A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\dfrac{R}{2}$, and the other mass, in a circular orbit of radius $\dfrac{3R}{2}$. The difference between the final and the initial total energies is
Answer: (C) $-\dfrac{GMm}{6R}$
Total energy of a mass $\mu$ in a circular orbit of radius $r$ is $E = -\dfrac{GM\mu}{2r}$.
Initial: $E_i = -\dfrac{GMm}{2R}$.
Final, two masses $\dfrac m2$:
$$E_f = -\frac{GM(m/2)}{2(R/2)} - \frac{GM(m/2)}{2(3R/2)} = -\frac{GMm}{2R} - \frac{GMm}{6R} = -\frac{2GMm}{3R}$$
Difference:
$$E_f - E_i = -\frac{2GMm}{3R} + \frac{GMm}{2R} = -\frac{GMm}{6R}$$
Solution by Sreeraj P, M.Sc Physics