9—Vector Calculus 1 227Solve forCand you have
C=−
4
3
πGρ 0 R^3 =−
4
3
πG
3 M
4 πR^3
R^3 =−GM
Put this all together and express the densityρ 0 in terms ofM andRto get
gr R
gr(r) =
{
−GM/r^2 (r > R)
−GMr/R^3 (r < R)
(9.39)
This says that outside the spherical mass distribution you can’t tell what its radiusRis. It
creates the same gravitational field as a point mass. Inside the uniform sphere, the field drops to zero
linearly toward the center. For a slight variation on how to do this calculation see problem9.14.
Non-uniform density
The density of the Earth is not uniform; it’s bigger in the center. The gravitational field even increases
as you go down below the Earth’s surface. What does this tell you about the density function?∇.~g=
− 4 πGρremains true, and I’ll continue to make the approximation of spherical symmetry, so this is
1
r^2
d
(
r^2 gr
)
dr
=
dgr
dr
+
2
r
gr=− 4 πGρ(r) (9.40)
That gravity increases with depth (for a little while) says
dgr
dr
=− 4 πGρ(r)−
2
r
gr> 0
Why> 0? Remember:gris itself negative andris measured outward. Sort out the signs. I can solve
for the density to get
ρ(r)<−
1
2 πGr
gr
At the surface,gr(R) =−GM/R^2 , so this is
ρ(R)<
M
2 πR^3
=
2
3
.^3 M
4 πR^3
=
2
3
ρaverage
The mean density of the Earth is 5. 5 gram/cm^3 , so this bound is 3. 7 gram/cm^3. Pick up a random
rock. What is its density?
9.9 Gravitational Potential
The gravitational potential is that functionV for which
~g=−∇V (9.41)
That such a function evenexistsis not instantly obvious, but it is a consequence of the second of the
two defining equations (9.36). If you grant that, then you can get an immediate equation forV by
substituting it into the first of (9.36).