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

In the process of formation of a body, work has to be done by an external agent bit by bit to build up the body. This energy of the external agent is stored as gravitational energy. This is called the self-gravitation potential energy or gravitational energy of mutual gravitational interaction.
a. Consider a sphere of any radius x.
Mass of the sphere $=4\pi /3$ $x^3\rho$ where $\rho =$ density of mass
Gravitational potential of the surface $=-4\pi /3$ G $\rho x^2$
This is also the work done in adding unit mass to the sphere by the external agent. When the thickness is increased by dx, mass added by the agent is $4\pi x^{2}dx\rho$.
Therefore, work done by the agent in increasing the surface from x to x is dx
$(-\frac{4\pi }{3}G\rho x^2)\left(4\pi x^{2}dx\rho \right)=\frac{16{\pi }^2}{3}G{\rho }^2{x}^{6}dx$
Therefore, total work done by the agent
$=-\frac{16{\pi }^{2}G{\rho }^2}{3}{\int }_0^R{x}^{4}dx=-\frac{16{\pi }^{2}G{\rho }^2{R}^5}{15}$
Now, $\rho =\frac{m}{\frac{4\pi }{3}{R}^3}=\frac{3m}{4\pi R^3}$
Therefore, U(self-energy)$=-\frac{16{\pi }^2{R}^{5}G}{15}\times \frac{9m^2}{16\pi 2R^6}$
$=-\frac{3}{5}\frac{G{M}^2}{R}$
b. Consider the shell when mass m has already been piled up by the agent.
Then, potential of the shell$=-GM/R$
This is also the work done by the agent in adding another unit mass to the shell. Therefore, the elementary work done in adding an elementary mass dm is $-Gm/R$ dm. Therefore, net work done by the agent in piling the entire mass
$=-{\int }_0^m\frac{GMdm}{R}=-\frac{G{m}^2}{2R}$
Therefore, U(self-potential energy)$=-\frac{1}{2}G{m}^2/R$.