Mathematical Tools for Physics

9—Vector Calculus 1 266

9.8 Applications to Gravity
The basic equations to describe the gravitational field in Newton’s theory are

∇.~g=− 4 πGρ, and ∇×~g= 0 (32)

In these equations, the vector field~gis defined by placing a (very small) test massmat a point and measuring
the gravitational force on it. This force is proportional tomitself, and the proportionality factor is called the
gravitational field~g. The other symbol used here isρ, and that is the volume mass 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.

ˆr

r

For the first example of solutions to these equations, take the case of a mass that is the source of a
gravitational field and that is spherically symmetric. The total mass isM and it occupies a sphere of radius
Rwith a uniform mass density. Whatever~gis, it has only a radial component,~g =grrˆ. 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)ˆr (33)

Now compute the divergence and the curl of this field. Use Eqs. ( 16 ) and ( 29 ) to get

∇.gr(r)ˆr=

1

r^2

d

(

r^2 gr

)

dr

and ∇×gr(r)ˆr= 0

The first equation says that the divergence of~gis proportional toρ.

1

r^2

d

(

r^2 gr

)

dr

=− 4 πGρ (34)