CHAPTER 4

Gravity and density inside the Earth

Revision date February 2016

4.1 Introduction . Gravitational forces are very important in the formation and evolution of a planetand the study of gravity on and above the Earth’s surface also allows us to infer something about thedistribution of mass within the Earth. The acceleration due to gravity varies with geographic location,because of Earth’s rotation and approximately elliptical shape, but also with topographic variations,and lateral density variations within the Earth. Gravitational potential in the form of the geoid alsoprovides information about density variations within the Earth, typically on a larger scale than that forthe acceleration due to gravity. This chapter discusses the various contributions to variations in gravity:we will define a reference state for an elliptical, rotating Earth, and then consider the residual gravityvariations and geophysical implications of the departures from this basic model.

4.2 Inverse Square Law . The most useful thing to remember in the study of gravity is the inverse-square law, known as Newton’s law of gravitation. Consider two point masses m1 and m2 separated bya distance r

CHAPTER 4Gravity and density inside the Earth

4.1 Introduction . Gravitational forces are very important in the formation and evolution of a planet

and the study of gravity on and above the Earth’s surface also allows us to infer something about the

distribution of mass within the Earth. The acceleration due to gravity varies with geographic location,

because of Earth’s rotation and approximately elliptical shape, but also with topographic variations,

and lateral density variations within the Earth. Gravitational potential in the form of the geoid also

provides information about density variations within the Earth, typically on a larger scale than that for

the acceleration due to gravity. This chapter discusses the various contributions to variations in gravity:

we will define a reference state for an elliptical, rotating Earth, and then consider the residual gravity

variations and geophysical implications of the departures from this basic model.

4.2 Inverse Square Law . The most useful thing to remember in the study of gravity is the inverse-

square law, known as Newton’s law of gravitation. Consider two point massesm1 andm2 separated by

a distance r

Fig. 4.1

The force on either mass, F , is

F =Gm1m2

r2(4.1)

G is called the “universal gravitational constant” and is probably the least accurately determined of the

fundamental constants. The presently accepted value is G = 6.673 ± .003 ! 10!11 m3 kg!1 s!2

How do we use the inverse-square law? Suppose we have a complicated body as shown in Figure 4.2

Fig. 4.2

Consider a small element of the body of mass !m. The gravitational force !f exerted by the mass !mon an observer of massm" at P is

!f =Gm"!m

b2(4.2)

The acceleration of themassm" due to the element !mwould be (remember force =mass! acceleration)

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Fig. 4.1

The force on either mass, F , isF =

Gm1m2

r2(4.1)

G is called the “universal gravitational constant” (or "Big-Gee" to its friends) and is probably the leastaccurately determined of the fundamental constants. The 2014 accepted value isG = 6.67408± .00031×10−11 m3 kg−1 s−2.

(Where does G come from? Originally from the Cavendish experiment of Henry Cavendish in 1798.He suspended lead weights on torsion fibers and was the first to directly measure the force of attractionbetween masses. His estimate was G = 6.74 ± .047 × 10−11 m3 kg−1 s−2 – within 1% of the currentvalue and within two standard errors.)

We can define a field due to m1 by dividing through by m2. We call this field gravity, or g, and it hasunits of acceleration:

g =F

m2=Gm1

r2

The SI units of acceleration are, of course, m/s2, but in gravity you will also find units of milligal(10−5 m/s2) and gravity units, or GU (10−6 m/s2) used.

How do we use the inverse-square law? Suppose we have a complicated body as shown in Figure 4.2Consider a small element of the body of mass δm. The gravitational force δf exerted by the mass δmon an observer of mass m′ at P is

δf =Gm′δm

b2(4.2)

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Fig. 4.1b The Cavendish experiment.

CHAPTER 4Gravity and density inside the Earth

4.1 Introduction . Gravitational forces are very important in the formation and evolution of a planet

and the study of gravity on and above the Earth’s surface also allows us to infer something about the

distribution of mass within the Earth. The acceleration due to gravity varies with geographic location,

because of Earth’s rotation and approximately elliptical shape, but also with topographic variations,

and lateral density variations within the Earth. Gravitational potential in the form of the geoid also

provides information about density variations within the Earth, typically on a larger scale than that for

the acceleration due to gravity. This chapter discusses the various contributions to variations in gravity:

we will define a reference state for an elliptical, rotating Earth, and then consider the residual gravity

variations and geophysical implications of the departures from this basic model.

4.2 Inverse Square Law . The most useful thing to remember in the study of gravity is the inverse-

square law, known as Newton’s law of gravitation. Consider two point massesm1 andm2 separated by

a distance r

Fig. 4.1

The force on either mass, F , is

F =Gm1m2

r2(4.1)

G is called the “universal gravitational constant” and is probably the least accurately determined of the

fundamental constants. The presently accepted value is G = 6.673 ± .003 ! 10!11 m3 kg!1 s!2

How do we use the inverse-square law? Suppose we have a complicated body as shown in Figure 4.2

Fig. 4.2

Consider a small element of the body of mass !m. The gravitational force !f exerted by the mass !mon an observer of massm" at P is

!f =Gm"!m

b2(4.2)

The acceleration of themassm" due to the element !mwould be (remember force =mass! acceleration)

101

Fig. 4.2

The acceleration of the massm′ due to the element δmwould be (remember force = mass× acceleration)

δg =δf

m′(4.3)

so

δg =Gδm

b2(4.4)

(remember that g is a vector so δg points from P towards δm. G is the gravitational constant.)This equation is the basic equation we use. We can find the total acceleration due to gravity at any pointfor any body by dividing up the body into small elements of mass δm, computing δg, then performingthe (vector) sum of all the contributions from all the elements. In the limit that the elements becomeinfinitesimally small, g can be found by integrating over the mass distribution. However, doing theintegration over a vector quantity is hard work. We can make life easier by introducing the potential.

A body in a gravitational field has potential energy. The gravitational field is an example of a“conservative” force field – no dissipative losses of energy occur when a body is moved from one pointto another. The differences in potential energy of a body between two points therefore depends onlyon the position of the points and not on the path along which the body has moved between them. Theusual definition of the potential energy of a body is that it is the work done in bringing it from infinity,where potential is defined to be zero, to its present position. In the case of gravity, in which bodiesalways attract, this results in a potential that is always negative. The “gravitational potential” is definedto be the gravitational potential energy per unit mass. To make matters clear, reconsider our two pointmasses, m1 and m2 separated by distance r. The gravitational potential due to mass m1 is defined by

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V = −Gm1

r(4.5)

so the gravitational potential energy of mass m2 at distance r from m1 is

EV = −Gm1m2

r(4.6)

Suppose we move mass m2 from position r1 to position r2. Its change in potential energy will be

∆EV = −Gm1m2

r1+Gm1m2

r2(4.7)

From the inverse square law, the acceleration due to gravity of mass m2 due to the mass m1 is

g =Gm1

r2(4.8)

so we can write the change in potential energy of 4.7 as

−m2g(r1)r1 +m2g(r2)r2 ' m2g(r2 − r1) (4.9)

where we have assumed that g(r1) ' g(r2) = g is roughly constant. This formula for the change inpotential energy of a body should be familiar to you as mgh.Consideration of 4.5 and 4.8 shows that

g = −∂V∂r

=Gm

r2(4.10)

In three dimensions we have the result g = −∇V .Note that g is a vector. We must remember that, on a spherically symmetric earth with radius defined

as outward, g points towards the origin of the coordinate system, so that its numerical value occurs inequations with a negative sign. As a consequence of this in some texts you will find the sign of thegravitational potential flipped so that it is always postive. However, if you do this the classic equationused to derive a field from a potential also has to have its sign flipped i.e., g = ∇V , which is bound tocause confusion, and so here we will hew to standard potential field theory conventions, even thoughwe might drop the negative sign when dealing with the vertical, or radial, component of gravity. Justremember that only differences in potential are physically meaningful and you must end up with gpointing in the right direction!

We can also generalize the idea of gravitational potential to a distribution of masses such as in figure4.2:

V = −G∫m

dm

r≈ −G

∑i

dmi

bi(4.11)

4.3 The Spherical Earth . Now, we will show how to compute the acceleration due to gravity within aspherically symmetric planet. For this we need Gauss’ Theorem, an extremely powerful tool in potentialfield theory.

Consider a closed surface, S, enclosing a point mass m (Figure 4.3b). At point P , we can form aninfinitesimal element of the surface S. A planar surface may be described by a vector whose magnitudeis the area of the surface and whose direction is the (outward) normal to the surface, so in the above caseds describes the surface element. Now we wish to introduce the concept of flux. The surface integralof a vector field F, ∫

S

F · ds

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m

Θ

Pds

g

S

m'g1

g2

Ω

r

d

Fig 4.3b. A closed surface S containing mass. Mass element m is inside S and mass element m′ isoutside the surface.

is called the flux of F through the surface S. Note that the dot product (x · y = x.y. cos θ, where θ is theangle between x and y) has the effect of taking the component of F perpendicular to the surface. Wewant to compute the total flux through the surface S surrounding the mass m:∫

S

g · ds =∫S

g. cos θds =∫S

G.m. cos θ.dsr2

Before performing the integration we note that

− cos θ.dsr2

= dΩ,

where dΩ is the solid angle subtended at m by the surface element ds. (The minus sign comes from thefact that θ is always > 90o). Now we have∫

S

g · ds =∫S

−G.m.dΩ = −4πGm.

We can ignore the contributions to the flux of all masses outside the surface S; referring to the figureagain it may be seen that masses outside the surface subtend two surface elements with identical solidangles but with flux of opposite signs, canceling. Thus we have the important result, Gauss’ Law, thatthe total flux through a closed surface is equal to −4πG times the mass enclosed by the surface. Theequation derived above is very useful. For example, we can show that outside a spherical distributionof mass, the gravitational field is the same as though the mass were all concentrated at a single, centralpoint:

Consider a spherically symmetric distribution of mass with centre O. If we take a point P a distancer from the mass’ centre, a sphere through P centred on O clearly has g perpendicular and equal over itsentire surface, so it is easy to compute the surface integral;∫

S

g · ds = −4πr2g

Applying Gauss’ Law this is equal to

−4πGM = −4πr2g

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r

g p

•

•

O

SS

A B

S

O.

Fig 4.3c A: Spherically symmetric mass centered on O. An observation of g is made at point P , adistance r from the center, and must be equal at every point on S, as well as perpendicular. B: For asymmetrical shell, we can include an inner surface to S, but since nothing has changed on the outersurface, g on the inner surface must be zero.

where we have set the total mass of the sphere to M . Thus the gravitational attraction anywhere outsidethe sphere is given by

g(r) = GM

r2(4.12)

which is the same expression for a point mass at O.We can use our Gauss’ Law result to obtain the gravitational force on a body of mass m at the surfaceof a spherically symmetric Earth:

F =GmM

a2= mg (4.13)

where M is the mass of the Earth and a is the radius of the Earth. g is the acceleration due to gravity ofthe body (remember that force = mass × acceleration).

The acceleration due to gravity at radius r < a inside the Earth (again assuming it is sphericallysymmetric) is

g(r) =G

r2

r∫0

4πx2ρ(x) dx (4.14)

where we have used the result that total mass within radius r is given by

M =

r∫0

4πx2ρ(x)dx (4.15)

and another result of Gauss’ Law that says that gravity inside a symmetric shell is zero everywhere.Equation (4.12) is a good approximation to the value of g on the Earth, but because the Earth is not

exactly spherical, there are small variations. The biggest variation is due to the equatorial bulge of theEarth. This bulge is caused by the Earth’s rotation. On long time scales the Earth behaves as a fluidand the spinning of the Earth causes it to have the shape of an oblate spheroid. The acceleration dueto gravity varies slightly with latitude because of this effect (and also, of course, the spinning itself).Superimposed on this are even smaller variations in g caused by topography, density inhomogeneities,etc. These small variations are called gravity anomalies. We can relate the measurements of g at thesurface to the density variations using Newton’s Law and so infer something about the structure causing

105

the anomalies. The gravity anomalies are what really interests us but to define an anomaly we mustfirst define a reference state. The natural reference state is the gravitational field of a rotating, laterallyhomogeneous, elliptical earth which we now consider in some detail.

4.4 Gravitational acceleration external to the rotationally disturbed Earth . The “flattening” of theEarth, f , is defined by

f =a− ca

= 3.353× 10−3 (4.16)

It should be clear to you from these numbers that the distortion of the Earth from a sphere (of meanradius, a, 6371 km) is very small, so the perturbation of g at the surface away from the value GM/a2

will also be very small. It should also be clear to you that g will be a weak function of latitude but nota function of longitude. This is because the rotationally distorted Earth is symmetric about the axis ofrotation. If we exaggerate the equatorial bulge we have a situation like

4.3 Gravitational acceleration external to the rotationally disturbed Earth . The “flattening” of the

Earth, f , is defined by

f =a ! c

a= 3.353 " 10!3 (4.16)

It should be clear to you from these numbers that the distortion of the Earth from a sphere (of mean

radius, a, 6371 km) is very small, so the perturbation of g at the surface away from the value GM/a2

will also be very small. It should also be clear to you that g will be a weak function of latitude but nota function of longitude. This is because the rotationally distorted Earth is symmetric about the axis of

rotation. If we exaggerate the equatorial bulge we have a situation like

Fig. 4.4

One consequence of the distortion is that g no longer points directly towards the center of the body.

Fig. 4.5

This exaggerated diagram shows that there will be more mass at the equator than the poles so g willbe deflected towards the bulge – it has a small tangential component as well as a radial component.

104

Fig. 4.4

One consequence of the distortion is that g no longer points directly towards the center of the body.

4.3 Gravitational acceleration external to the rotationally disturbed Earth . The “flattening” of the

Earth, f , is defined by

f =a ! c

a= 3.353 " 10!3 (4.16)

It should be clear to you from these numbers that the distortion of the Earth from a sphere (of mean

radius, a, 6371 km) is very small, so the perturbation of g at the surface away from the value GM/a2

will also be very small. It should also be clear to you that g will be a weak function of latitude but nota function of longitude. This is because the rotationally distorted Earth is symmetric about the axis of

rotation. If we exaggerate the equatorial bulge we have a situation like

Fig. 4.4

One consequence of the distortion is that g no longer points directly towards the center of the body.

Fig. 4.5

This exaggerated diagram shows that there will be more mass at the equator than the poles so g willbe deflected towards the bulge – it has a small tangential component as well as a radial component.

104

Fig. 4.5

This exaggerated diagram shows that there will be more mass at the equator than the poles so g willbe deflected towards the bulge – it has a small tangential component as well as a radial component.Consider the position P , we have

106

Consider the position P , we have

Fig. 4.6

(We are presently using the convention that g is positive if directed inward.) The tangential componentof g is !gt. The radial component can be written as the “expected” component (ge = GM/r2) plus an

“anomalous” component (!gr), or

gr =GM

r2+ !gr (4.17)

where

!gr ! GM

r2

We now have

g = (g2r + !g2

t )12

=!(GM

r2+ !gr)2 + !g2

t

" 12

=!(GM

r2)2 + 2!gr

GM

r2+ !g2

r + !g2t

" 12

From the discussion it should be clear to you that terms like !g2r are much smaller than terms like

2!grGM/r2 so a good approximation to g is

g =!(GM

r2)2 + 2!gr

GM

r2

" 12

This can be rewritten as

g =GM

r2

!1 +

2!grr2

GM

" 12

" GM

r2

!1 +

!grr2

GM

"=

GM

r2+ !gr = gr (4.17)

105

Fig. 4.6

(We are presently using the convention that g is positive if directed inward.) The tangential componentof g is δgt. The radial component can be written as the “expected" component (ge = GM/r2) plus an“anomalous" component (δgr), or

gr =GM

r2+ δgr (4.17)

where

δgr GM

r2

We now have

g = (g2r + δg2

t )12

=[(GM

r2+ δgr)2 + δg2

t

] 12

=[(GM

r2)2 + 2δgr

GM

r2+ δg2

r + δg2t

] 12

From the discussion it should be clear to you that terms like δg2r are much smaller than terms like

2δgrGM/r2 so a good approximation to g is

g =[(GM

r2)2 + 2δgr

GM

r2

] 12

This can be rewritten as

g =GM

r2

[1 +

2δgrr2

GM

] 12

' GM

r2

[1 +

δgrr2

GM

]=GM

r2+ δgr = gr (4.17)

where we used the binomial expansion (recall: (1+x)n = 1+nx+ ...). This expression for g is correct tofirst order in the small quantities δgr and δgt. To this accuracy, we see that we can neglect the transverse

107

component, δgt, and we only have to worry about δgr, i.e., to first order g still points towards the centerof the body.δgr is a function of latitude. The full solution for the latitude dependence requires the use of spherical

harmonic analysis and we only outline the method here. It also turns out to be easiest to use thegravitational potential, V , (see equation 4.5) because it is a scalar. We can always get the accelerationdue to gravity by differentiating the gravitational potential. From equation 4.5, we have that, outside aspherical earth, the gravitational potential is just given by

V = −GMr

(4.18)

(Note that equation 4.15 for g outside a spherical earth can be obtained by differentiating the aboveequation). It is also easy to show that, outside a spherical Earth,

1r2

d

dr

(r2dV

dr

)= ∇2V = 0 (4.19)

(in spherical symmetry.) The equation ∇2V = 0 is called Laplace’s equation and also holds outside anon-spherical Earth. In the case of a rotationally distorted Earth, the potential V depends upon both rand θ (the colatitude: 90 - latitude = 0 at the north pole and 180 at the south pole). In this case, thesolution to Laplace’s equation can be written as a series expansion:

V (r, θ) = −GMr

[J0P0( cos θ)− a

rJ1P1( cos θ)− a2

r2J2P2( cos θ)− a3

r3J3P3( cos θ) · · ·] (4.20)

(a is the equatorial radius of the Earth). In this expansion, the J’s are empirically determined coefficientsand the P ’s are “Legendre polynomials.” The first few are

P0( cos θ) = 1

P1( cos θ) = cos θ

P2( cos θ) = 12 (3 cos 2θ − 1)

P3( cos θ) =cos θ

2(5 cos 2θ − 3)

where we used the binomial expansion. This expression for g is correct to first order in the smallquantities !gr and !gt. To this accuracy, we see that we can neglect the transverse component, !gt, and

we only have to worry about !gr, i.e., to first order g still points towards the center of the body.!gr is a function of latitude. The full solution for the latitude dependence requires the use of

spherical harmonic analysis and we only outline the method here. It also turns out to be easiest to use the

gravitational potential, V , (see equation 4.5) because it is a scalar. We can always get the accelerationdue to gravity by differentiating the gravitational potential. From equation 4.5, we have that, outside a

spherical earth, the gravitational potential is just given by

V = !GM

r(4.18)

(Note that equation 4.15 for g outside a spherical earth can be gotten by differentiating the aboveequation). It is also easy to show that, outside a spherical Earth,

1r2

d

dr

!r2 dV

dr

"= "2V = 0 (4.19)

(in spherical symmetry.) The equation "2V = 0 is called Laplace’s equation and also holds outside anon-spherical Earth. In the case of a rotationally distorted Earth, the potentialV depends upon both r and" (the colatitude). In this case, the solution to Laplace’s equation can be written as a series expansion:

V (r, ") = !GM

r[J0P0( cos ") !

a

rJ1P1( cos ") !

a2

r2J2P2( cos ") !

a3

r3J3P3( cos ") · · ·] (4.20)

(a is the equatorial radius of the Earth). In this expansion, the J’s are empirically determined coefficientsand the P ’s are “Legendre polynomials.” The first few are

P0( cos ") = 1

P1( cos ") = cos "

P2( cos ") = 12 (3 cos 2" ! 1)

P3( cos ") =cos "

2(5 cos 2" ! 3)

Fig. 4.7

These functions, plotted in Figure 4.7, are progressively wigglier functions of the colatitude " and wecan choose the J’s so that the sum can approximate very accurately any observed dependence of V on

106

Fig. 4.7

These functions, plotted in Figure 4.7, are progressively wigglier functions of the colatitude θ and wecan choose the J’s so that the sum can approximate very accurately any observed dependence of V onθ. Let us consider the first few terms in the above expansion. At very large distances from the Eartha/r → 0 so

108

V (r, θ) = −GMrJ0 for r a

As you go away from the Earth it should be clear to you that the Earth will appear as a point mass so Vshould have the form −GM/r, hence we can conclude that that J0 must be 1. Now consider the shapeof P1( cos θ) as a function of colatitude (Fig. 4.7). This function is antisymmetric about the equator –our equatorial bulge is symmetric about the equator. This term isn’t going to help us fit the gravitationalpotential because it doesn’t have the right geographical dependence – we can therefore set J1 = 0. (Ifwe choose the center of mass as the origin of our coordinate system then J1 = 0 by definition.)

Now let us examine P2( cos θ) (Fig. 4.7). This function is symmetric about the equator and hasroughly the shape of an equatorial bulge. We would therefore expect J2 to be important. In fact, whenwe fit the observed latitude dependence of V we find that J2 is over 500 times bigger than any of theother J’s so to a very good approximation we have

V (r, θ) = −GMr

(1− a2

r2J2P2( cos θ)

)so substituting for P2

= −GMr

(1− a2

r2J2

2(3 cos 2θ − 1)

)

= −GMr

(1− a2

r2J2

2(3 sin 2λ− 1)

)(4.21)

where λ is the latitude. (θ = π2 − λ so cos θ = sinλ)

To first order we found that g still points radially inwards so we can calculate g (the acceleration dueto gravity) by using the relationship g = −dV/dr, i.e.,

g(r, λ) =GM

r2− 3GMa2

2r4J2(3 sin 2λ− 1) =

GM

r2

(1− 3a2

2r2J2(3 sin 2λ− 1)

)(4.22)

where we have defined g as positive downwards (unlike r!). This equation gives g at a radius r abovethe center of the Earth and the values of the constants used are

GM = 3.986005× 1014 m3 s−2

a = 6378.139km (the equatorial radius)J2 = 1.08270× 10−3

(Note that J2 is still much less than 1 even though it is much larger than any of the other J’s.) Thisequation gives us the gravitational effect of the equatorial bulge but this isn’t the only effect. Since weare on a rotating Earth, there will be a centrifugel acceleration term. Before we look at this, we takea short diversion from developing an equation for gravity on the rotationally distorted Earth to look atmoments of inertia and how they are sensitive to the internal density distribution of a planet.

4.5 Centrifugal acceleration . Earlier, we developed an equation for the gravitational acceleration ofthe Earth which took into account the mass of the Earth and the departure from spherical symmetrycaused by the equatorial bulge. This equation cannot be used on the surface of the Earth because theEarth is rotating. An object on the Earth’s surface is subject to centrifugal acceleration. Consider thefollowing diagram

109

following diagram

Fig. 4.12

The centrifugal acceleration, g!, is given by

g! = !2s (4.28)

We want the component of g! in the radial direction. g! points outwards in a direction perpendicular to

the rotation axis so the radial component is g! cos!. From the diagram you can also see that s = r cos!so the radial component outward of g! is given by

!2r cos 2! (4.29)

Wehave been using the sign convention that a positive value of g points radially inwards so the centrifugalacceleration acts to counteract this, i.e., on the surface of the Earth

g(r, !) =GM

r2! 3GMa2

2r4J2(3 sin 2! ! 1) ! !2r cos 2!

or (to emphasize which terms are small)

g(r, !) =GM

r2

!1 ! 3J2

2a2

r2(3 sin 2! ! 1) ! !2a3

GM

r3

a3cos 2!

"(4.30)

Note that the term !2a3/GM is 3.46775 " 10!3 and so is small relative to 1. We can also redefine

the gravitational potential at the Earth’s surface to include the centrifugal potential. This is sometimes

called the geopotential and is given the symbol U instead of V . We get U by integrating the expression

for g – the first two terms are the same as for V but now we have an extra term:

U(r, !) = !GM

r+

GMa2J2

2r3(3 sin 2! ! 1) ! 1

2!2r2 cos 2!

or

U(r, !) = !GM

r

!1 ! J2

2a2

r2(3 sin 2! ! 1) +

12

!2a3

GM

r3

a3cos 2!

"(4.31)

112

Fig. 4.12

The centrifugal acceleration, gω, is given by

gω = Ω2s (4.23)

We want the component of gω in the radial direction. gω points outwards in a direction perpendicular tothe rotation axis so the radial component is gω cosλ. From the diagram you can also see that s = r cosλso the radial component outward of gω is given by

gωr = Ω2r cos 2λ (4.24)

We have been using the sign convention that a positive value of g points radially inwards so the centrifugalacceleration acts to counteract this, i.e., on the surface of the Earth

g(r, λ) =GM

r2− 3GMa2

2r4J2(3 sin 2λ− 1)− Ω2r cos 2λ

or (to emphasize which terms are small)

g(r, λ) =GM

r2

(1− 3J2

2

(a

r

)2

(3 sin 2λ− 1)− Ω2a3

GM

(r

a

)3

cos 2λ

)(4.25)

Note that the term Ω2a3/GM is 3.46775 × 10−3 and so is small relative to 1 (but not with respect toJ2!). We can also redefine the gravitational potential at the Earth’s surface to include the centrifugalpotential. This is sometimes called the geopotential and is given the symbol U instead of V . We get Uby integrating the expression for g with respect to r – the first two terms are the same as for V but nowwe have an extra term:

U(r, λ) = −GMr

+GMa2J2

2r3(3 sin 2λ− 1)− 1

2Ω2r2 cos 2λ

or

U(r, λ) = −GMr

(1− J2

2

(a

r

)2

(3 sin 2λ− 1) +12

Ω2a3

GM

(r

a

)3

cos 2λ

)(4.26)

110

4.6 Moments of inertia . The gravity field g(r, λ) can be measured accurately from tracking the orbitsof artificial satellites, allowing the measurement of J2. It turns out that J2 is related to the momentsof inertia of the planet and is helpful in constraining the density distribution. What are "moments ofinertia"? Consider an arbitrary body as in Fig. 4.8.

from tracking the orbits of artificial satellites, allowing the measurement of J2. What are ”moments of

inertia”? Consider an arbitrary body as in Fig. 4.8.

Fig. 4.8

If we choose an axis going through the body then the moment of inertia of an element of mass !m at a

perpendicular distance x from the axis is

x2!m

The total moment of inertia of the body about the axis is obtained by integrating this expression over the

entire mass distribution of the body. As we shall see, the moments of inertia of a planet are observable

quantities and can be used to place constraints on how density varies inside the planet.

To see how this might be done we consider a planet which is spherically symmetric, i.e., density is

just a function of radius. Visualize a cartesian coordinate systemwith the z axis aligned with the rotationaxis:

Fig. 4.9

The principalmoments of inertia of the planet are themoments of inertia about thex, y, and z axes. Theseare called A, B and C respectively. For a spherically symmetric planet it is clear that A = B = C. Weshall compute C. It is useful to use a spherical polar coordinate system which is related to the cartesian

108

Fig. 4.8

If we choose an axis going through the body then the moment of inertia of an element of mass δm at aperpendicular distance x from the axis is

x2δm

The total moment of inertia of the body about the axis is obtained by integrating this expression over theentire mass distribution of the body. As we shall see, the moments of inertia of a planet are observablequantities and can be used to place constraints on how density varies inside the planet.

To see how this might be done we consider a planet which is spherically symmetric, i.e., densityis just a function of radius. Visualize a cartesian coordinate system with the z axis aligned with therotation axis:

from tracking the orbits of artificial satellites, allowing the measurement of J2. What are ”moments of

inertia”? Consider an arbitrary body as in Fig. 4.8.

Fig. 4.8

If we choose an axis going through the body then the moment of inertia of an element of mass !m at a

perpendicular distance x from the axis is

x2!m

The total moment of inertia of the body about the axis is obtained by integrating this expression over the

entire mass distribution of the body. As we shall see, the moments of inertia of a planet are observable

quantities and can be used to place constraints on how density varies inside the planet.

To see how this might be done we consider a planet which is spherically symmetric, i.e., density is

just a function of radius. Visualize a cartesian coordinate systemwith the z axis aligned with the rotationaxis:

Fig. 4.9

The principalmoments of inertia of the planet are themoments of inertia about thex, y, and z axes. Theseare called A, B and C respectively. For a spherically symmetric planet it is clear that A = B = C. Weshall compute C. It is useful to use a spherical polar coordinate system which is related to the cartesian

108

Fig. 4.9

The principal moments of inertia of the planet are the moments of inertia about the x, y, and z axes.These are called A,B and C respectively. For a spherically symmetric planet it is clear that A = B = C.We shall compute C. It is useful to use a spherical polar coordinate system which is related to thecartesian coordinate system in the following way.

111

coordinate system in the following way.

Fig. 4.10

Note that

x = r sin ! cos"; y = r sin ! sin" and z = r cos !

An element of volume

dV = dxdydz in a cartesian coordinate system

= r2 sin !drd!d" in a spherical coordinate system

Now reconsider the planet.

Fig. 4.11

The point P is a perpendicular distance s from the z axis. Our definition of the moment of inertia aboutthe z axis gives

C =!

M

s2 dm

109

Fig. 4.10

Note that

x = r sin θ cosφ; y = r sin θ sinφ and z = r cos θ

An element of volume

dV = dxdydz in a cartesian coordinate system

= r2dr. sin θdθ.dφ in a spherical coordinate system

Now reconsider the planet.

coordinate system in the following way.

Fig. 4.10

Note that

x = r sin ! cos"; y = r sin ! sin" and z = r cos !

An element of volume

dV = dxdydz in a cartesian coordinate system

= r2 sin !drd!d" in a spherical coordinate system

Now reconsider the planet.

Fig. 4.11

The point P is a perpendicular distance s from the z axis. Our definition of the moment of inertia aboutthe z axis gives

C =!

M

s2 dm

109

Fig. 4.11

The point P is a perpendicular distance s from the z axis. Our definition of the moment of inertia aboutthe z axis gives

C =∫M

s2 dm

where the integral is over the entire mass distribution and dm is the mass of an element of volume dVat the point P . We have

dm = ρ(r) dV

112

and also

s2 = x2 + y2 = r2 sin 2θ( cos 2φ+ sin 2φ)

= r2 sin 2θ

So now we can write

C =∫V

r2 sin 2θρ(r) dV

where the integral is over the volume of the planet. This integral can be written as an integral over r(from 0 to a), over θ (from 0 to π) and over φ (from 0 to 2π) by substituting in the expression for dV :

C =

2π∫0

π∫0

a∫0

ρ(r)r4 sin 3θ drdθdφ =

a∫0

ρ(r)r4dr

π∫0

sin 3θdθ

2π∫0

dφ

We can do these integrals separately. You will note that nothing in the integrand (ρ(r)r4 sin 3θ) dependsupon φ, the longitude, so the dφ integral is just 2π and we can write

C = 2π

a∫0

ρ(r)r4 dr

π∫0

sin 3θ dθ

The last integral is simple to do if we let f = cos θ; we can write df/dθ = − sin θ and f varies from 1 to−1 as θ varies from 0 to π, i.e., df = − sin θdθ so

π∫0

sin 3θ dθ = −−1∫1

sin 2θ df = −−1∫1

(1− cos 2θ) df = −−1∫1

(1− f2) df = −[f − f3

3

]−1

1

=43

Finally, we have

C =8π3

a∫0

ρ(r)r4 dr (4.27)

This equation tells us something about the density distribution within the planet. Let us suppose thata planet has a uniform density, ρ, then

C =8π3ρ

a∫0

r4 dr =8π15ρa5 (4.28)

The datum that is used is usually not C but C/Ma2 which is a dimensionless number (M is the mass ofthe planet). Now

M =43πa3ρ

So for a uniform density planet

C

Ma2=

8π15ρa5 3

4πa3ρ

1a2

=25

= 0.4

113

If the density is greater near the center of the planet we find that C/Ma2 < 0.4. For example C/Ma2 forthe Earth is .3308 but for the moon C/Ma2 = 0.392. Clearly the moon is almost a uniform density body.

How might we measureC? On a rotationally distorted Earth we haveA = B (because of the symmetryabout the rotation axis) but now A 6= C. A straightforward, but tedious, calculation allows J2 to be castin terms of the principal moments of inertia of the body (the details are in section 5.2 of Turcotte andSchubert). We find that

J2 =C −AMa2

(4.29)

This equation allows us to estimate C − A because J2 is measured. The gravitational attractions ofthe Sun and the Moon, acting on the equatorial bulge of the Earth, causes a precession of the axis ofrotation. From the rate of precession we can find the “dynamical ellipticity”, H, which is given by

H =C −AC

(4.30)

H is estimated to be 1/305.51 and by combining equations we have

C =J2

HMa2 = .3308Ma2 (4.31)

4.7 The geoid . An “equipotential surface” is a surface on which U is a constant. The vector g willbe normal to any such surface, thus U defines the local horizontal. Apart from the effects of wind andcurrents, the sea surface is an equipotential surface and the reference equipotential surface which definessea level is called the geoid. We can get an expression for this reference surface accurate to first orderin the following way.

Recall our expression for the geopotential:

U(r, λ) = −GMr

(1− J2

2a2

r2(3 sin 2λ− 1) +

12

Ω2a3

GM

r3

a3cos 2λ

)If we evaluate U at the equator we have r = a and λ = 0 so

Ueq = −GMa

(1 +

J2

2+

12

Ω2a3

GM

)If we evaluate U at the pole we have r = c and λ = π/2 so

Upole = −GMc

(1− J2(

a

c)2)

Setting Upole = Ueq (for an equipotential surface) we have

1 +12J2 +

12a3Ω2

GM=a

c[1− J2(

a

c)2]

It is conventional to eliminate c and use the “flattening” f . Remember f = (a − c)/a so c = a(1 − f).You should also recall that both J2 and f are much less than one so we have

1 +J2

2+

12a3Ω2

GM=

a

a(1− f)

[1− J2(

a

c)2]

=[1− J2(

a

c)2]

(1− f)−1

'[1− J2(

a

c)2]

[1 + f ]

114

' 1− J2(a

c)2 + f

= 1− J2

(1− f)2+ f

' 1− J2(1 + 2f) + f

' 1− J2 + f

(We have neglected products like fJ2 and f2 because they are small.) Rearranging gives

f ' 32J2 +

12a3Ω2

GM(4.32)

Using the value of a3Ω2/GM = 3.46775 × 10−3, we can compute f and find that it is 3.3579 × 10−3.Including higher order terms gives f = 3.35282 × 10−3 – this should demonstrate to you that we dopretty well by just keeping first order terms. It is now possible to show that the equipotential surfacer0 that corresponds to the geoid is given to first order by r0 = a(1 − f sin 2λ). This is the subject of ahomework problem. r0 is almost a spherical surface (remember that f is much less than one). Whenbetter than first order accuracy is required we use a spheroid (an ellipsoid of revolution) to define thereference geoid. Observed departures of the actual geoid from the reference geoid are called “geoidanomalies.” The reference geoid is given to second order by

r0 = a

[1 +

(2f − f2)(1− f)2

sin 2λ

]− 12

(4.33)

If we plot the actual geoid (the equipotential surface which most closely coincides with the sea surface)and the reference geoid we may have the situation

If we plot the actual geoid (the equipotential surface which most closely coincides with the sea surface)

and the reference geoid we may have the situation

Fig. 4.13

The difference between these, !N , is a geoid anomaly (also called the geoid height) and is measuredin meters. A map of these is shown in Figure 4.14, where it can be seen that the geoid anomaly

ranges from about -105 m near the southern tip of India to 73 m near Indonesia. Remember that since

f ! 1300 the departure of the reference ellipsoid from a sphere is about 20 km. The causes of these

geoid anomalies are not well-understood though positive geoid anomalies tend to be associated with

subducting slabs. Recent work has shown that most of the large scale geoid anomalies can be explained

by density anomalies in the lower mantle and topography on the core mantle boundary (caused probably

by temperature and chemical variations due to convection) – see the accompanying powerpoint.

Geoid height is directly related to local geopotential as follows: local gravity is given by

g = "r!U

!r.

Clearly we have |g| = g = !U!r . Remember by our definition U is negative, and increases as we move

away from Earth. Suppose we call the reference geoid UR then the observed geoid U0 can be computed

in terms of the properties of the reference geoid using a Taylor series expansion:

U0 = UR +!

!UR

!r

"

r0

!N

Now the radial derivative of the potential is the acceleration due to gravity so we can define a reference

gravity field, gR, (the acceleration due to gravity on the reference geoid) such that

gR =!

!UR

!r

"

r0

An anomaly in the potential !U is defined as

!U = UR " U0 = "gR!N,

and the geoid anomaly !N is

!N = "UR " U0

gR

Since g varies by less than a percent the above is an excellent approximation, that is geoid height issimply proportional to the negative of the local geopotential anomaly.

A reference gravity field has been defined so that potential anomalies can be calculated unambiguously

and so that departures of g from the reference field (gravity anomalies) can be unambiguously calculated.Gravity anomalies are the differences between the measured values of g on the reference geoid and thereference field gR which is given by

114

Fig. 4.13

The difference between these, ∆N , is a geoid anomaly (also called the geoid height) and is measuredin meters. A map of these is shown in Figure 4.14, where it can be seen that the geoid anomalyranges from about -105 m near the southern tip of India to 73 m near Indonesia. Remember that sincef ≈ 1/300 the departure of the reference ellipsoid from a sphere is about 20 km. The causes of thesegeoid anomalies are not well-understood though positive geoid anomalies tend to be associated withsubducting slabs. Recent work has shown that most of the large scale geoid anomalies can be explainedby density anomalies in the lower mantle and topography on the core mantle boundary (caused probablyby temperature and chemical variations due to convection) – see the accompanying powerpoint.

Geoid height is directly related to local geopotential as follows: local gravity is given by

g = −r∂U

∂r.

Clearly we have |g| = g = ∂U∂r . Remember by our definition U is negative, and increases as we move

away from Earth. Suppose we call the reference geoid UR then the observed geoid U0 can be computedin terms of the properties of the reference geoid using a Taylor series expansion:

115

Figure 4.14 Top: Geoid height anomaly, positive if the geoid height is above the reference spheroid.(Plate 6 of Fowler, from Bowin, Rev. Geophys., 2000). Bottom: Gravity anomaly, Plat 7 of Fowler.

116

From Davies, Dynamic Earth, (1999).

Earth Topography

Figure 4.15 Earth’s Topography

117

U0 = UR +(∂UR∂r

)r0

∆N

Now the radial derivative of the potential is the acceleration due to gravity so we can define a referencegravity field, gR, (the acceleration due to gravity on the reference geoid) such that

gR = −(∂UR∂r

)r0

An anomaly in the potential ∆U is defined as

∆U = UR − U0 = −gR∆N,

and the geoid anomaly ∆N is

∆N = −UR − U0

gR

Since g varies by less than a percent the above is an excellent approximation, that is geoid height issimply proportional to the negative of the local geopotential anomaly.

A reference gravity field has been defined so that potential anomalies can be calculated unambiguouslyand so that departures of g from the reference field (gravity anomalies) can be unambiguously calculated.Gravity anomalies are the differences between the measured values of g on the reference geoid and thereference field gR which is given by

gR = 9.78031846(1 + .005278895 sin 2λ+ .000023462 sin 4λ) (4.34)

This equation includes higher order terms but a close approximation to this can be found by evaluatingequation 4.31 on the reference geoid, r = r0.

(This is the 1967 formula as quoted by Fowler. There is a 1980 update

gR = 9.780327(1 + .0053024 sin 2λ− 0.0000058 sin 22λ) (4.35)

which differs by less than a milligal from the 1967 formula.)The long wavelength geoid can be found by the same procedures we discussed for J2. Careful records

are kept of artificial satellite orbits, whose positions in space are triangulated from the ground via theirradio time-of-flight delays. Then the equations of motion for a small body in a gravitational field aresolved, taking account of things like corrections to the orbit from perturbations due to the gravitationalattractions of the sun and moon. The results provide maps like the one in Figure 4.14, where you seethe deviations of the geoid (top) from that expected for a rotating hydrostatic earth, to a resolution ofabout 2000 km. One way to think about the differences between the geoid and gravity signal is that tofirst order the gravitational acceleration is proportional to 1

r2 , while the geoid is a measure of potentialproportional to 1

r . Thus one can think of the geoid as being more sensitive to density anomalies at greaterdepth in the Earth: this is reflected in the enhancement of long wavelength anomalies relative to thegravity signal. More recently the satellites Geosat and Seasat have been equipped with radar altimeterscapable of measuring the satellite altitude to an accuracy of a few centimeters above a mean ocean(the altimeters have a footprint several km2 in area). These measurements provide short wavelengthinformation (scales less than 500km), mostly about submarine topography, since the geoid is elevatedover topographic highs, and can be used to make high resolution gravity anomaly maps. The bottompart of Figure 4.14 shows such measurements derived from radar. For comparison global topographyis shown in Figure 4.15 along with a plot of its global distribution in statistical terms. In contrast togravity, the long-wavelength signal in the geoid shows less obvious correlations with surface signals

118

because of isostatic compensation – elevated areas are supported at depth by low density material. Forexample there is little expression of the midocean ridges in the geoid.

We have spent a lot of time defining reference surfaces and gravity fields so that we can meaningfullydefine geoid and gravity anomalies. We shall consider these in the next section.

4.8 Gravity anomalies – The effect of buried density anomalies . The computation of a gravityanomaly caused by a buried density anomaly is relatively straightforward using equation 4.4. Usuallywe have to resort to numerical evaluation for arbitrarily shaped bodies but for some simple geometriesit is possible to compute analytically the gravity anomaly. Consider a buried sphere of radius R with auniform density anomaly ∆ρ (the density anomaly is the density difference between the sphere and itssurroundings).

gR = 9.78031846(1 + .005278895 sin 2! + .000023462 sin 4!) (4.34)

This equation includes higher order terms but a close approximation to this can be found by evaluating

equation 4.31 on the reference geoid, r = r0.

The long wavelength geoid can be found by the same procedures we discussed for J2. Careful

records are kept of artificial satellite orbits, whose positions in space are triangulated from the ground

via their radio time-of-flight delays. Then the equations of motion for a small body in a gravitational

field are solved, taking account of things like corrections to the orbit from perturbations due to the

gravitational attractions of the sun and moon. The results provide maps like the one in Figure 4.14,

where you see the deviations of the geoid (top), and gravity (middle) from that expected for a rotating

hydrostatic earth, to a resolution of about 2000 km. One way to think about the differences between the

geoid and gravity signal is that to first order the graviational acceleration is proportional to 1r2 , while

the geoid is a measure of potential proportional to 1r . Thus one can think of the geoid as being more

sensitive to density anomalies at greater depth in the Earth: this is reflected in the enhancement of

long wavelength anomalies relative to the gravity signal. More recently the satellites Geosat and Seasat

have been equipped with radar altimeters capable of measuring the satellite altitude to an accuracy of

a few centimeters above a mean ocean (the altimeters have a footprint several km2 in area). These

measurements provide short wavelength information (scales less than 500km), mostly about submarine

topography, since the geoid is elevated over topographic highs, and can be used to make high resolution

gravity anomalymaps. The bottom part of Figure 4.14, (c) shows suchmeasurements derived from radar.

For comparison global topography is shown in Figure 4.15 along with a plot of its global distribution

in statistical terms. In contrast to (c), the long-wavelength signal in (a) and to a lesser extent in (b)

shows less obvious correlations with surface signals because of isostatic compensation - elevated areas

are supported at depth by low density material. For example there is little expression of the midocean

ridges in the geoid.

We have spent a lot of time defining reference surfaces and gravity fields so that we can meaningfully

define geoid and gravity anomalies. We shall consider these in the next section.

4.7 Gravity anomalies – The effect of buried density anomalies . The computation of a gravity anomaly

caused by a buried density anomaly is relatively straightforward using equation 4.4. Usually we have to

resort to numerical evaluation for arbitrarily shaped bodies but for some simple geometries it is possible

to compute analytically the gravity anomaly. Consider a buried sphere of radiusRwith a uniform densityanomaly !" (the density anomaly is the density difference between the sphere and its surroundings).

Fig. 4.16

116

Fig. 4.16

For an observer at position x on the surface a distance r from the center of the anomaly, the accelerationdue to gravity caused by the anomalous mass is

gm =G

r243πR3∆ρ (from δg =

Gδm

r2)

and points directly towards the center of the anomalous mass. The contribution to the dominant verticalacceleration due to gravity is gm cos θ. To first order, what we measure is this perturbation to the verticalcomponent of g (as shown in the first section) so our gravity anomaly is

∆g = gm cos θ

cos θ can be put in terms of x (the distance from the point on the surface above the anomaly) and b (thedepth of burial of the sphere), i.e.,

cos θ =b

r=

b

(x2 + b2)1/2(r2 = x2 + b2)

so

∆g =G

r243πR3∆ρ

b

(x2 + b2)1/2

=43GπR3∆ρ

b

(x2 + b2)3/2

If we sketch this as a function of position above the anomaly we find that, for a positive ∆ρ it lookssomething like:

119

Horizontal offset

∆

g

Figure 4.17. Left: Anomaly due to a buried sphere. Right: A given gravity anomaly can be causedby an infinite number of different bodies at depths from the surface up to a maximum depth.

If ∆ρ is negative, ∆g is also negative. This is the kind of gravity anomaly found over salt domes and anexample is given in Fig. 4.18.

The computation of gravity anomalies over some other simple bodies is the subject of some homeworkquestions. While it is relatively simple to compute the gravity anomaly caused by a buried densityanomaly, the inverse problem (i.e., inferring the density anomaly from a measured gravity anomaly) isvery difficult. The main problem is that different density anomalies can give rise to exactly the samegravity anomalies. In the above example we can trade off R and ∆ρ so that R3∆ρ stays exactly the sameand ∆g will not be changed. Even if we are willing to assert that we know the geometry of the body(in this case spherical) we do not have a unique answer and we have to supply more information (in thiscase a value for ∆ρ is usually assumed).

body (in this case spherical) we do not have a unique answer and we have to supply more information

(in this case a value for !! is usually assumed).

Fig. 4.18 a) Contour map (.01 mm s!2 contours) of the surface gravity anomaly over a salt domenear the outer edge of the continental shelf off Texas. b) Measurements of gravity on section AA

compared with the theoretical prediction of a spherical body.

4.8 The effect of topography on gravity – Bouguer anomaly . The effect of general topography again

must be treated numerically (sometimes called a terrain correction) but, if the topography is slowly

varying (i.e., it has a shallow slope) we can derive an approximate expression for the gravitational effect

of the topography.

We consider a cylindrical disc of material of radius R and thickness h. An observer is located adistance b above the upper surface of the disc. The density in the disc is assumed to be just a functionof depth, i.e., ! = !(z)

Fig 4.19

Because of the symmetry of the disc we know that the net gravitational attraction will be vertically

119

Fig. 4.18 a) Contour map (.01 mm s−2 contours) of the surface gravity anomaly over a salt domenear the outer edge of the continental shelf off Texas. b) Measurements of gravity on section AAcompared with the theoretical prediction of a spherical body.

4.9 The effect of topography on gravity – Bouguer anomaly . The effect of general topography againmust be treated numerically (sometimes called a terrain correction) but, if the topography is slowlyvarying (i.e., it has a shallow slope) we can derive an approximate expression for the gravitational effectof the topography.

The Bouguer correction makes the approximation that the Earth’s surface is flat and at the sameelevation as the station being considered, i.e. a slab of infinite extent (deviations from this approximationare corrected, if necessary, by making a terrain correction for departure from a simple slab, but these

120

hdm

P

R = r2 + z2

•

Figure 4.19: To compute the Bouguer effect, we integrate the effect of an infinite slab of verticalthickness h on the point P , on the surface of the slab. This is most easily accomplished in a cylindricalcoordinate system.

dz

drr

dΘ

Z

Θ

rdΘ

Figure 4.20: Volume element in cylindrical coordinates.

effects are small except for extreme topography).We can replace the integration over mass in Equation 4-11 with an integration over density

U = −G∫m

dm

R= −G

∫V

ρ(v)R

dv

In cylindrical coordinates, the volume element is now r.dr.dθ.dz. We will assume that ρ is constant so

U = −Gρc∫V

r

Rdrdθdz = −Gρc

∫V

r

(r2 + z2)12drdθdz

Turns out this is a case where we are better off differentiating now rather than later:

gz = −dUdz

= −Gρc∫V

zr

(r2 + z2)32drdθdz

so using integration limits which define an infinte slab we have (as usual the trick to this is to do theintegration in the right order)

gz = −Gρc

h∫0

dz

∞∫0

dr

2π∫0

zr

(r2 + z2)32dθ

Nothing depends on θ so

gz = −2πGρc

h∫0

dz

∞∫0

ρ(z)zr

(r2 + z2)32dr

121

= 2πGρc

h∫0

dzz

(r2 + z2)12

∣∣∣∣∞r=0

= −2πGρc

h∫0

dz

(0− z

(z2)12

)

gz = 2πGρc

h∫0

dz = 2πGρch (4.36)

This is called the Bouguer gravity formula. If topography has a height h and a density ρ its contributionto g will be given by

∆g = 2πGρh (4.37)

(Note that we have taken h as positive downwards.) We can therefore use this equation to correctgravity measurements for the effect of topography. Since we are trying to remove the effect of the slab,this correction is subtracted when the station is above the datum. The average crustal density is about2670 kg/m3 so the Bouguer correction is about 10−6 (m/s2)/m, or 0.1 mgal/m.

An additional effect must also be considered and that is the fact that if we make a gravity measurementat various elevations, we are at different distances from the center of mass of the Earth and so g changes.We can also correct for this effect.

Suppose we make a gravity measurement at a particular position on the surface of the Earth. Tocalculate the gravity anomaly we would first subtract out the reference gravity field, gR, (which has alatitude dependence) evaluated on the reference geoid which is at r0. We may calculate the effect ofelevation in the following way. To zeroth order we have (taking g positive down)

g =GM

r2

so the effect of changing distance from the center of a spherical earth is:

δg

δr= −2

GM

r3= −2

g

rtherefore the anomaly caused by elevation is approximated by:

∆g = −2hr0gR (4.38)

i.e., gravity is reduced by an amount |∆g| when height increases by h. If we add ∆gh = −∆g toour measurement we have corrected for the effect of elevation. This correction is called the free-aircorrection and a gravity anomaly defined by

∆gfa = gobs − gR + ∆gh (4.39)

is called a free-air gravity anomaly. If we make a further correction for the effect of topography usingthe Bouguer gravity formula we have a new gravity anomaly defined by

∆gB = ∆gfa − 2πGρch (4.40)

This is called a Bouguer gravity anomaly.By making all these corrections we are obviously trying to take out all the effects of near surface

structure so that we can see the effect of density anomalies deeper in the Earth. Sometimes correctingfor topography causes large Bouguer gravity anomalies while the free-air anomaly is quite small. Thiseffect is caused by isostatic compensation.

122

4.10 Compensation and isostatic geoid anomalies . The Bouguer gravity formula removes the effectof topography as if the topography were uncompensated. Short wavelength features can be supportedby the lithosphere and so the Bouguer correction effectively accounts for the gravitational attraction ofthe anomalous mass. Long wavelength topography (e.g., a mountain range) causes the lithosphere to“sag" into the mantle and so such “compensated" features have low density roots associated with them.This results in a large negative Bouguer anomaly:

so the effect of changing distance from the center of a spherical earth is:

!g

!r= !2

GM

r3= 2

g

r

therefore the anomaly caused by elevation is approximated by:

!gh =2h

r0gR

i.e., gravity is reduced by an amount !gh where

!gh =2hgR

r0(4.37)

If we add !gh to our measurement we have corrected for the effect of elevation. This correction is

called the free-air correction and a gravity anomaly defined by

!gfa = gobs ! gR + !gh (4.38)

is called a free-air gravity anomaly. If we make a further correction for the effect of topography using

the Bouguer gravity formula we have a new gravity anomaly defined by

!gB = !gfa ! 2"G#ch (4.39)

This is called a Bouguer gravity anomaly.

By making all these corrections we are obviously trying to take out all the effects of near surface

structure so that we can see the effect of density anomalies deeper in the Earth. Sometimes correcting

for topography causes large Bouguer gravity anomalies while the free-air anomaly is quite small. This

effect is caused by isostatic compensation.

4.9 Compensation and isostatic geoid anomalies . The Bouguer gravity formula removes the effect of

topography as if the topography were uncompensated. Short wavelength features can be supported by

the lithosphere and so the Bouguer correction effectively accounts for the gravitational attraction of the

anomalous mass. Long wavelength topography (e.g., a mountain range) causes the lithosphere to “sag”

into the mantle and so such “compensated” features have low density roots associated with them. This

results in a large negative Bouguer anomaly:

Fig 4.21

122

positive FA anomaly

small FA anomaly

Fig 4.21

Let us suppose that topography is perfectly isostatically compensated. In that case we know that thetotal mass in vertical columns of material must be equal. In our picture the total mass in a columnof material under A is the same as the total mass in a column under B. This means that the free-airanomaly must be zero i.e., the only variation in g is due to the fact that at position B we are further fromthe center of the Earth than at position A. The condition for isostatic compensation can be written as

h∫0

∆ρ(z)dz = 0 (4.41)

where ∆ρ is the difference in density between two columns as a function of depth and h is the depth ofcompensation. This equation tells us something about lateral variation in density as a function of depth.It turns out that the geoid anomalies associated with isostatic compensation give us more informationabout ∆ρ(z).

To compute the geoid anomaly over topography we take an approach similar to computing theBouguer effect. Starting again with

U = −G∫V

ρ(z)r

Rdrdθdz

where we allow ρ do vary with depth, we have the change in potential due to a change in density

∆U = −G∫V

∆ρ(z)r

Rdrdθdz

We won’t use an infinite slab this time (you’ll see soon why), but integrate over a disk of radius Ro

∆U = −Gh∫

0

∆ρ(z)dz

Ro∫0

dr

2π∫0

r

(r2 + z2)12dθ

123

= −2πG

h∫0

dz

Ro∫0

∆ρ(z)r

(r2 + z2)12dr

= −2πG

h∫0

dz∆ρ(z)(r2 + z2)12

∣∣∣∣Ro

r=0

= −2πG

h∫0

∆ρ(z)dz[(R2

o + z2)12 − z

]Now you can see why we didn’t integrate r to infinity, but we can say that Ro z so that

∆U ≈ −2πG

h∫0

∆ρ(z)(Ro − z)dz

= −2πG[Ro

h∫0

∆ρ(z)dz −h∫

0

∆ρ(z)zdz]

For perfect isostatic compensation, we have

h∫0

∆ρ(z)dz = 0

so

∆U = 2πG

h∫0

z∆ρ(z)dz (4.42)

Geoid anomalies (∆N) are measured in meters and are related to anomalies in potential by using theequation developed earlier, i.e.,

∆N = −∆UgR

= −2πGgR

h∫0

z∆ρ(z)dz (4.43)

where gR is gravity on the reference geoid. We can use this equation to tell us about ∆ρ(z) – in fact itcan be used to distinguish between different models of compensation.

Isostatic compensation can be achieved in several ways. We discuss two simplified models but inreality compensation is probably achieved by a complicated mixture of these models (and of othereffects).

4.11 Airy compensation . Here isostatic compensation is achieved by varying the thickness of a constantdensity crust:

The thickness of continental crust with zero elevation (i.e., the surface at sea level) is H. Topographyof height h is associated with a low density root of thickness b. Using isostasy we have

ρcH + ρmb = ρc(h+H + b)

which we can rearrange to give

b(ρm − ρc) = ρch (4.44)

124

so

!U = !2!G

!"

#R

h$

0

!"(z)dz !h$

0

!"(z)(z + b)dz

%&

'

= !2!G

!"

#R

h$

0

!"(z)dz !h$

0

!"(z)zdz ! b

h$

0

!"(z)dz

%&

'

For perfect isostatic compensation, we have

h$

0

!"(z)dz = 0

so

!U = 2!G

h$

0

z!"(z)dz (4.41)

Geoid anomalies (!N ) are measured in meters and are related to anomalies in potential by using theequation developed earlier, i.e.,

!N = !!U

gR= !2!G

gR

h$

0

z!"(z)dz (4.42)

where gR is gravity on the reference geoid. We can use this equation to tell us about !"(z) – in fact itcan be used to distinguish between different models of compensation.

Isostatic compensation can be achieved in several ways. We discuss two simplified models but in

reality compensation is probably achieved by a complicated mixture of these models (and of other

effects).

4.10 Airy compensation . Here isostatic compensation is achieved by varying the thickness of a constant

density crust:

Fig 4.22

The thickness of continental crust with zero elevation (i.e., the surface at sea level) isH . Topographyof height h is associated with a low density root of thickness b. Using isostasy we have

"cH + "mb = "c(h + H + b)

124

Fig 4.22

(If the topography is negative and covered with sea water we have a slightly different expression.) Thegeoid anomaly associated with compensated positive topography can be found by applying equation4.43. If we choose sea level as our origin and note that ∆ρ(z) is ρ(z) in column A minus ρ(z) in columnB we have

∆N = −2πGgR

0∫−h

ρczdz +

H∫0

(ρc − ρc)zdz +

H+b∫H

(ρc − ρm)zdz

where we are integrating downwards starting at −h

= −2πGgR

[−ρch

2

2+(ρc − ρm

2

)[(H + b)2 −H2]

]

= +πG

gR

[ρch

2 + (ρm − ρc)(2Hb+ b2)]

we can now eliminate b using equation 4.44 giving

∆N =πG

gRρc

[2Hh+

ρmh2

ρm − ρc

](4.45)

This kind of model can explain the geoid anomaly observed at continental margins quite well – anexample is shown in Figure 4.23. Passive continental margins are close to Airy isostatic equilibrium.

4.12 Pratt compensation . An alternative way to achieve isostatic compensation is to have horizontalvariations in density over some prescribed depth range, W .

Fig 4.24

This kind of model can explain the geoid anomaly observed at continental margins quite well – an

example is shown in Figure 4.23. Passive continental margins are close to Airy isostatic equilibrium.

4.11 Pratt compensation . An alternative way to achieve isostatic compensation is to have horizontal

variations in density over some prescribed depth range,W .

Suppose the reference density corresponding to no topography is !0 and suppose the pth column has

positive topography of height h and density !p. Then

!0W = !p(W + h)

which we can rearrange to give !p

!p =!0W

W + h(4.45)

(A slightly different expression arises for negative topography covered by sea water.) The geoid anomaly

is

!N = !2"G

gR

h!

0

z!!(z)dz

Here !! is the density difference betweeen the pth column and the reference density. Thus

!N = !2"G

gR

"#

$

0!

!h

!pzdz +W!

0

(!p ! !0)zdz

%&

'

= !2"G

gR

(!!ph2

2+ (!p ! !0)

W 2

2

)

Eliminating !p using equation 4.47 gives

!N ="G

gR!0Wh (4.46)

In this case !N is proportional to h. One place where this analysis may be appropriate is over a hotspot such as Hawaii. There is a nearly circular region of elevated topography surrounding the Hawaiian

islands with a radius of about 700 km and a maximum elevation of about 1.5 km. This is called the

Hawaiian swell. We find that W " 100 km best fits the observed geoid anomaly (Figure 4.25). Notethat the Airy model completely fails to explain the observations.

126

Fig 4.24

125

which we can rearrange to give

b =!ch

!m ! !c(4.43)

(If the topography is negative and covered with sea water we have a slightly different expression.) The

geoid anomaly associated with compensated positive topography can be found by applying equation

4.44. If we choose sea level as our origin and note that!!(z) is !(z) in column Aminus !(z) in columnB we have

!N = !2"G

gR

!

"0#

!h

!czdz +H#

0

(!c ! !c)zdz +H+b#

H

(!c ! !m)zdz

$

%

= !2"G

gr

&!!ch2

2+

'!c ! !m

2

([(H + b)2 ! H2]

)

= +"G

gr

*!ch

2 + (!m ! !c)(2Hb + b2)+

we can now eliminate b using equation 4.44 giving

!N ="G

gR!c

&2Hh +

!mh2

!m ! !c

)(4.44)

Fig 4.23 a) Observed geoid anomaly across the Atlantic continental margin of North America

at 40.5" N compared with the predicted anomaly assuming Airy compensation and the densitydistribution shown in (b).

125

Fig 4.23 a) Observed geoid anomaly across the Atlantic continental margin of North Americaat 40.5 N compared with the predicted anomaly assuming Airy compensation and the densitydistribution shown in (b).

Suppose the reference density corresponding to no topography is ρ0 and suppose the pth column haspositive topography of height h and density ρp. Then

ρ0W = ρp(W + h)

which we can rearrange to give ρp

ρp =ρ0W

W + h(4.46)

(A slightly different expression arises for negative topography covered by sea water.) The geoid anomalyis

∆N = −2πGgR

h∫0

z∆ρ(z)dz

Here ∆ρ is the density difference betweeen the pth column and the reference density. Thus

∆N = −2πGgR

0∫−h

ρpzdz +

W∫0

(ρp − ρ0)zdz

= −2πG

gR

[−ρph2

2+ (ρp − ρ0)

W 2

2

]Eliminating ρp using equation 4.46 gives

∆N =πG

gRρ0Wh (4.47)

126

In this case ∆N is proportional to h. One place where this analysis may be appropriate is over a hotspot such as Hawaii. There is a nearly circular region of elevated topography surrounding the Hawaiianislands with a radius of about 700 km and a maximum elevation of about 1.5 km. This is called theHawaiian swell. We find that W ≈ 100 km best fits the observed geoid anomaly (Figure 4.25). Notethat the Airy model completely fails to explain the observations.

Fig. 4.25. Dependence of the observed geoid anomalies on bathymetry across the Hawaiian Swell

and across the Bermuda Swell compared with the predicted dependence for crustal thickening (Airy

compensation) and Pratt compensation with various depths of compensation.

127

Fig. 4.25. Dependence of the observed geoid anomalies on bathymetry across the Hawaiian Swelland across the Bermuda Swell compared with the predicted dependence for crustal thickening (Airycompensation) and Pratt compensation with various depths of compensation.

127