Page 346 - Physical Principles of Sedimentary Basin Analysis

328 Gravity and gravity anomalies

when the density is zero. We have already seen that the integral (10.22) gives the potential
U, and Exercise 10.6 shows that it is a solution of Poisson’s equation. One might won-
der why we need Poisson’s equation for the potential when we already have a solution.
The reason is that we don’t always know the density distribution, which is required in the
solution (10.22), but instead we might for example know the potential along the xy-plane.
The potential above the plane is then found as a solution of Laplace’s equation using the
potential at the xy-plane as a boundary condition.
Note 10.2 Newton’s law of gravitation applies to point masses, and a point mass is a finite
mass placed at a point, which by definition has no volume. The normal definition of density
doesn’t make sense when something has a zero volume. Yet it is still possible to work with
a point mass as if it has a density, by using the Dirac δ-function. The δ-function has two
properties: δ(r) is zero everywhere except at r = 0, and the integral over any region V that

includes r = 0is δ(r) dV = 1. The “density” of a point mass M placed in position r 0
V
is therefore
(r) = M δ(r − r 0 ). (10.49)
The integral (10.22) gives the potential in position r, and we have

δ(r − r 0 ) GM

U(r) =−GM dV (r ) =− (10.50)

V |r − r| |r − r 0 |
where V is any volume sufficiently large to include both positions r and r 0 .Wenow
have that the density (10.49) gives the potential (10.50). The potential is also a solution
of Poisson’s equation (10.47) with the “density” of a point mass, and we therefore have

2 1
∇ =−4πδ(r − r 0 ). (10.51)
|r − r 0 |
That this really is the case is checked in Exercise 10.5.

Exercise 10.5
(a) Show that
1

2
∇ = 0 (10.52)
r
for r = 0 by differentiation.
(b) Show that

2 1
∇ =−4π (10.53)
r
for r = 0 using a volume integral that covers r = 0.
2
Solution: (a) A direct calculation shows that ∇ (1/r) = 0for r = 0, where the distance
2
2
2
2 1/2
2
from the origin is r = (x + y + z ) . We get that dr/dx = x/r and that d r/dx =
2
2
2
2
5
−(r − 3x )/r , and similar calculations of d r/dy 2 and d r/dz 2 then give that
2
∇ (1/r) = 0.