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rG

A = 1=4 . Note that this result is independent of the radius R that you have used.

Problem f: Show that the Green's function is given by:

G(r r0 ) = 41 1 r0 : (17.8)

;

jr ; j

With (17.4) this implies that the gravitational potential is given by:

Z (r0 ) dV 0 :

V (r) = (17.9)

r0

;G

jr ; j

This general expression is useful for a variety of di erent purposes, we will make exten-

sive use of it. By taking the gradient of this expression one obtains the gravitational

acceleration g. This acceleration was also derived in equation (6.5) of section 6.2 for the

special case of a spherically symmetric mass distribution. Surprisingly it is a non-trivial

calculation to derive (6.5) by taking the gradient of (17.9).

17.2 Upward continuation in a at geometry

Suppose one has body with variable mass in two dimensions and that the mass density is

only nonzero in the half-space z < 0. In this section we will determine the gravitational

potential V above the half-space when the potential is speci ed at the plane z = 0 that

forms the upper boundary of this body. The geometry of this problem is sketched in

gure 17.2. This problem is of relevance for the interpretation of gravity measurements

CHAPTER 17. POTENTIAL THEORY

244

V(x,z) z>0

V(x,z=0)

z=0

Figure 17.2: Geometry of the upward continuation problem. A mass anomaly (shaded)

leaves an imprint on the potential at z = 0. The upward continuation problem states how

the potential at the surface z = 0 is related to the potential V (x z) at greater height.

taken above the Earth's surface using aircraft or satellites orbits because the rst step

in this interpretation is to relate the values of the potential at the Earth's surface to the

measurements taken above the surface. This process is called upward continuation.

Mathematically the problem can be stated this way. Suppose one is given the function

V (x z = 0), what is the function V (x z)? When we know that there is no mass above the

surface it follows from (17.1) that the potential satis es:

2 V (r) = 0 for z > 0 : (17.10)

r

It is instructive to solve this problem by making a Fourier expansion of the potential in

Z1

the variable x:

v(k z)eikx dk :

V (x z) = (17.11)

;1

Problem a: Show that for z = 0 the Fourier coe cients can be expressed in the known

value of the potential at the edge of the half-space:

1 Z 1 V (x z = 0)e;ikx dx :

v(k z = 0) = 2 (17.12)

;1

Problem b: Use Poisson's equation (17.10) and the Fourier expansion (17.11) to derive

that the Fourier components of the potential satisfy for z > 0 the following di eren-

tial equation:

@ 2 v(k z) k2v(k z) = 0 : (17.13)

@z2 ;

Problem c: Show that the general solution of this di erential equation can be written as

v(k z) = A(k) exp (+ z) + B(k) exp (; z) and explain why the absolute values

jkj jkj

of the wave number k in the exponent can be taken .

the coe cient A(k)

Since the potential must remain nite at great height (z ! 1)

must be equal to zero. Setting z = 0 shows that B(k) = v(k z = 0), so that the potential

is given by: Z1

v(k z = 0)eikx e;jkjz dk :

V (x z) = (17.14)

;1

This expression is interesting because it states that the di erent Fourier components of

the potential decay as exp (; z) with height.

jkj

17.2. UPWARD CONTINUATION IN A FLAT GEOMETRY 245

Problem d: Explain that this implies that the short-wavelength components in the po-

tential eld decay must faster with height than the long-wavelength components.

The decrease of the Fourier components with the distance z to the surface is bad news when

one wants to infer the mass-density in the body from measurements of the potential or from

gravity at great height above the surface because the in uence of mass perturbations on

the gravitational eld decays rapidly with height, the measurement of the short-wavelength

component of the potential at great height therefore carries virtually no information about

the density distribution within the Earth. This is the reason why gravity measurements

from space are preferably carried out using satellites in low orbits rather than in high

orbits. Similarly, for gravity surveys at sea a gravity meter has been developed that is

towed far below the sea surface 70]. The idea is that by towing the gravity meter closer

to the sea-bottom, the gravity signal generated at the sub-surface for short wavelength

su ers less from the exponential decay due to upward continuation.

Problem e: Take the gradient of (17.14) to nd the vertical component of the gravity

eld. Use the resulting expression to show that the gravity eld g is less sensitive

to the exponential decay due to upward continuation than the potential V .

This last result is the reasons why satellites in low orbits are used for measuring the Earth's

gravitational potential and why for satellites in high orbits one measures gravity. In fact,

presently a space-born gradiometer 50] is presently being developed. This instrument

measures the gradient of the gravity vector by monitoring the di erential motion between

two masses in the satellite. Taking the gradient of the gravity leads to another factor of k

in the Fourier expansion so that the e ects of upward continuation are further reduced.

We will now explicitly express the potential at height z to the potential at the surface

z = 0.

Problem f: Insert expression (17.12) in (17.14) to show that the upward continuation of

the potential is given by:

Z1

H(x x0 z)V (x0 z = 0)dx0

V (x z) = (17.15)

;1

;

1 Z 1 e;jkjz e;ikxdk :

with

H(x z) = 2 (17.16)

;1

Note that expression (17.15) has exactly the same structure as equation (11.57) of section

11.7 for a time-independent linear lter. The only di erence is that the variable x now plays

the role the role of the variable t in expression (11.57). This means that we can consider

upward continuation as a linear ltering operation. The convolutional lter H(x z) maps

the potential from the surface z = 0 onto the potential at height z.

Problem g: Show that this lter is given by:

z

H(x z) = 1 z2 + x2 : (17.17)

CHAPTER 17. POTENTIAL THEORY

246

Problem h: Sketch this lter as a function of x for a large value of z and a small value

of z.

Problem i: Equation (17.15) implies that H(x z) = (x), with (x) the Dirac delta

function. Convince yourself of this by showing that H(x z) becomes more and more

peaked round x = 0 when z 0 and by proving that for all values of z the lter

R1 !

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