Mathematical Tools for Physics

9—Vector Calculus 1 268

9.9 Gravitational Potential
The gravitational potential is that functionV for which

~g=−∇V (36)

That such a function evenexistsis not instantly obvious, but it is a consequence of the second of the two defining
equations ( 32 ). If you grant that, then you can get an immediate equation forV by substituting it into the first
of ( 32 ).
∇.~g=−∇.∇V =− 4 πGρ, or ∇^2 V = 4πGρ (37)

This is a scalar equation instead of a vector equation, so it will often be easier to handle. Apply it to the same
example as above, the uniform spherical mass.
The Laplacian,∇^2 is the divergence of the gradient, so to express it in spherical coordinates, combine
Eqs. ( 24 ) and ( 27 ).

∇^2 V =

1

r^2

∂

∂r

(

r^2

∂V

∂r

)

+

1

r^2 sinθ

∂

∂θ

(

sinθ

∂V

∂θ

)

+

1

r^2 sin^2 θ

∂^2 V

∂φ^2

(38)

Because the mass is spherical it doesn’t change no matter how you rotate it so the same thing holds for
the solution,V(r). Use this spherical coordinate representation of∇^2 and for this case theθandφderivatives
vanish.
1
r^2

d

dr

(

r^2

dV

dr

)

= 4πGρ(r) (39)

I changed from∂todbecause there’s now only one independent variable. Just as with Eq. ( 34 ) I’ll divide this
into two cases, inside and outside.

Outside:

1

r^2

d

dr

(

r^2

dV

dr

)

= 0, so r^2

dV

dr

=C

Continue solving this and you have

dV

dr

=

C

r^2

−→V(r) =−

C

r

+D (r > R) (40)