9—Vector Calculus 1 226
density,dm/dV of the matter that is generating the gravitational field. Gis Newton’s gravitational
constant:G= 6. 67 × 10 −^11 N.m^2 /kg^2.
rˆ
r
For the first example of solutions to these equations, take the case of a spherically symmetric
mass that is the source of a gravitational field. Assume also that its density is constant inside; the
total mass isM and it occupies a sphere of radiusR. Whatever~gis, it has only a radial component,
~g=grˆr. Proof: Suppose it has a sideways component at some point. Rotate the whole system by
180 ◦about an axis that passes through this point and through the center of the sphere. The system
doesn’t change because of this, but the sideways component of~gwould reverse. That can’t happen.
The componentgrcan’t depend on eitherθorφbecause the source doesn’t change if you rotate
it about any axis; it’s spherically symmetric.
~g=gr(r)rˆ (9.37)
Now compute the divergence and the curl of this field. Use Eqs. (9.16) and (9.33) to get
∇.gr(r)rˆ=
1
r^2
d
(
r^2 gr
)
dr
and ∇×gr(r)rˆ= 0
The first equation says that the divergence of~gis proportional toρ.
1
r^2
d
(
r^2 gr
)
dr
=− 4 πGρ (9.38)
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: CandC′. The latter has to be zero, because
C′/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. (9.38) 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
