Mathematical Tools for Physics

9—Vector Calculus 1 267

Outside the surfacer=R, the mass density is zero, so this is

1

r^2

d

(

r^2 gr

)

dr

= 0, implying r^2 gr=C, and gr=

C

r^2
whereCis some as yet undetermined constant. Now do this inside.

1

r^2

d

(

r^2 gr

)

dr

=− 4 πGρ 0 , where ρ 0 = 3M/ 4 πR^3

This is
d

(

r^2 gr

)

dr

=− 4 πGρ 0 r^2 , so r^2 gr=−

4

3

πGρ 0 r^3 +C′,

or gr(r) =−

4

3

πGρ 0 r+

C′

r^2
There are two constants that you have to evaluate:C andC′. The latter has to be zero, becauseC′/r^2 → ∞
asr→ 0 , and there’s nothing in the mass distribution that will cause this. As for the other, note thatgrmust
be continuous at the surface of the mass. If it isn’t, then when you try to differentiate it in Eq. ( 34 ) you’ll be
differentiating a step function and you get an infinite derivative there (and the mass density isn’t infinite there).

gr(R−) =−

4

3

πGρ 0 R=gr(R+) =

C

R^2

Solve 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)

(35)

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 problem 14.