Question

If the proper potential energy of gravitational interaction of matter forming a thin uniform spherical layer of mass $m$ and radius $R$ is given as $U=-\frac{\gamma m^2}{xR}$. Find $x$

Answer 1

Since the each pint on the sphere is at a constant radius, every point is at a constant potential. Therefore, potential of each point will be $\rho =-\frac{\gamma m}{R}$. Now the energy of this point with another small mass $dm$ will be $dE=\rho dm=-\frac{\gamma m}{R}dm$

Therefore, total gravitational potential energy of interaction will be,

$E=\frac{1}{2}{\int }_0^{m}dE=\frac{1}{2}-\frac{\gamma m}{R}{\int }_0^{m}dm=-\frac{\gamma m^2}{2R}$ since while integrating, we have taken same

interaction twice and hence a factor of $\frac{1}{2}$ is required.