Question

Calculate the self-gravitational potential energy of matter forming (a) a thin uniform shell of mass M and radius R, and (b) a uniform sphere of mass m and radius R.

Answer 1

Hint: Gravitational self potential energy is the amount of work done to create an object of mass M formed by joining small masses of mass dm which were brought from infinity. Clearly, integration has to be performed by considering the potential created by for small mass m first.
Formula used:
Gravitational potential due to a mass m is:
$V={\displaystyle \frac{-Gm}{r}}$

Complete answer:
(a) Consider a spherical shell of radius R. Let initially, a mass of m be equally distributed over this giving a potential of:
$V={\displaystyle \frac{-Gm}{R}}$
on its surface.
We now bring a mass dm from infinity and add to this. Therefore, the work done in bringing the mass dm to a point on the surface of the sphere,
$dW=Vdm={\displaystyle \frac{-Gmdm}{R}}$
To find out the total work done in creating a mass of M, we integrate from 0 to M.
$U={\int }_0^M{\displaystyle \frac{-Gm}{R}}dm={\displaystyle \frac{-G}{R}}{\left({\displaystyle \frac{m^2}{2}}\right)}_0^M={\displaystyle \frac{-G{M}^2}{2R}}$

This is nothing but the gravitational self potential energy of the spherical shell of mass M and radius R.

(b) For the case of a solid sphere of radius R, there is a uniform mass distribution inside the sphere so it is not as simple as for the case of a shell. The density of the sphere has to be:
$d={\displaystyle \frac{M}{volume}}={\displaystyle \frac{M}{(4/3)\pi R^3}}$

Now, in case of solid sphere, we perform mass additions in the form of shells of radius r, thickness dr so, we can write:
$dm=d\times 4\pi r^{2}dr$,
assuming the volume of the shell to be simply $4\pi r^{2}dr$.
Before adding this shell there was an initial mass m in the system constituting a sphere of radius r which is written as:
$m=d\times {\displaystyle \frac{4}{3}}\pi r^3$ .
Work done in addition of the shell to this mass m is:
$dW=Vdm={\displaystyle \frac{-Gmdm}{r}}$ .
Upon substituting for m and dm we get:
$dW={\displaystyle \frac{-16}{3}}G{\pi }^2{d}^2{r}^{4}dr$ .
Integrating on r from the limits 0 to R, we get:
$U={\displaystyle \frac{-16}{3}}G{\pi }^2{d}^2{\int }_0^R{r}^{4}dr$
$U={\displaystyle \frac{-16}{3}}G{\pi }^2{d}^2{\left({\displaystyle \frac{r^5}{5}}\right)}_0^R$
Upon keeping the limits and substituting the value of d, we get the:
$U={\displaystyle \frac{-16}{3}}G{\pi }^2{\left({\displaystyle \frac{M}{(4/3)\pi R^3}}\right)}^2\left({\displaystyle \frac{R^5}{5}}\right)={\displaystyle \frac{-3G{M}^2}{5R}}$ .

This is the required gravitational self potential energy for the case of a solid sphere.
So, for the case of (a) a thin uniform shell we have ${\displaystyle \frac{-G{M}^2}{2R}}$ and for the case of (b) a uniform sphere of mass m and radius R we have ${\displaystyle \frac{-3G{M}^2}{5R}}$.

Note:
Consider the case of electric potential, work done on a charge to bring it from infinity to a point in the vicinity of another charge with potential V is qV. Gravitational force is the force acting between masses therefore we wrote work to be Vdm. One can perform dimensional analysis if any confusion is present regarding this formula.