The Earth's Gravitational Field

The earth is not a point!

We want to show that if the center of the earth is located at the origin of a 3-dimensional coordinate system, then, to a good approximation, its gravitational field in outer space has the form

F(r) = -(GM/r

) u

(1)

where F(r) is the gravitational force on a unit mass with position vector r, G is a positive constant, M is the mass of the earth, r is the magnitude of r (that is, r = ||r||), and u is a unit vector in the direction of r. We take it as given (by experiment or whatever) that such an inverse square law holds for the gravitational field of a small body which can be idealized as a point, but the earth is not a point. Indeed, most space missions take place near the earth, where astronauts have noticed that the earth looks very large, very beautiful, and not at all like a point!

Nevertheless, the gravitational field of the earth behaves like that of a point, at least in outer space (inside the earth it's a different story). To start with, we need to prove the following result.

If a closed surface S encloses a point mass m, then the total inward flux of the gravitational field of m through S is 4Gm.

The fact that the flux is inward makes sense, since the gravitational field of the mass m is attractive and points toward the mass, not away from it. "Encloses" means that a small sphere S' centered at m with radius r can be contained inside S (see the figure below).

As mentioned above, we are assuming that the gravitational field F of the point mass m has magnitude ||F||=Gm/r at a distance r from m, and F points toward m. In this case, the symmetry makes the flux (red arrows) through the sphere S' easy to compute. It's just (surface area of S')(||F||) = (4r)(Gm/r) = 4Gm. This value is just what we wanted, but what about the flux (green arrows) through the arbitrary surface S?

The key is to remember that the divergence of an inverse square field is always 0, that is, div F = 0 everywhere, except where the mass m is located. In particular, div F = 0 throughout the volume V between the surfaces S' and S. But the divergence theorem tells us that the total flux leaving V is equal to the triple integral of div F taken over all of V. Since div F = 0, we conclude that the total flux leaving V is 0. This means that

0 = (total flux leaving V) = (flux leaving V through S') - (flux entering V through S)

and hence

(flux entering V through S) = (flux leaving V through S') = 4

Gm.

Therefore, the total inward flux through S is 4Gm, which is what we wanted to show.

Getting back to the gravitational field of the earth, let us assume that the mass distribution of the earth is spherically symmetric, which is approximately true. The spherical symmetry implies that any force vector F points directly towards the center of the earth, and thus F is parallel to -u, as indicated by equation (1). Also, the spherical symmetry implies that the magnitude F of F depends only on the distance r from the center of the earth, so we can write ||F|| = F(r). This means that F has the form

F(r) = -F(r) u,

so in view of equation (1), our problem boils down to showing that

F(r) = GM/r

(2)

We will idealize the earth as consisting of a spherically symmetric arrangement of n point masses m, m,..., m (like atoms or molecules), where n is a very large number. The point masses have gravitational fields F, F,..., F. It is well-known from experiment that the effects of the fields add; that is, the field F of the earth is just F = F + F + ... + F, and the mass M of the earth is of course M = m + m + ... + m. It follows that the inward flux of F through a sphere of radius r concentric with the earth is the sum of the flux contributions of the fields F, F,..., F. By our previous result, each field F makes a contribution of 4Gm and thus the total inward flux through the sphere is