Theorem. It relates the variation of g from poles to equator to

i. rotation of the earth (via m)

ii. flattening of the earth (via f )

f m g

g g

e

e p −=−

2

5



The Geoid

The geoid is the surface of equipotential. The geoid of the

real earth is of course a highly complex surface, unlike that

of the reference ellipsoid.



Note the presence of a low geoid in the Indian Ocean and

of a high geoid in the Western Pacific. These are a result of

the internal mass distribution inside the earth.



An excess mass above or below the geoid will raise the geoid

while a mass deficiency will lower the geoid.



The differences in height between the geoid and the reference

ellipsoid are called geoid undulations. Note that the direction of

the vertical is the direction normal to the geoid.

The potential of the real earth, which would define the geoid,cannot be written in terms of r and θ only, but in the solution to

Laplace’s equation we would have to use the whole spherical

harmonic series, with many more coefficients than J1, J2, J3,…. .

The whole solution would be

( )∑ ∑ +

−=

∞=

=

=

=

n

n

nm

mnmnmnm

n

P mS mC r

a

r

GM V

0 0

)(cossincos θ φ φ

where φ is the longitude , Pnm(cosθ) are the associated Legendrepolynomials and the coefficients Cnm and Snm are calculated

experimentally to a very high degree of accuracy by satellite

geodesy.

The Principle of Isostasy

At a certain level below the earth’s surface, called the level of compensation, the pressures are all hydrostatic, ie the pressure due to

the weight of material above this level is everywhere the same. This

must mean that if there is an excess of mass above the surface, such asa large mountain chain, this must be compensated by some form of

mass deficiency below the surface. This is called isostaticcompensation.

In 1855, J.H.Pratt and Sir George Airy proposed two different

mechanisms to describe how this could take place. In each case we

require the presence of a rigid layer – the lithosphere – “floating” on adenser substratum - the astenosphere. We can also consider these two

layers as the crust and the mantle, without serious error.

Airy’s Hypothesis



This assumes the rigid and the lower layer to have uniform densities ρc

and ρm . Isostatic compensation is achieved by mountains having deep

roots.

At the level of compensation, we have:

cr t h

mr

ct ρ ρ ρ )

1(

1++=+

)(

)(

11

11

cm

c

ccm

hr

hr

ρ ρ

ρ

ρ ρ ρ

−=

=−

depth of root

Similarly, show that a thin ocean crust would have an “anti-root”

given by

)(

)(3

cm

wcd r

ρ ρ

ρ ρ

−−

= thickness of anti-root

Pratt’s Hypothesis

h1

r 1

t

d

r 3

ρc

ρm

Sea level

ρc= density of crust

ρm

= density of mantle



Pratt assumed that the base of the crust is at a constant level and

that isostatic equilibrium is achieved by allowing the crust to have

lateral variations in density. Thus a high mountain would be

compensated by assigning a low density to that “column” and

ocean basins being underlain by material of higher density.

Show that in this model, the density of, and beneath mountain h1 is

+

= Dh

Dc

1

1 ρ ρ

And that the material underneath the oceans has density

d D

d D wcd

−

−=

ρ ρ ρ

In the real earth, isostatic compensation is achieved in both ways,

and sometimes as a combination of both. For example, we know

from seismic studies that mountains have roots, and therefore the

Airy model is obeyed. Similarly, there are regions where hot

h1 h

2

ρc

ρ1 ρ

2 ρd

dD

ρm



For (i), the difference in g is approximately

−

R

h g

210 , where g 0 is

the IGF value at the surface of the ellipsoid, R is the earth’s radius

and h is the elevation. This is called the free air correction, and

has to be added to the observed value. The remaining difference iscalled the free-air gravity anomaly:

00

21 g

R

h g g g obs F −

−+=

If the measurement were performed above an ocean, the free-air

anomaly would be zero, but because of the mass of the mountain

((ii)), the free-air anomaly should still be positive. However, if the

mass of the mountain is isostatically compensated, the “root”

should reduce the free air anomaly to almost zero. Thus the free-air anomaly may be used to indicate whether a structure is

isostatically compensated or not.

(There are several other corrections which need to be made to a

measured value of g but these will be dealt with in the second part

of the course.)

Elastic Rebound

Imagine that a large load is suddenly placed upon a section of crust

that was previously in isostatic equilibrium. The equilibrium will

now be destroyed. How does the earth react to this?

Although we say that the lithosphere is rigid, on a geological

timescale it behaves elastically and is able to bend. Similarly on avery short seismic timescale (during the passage of a seismic

wave) the mantle behaves as a solid, but on a geological timescale

it behaves as a viscous fluid. Thus the lithosphere reacts to a load

by bending downwards, in so doing, “pushing out” the underlying

mantle, and settling to a new state of isostatic equilibrium. If the



load is removed, eg the melting of an ice sheet, the equilibrium is

again destroyed and the lithosphere begins to “rebound” upwards

to the original state.

Measurements of these rates of rebound eg in Fennoscandia enablethe elastic properties of the lithosphere and mantle to be estimated.

In Finland and Scandinavia, the ice sheet melted about 10000 years

ago. The ice was about 2.5km thick and had an area of about 4 x

106 km2. The lithosphere is still moving upward at the rate of

about 9mm per year, and it is estimated that about 30m of uplift

remains.



Elastic bending of lithosphere on a viscous astenosphere when

subjected to a load eg a seamount.

Gravitational Potential of Earth

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