Question

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 $\frac{R}{2}$, and the other mass, in a circular orbit of radius $\frac{3R}{2}$. The difference between the final and initial total energies is :

• A. $ - \frac{GM m}{2R}$
• B. $ + \frac{GM m}{6R}$
• C. $ \frac{GM m}{2R}$
• D. $ - \frac{GM m}{6R}$

Answer 1

Answer: (D)

Initial energy $=\frac{-G M m}{R}+\frac{1}{2} \frac{G M m}{R}=\frac{-G M m}{2 R}$
Final Energy $=\frac{-G M(m / 2)}{R / 2}-\frac{G M(m / 2)}{3 R / 2}+\frac{1}{2} \frac{G M(m / 2)}{R / 2}+\frac{1}{2} \frac{G M(m / 2)}{3 R / 2}$
$=\frac{-G M(m / 2)}{2 R / 2}-\frac{G m(m / 2)}{2 \times 3 R / 2}$
Therefore, $\Delta E= $ Final energy $-$ Initial energy
$=\frac{-G M(m / 2)}{2 R / 2}-\frac{G M(m / 2)}{2 \times 3 R / 2}-\left(\frac{-G M m}{2 R}\right)$
$\Rightarrow \Delta E=\frac{-G M m}{6 R}$