Question

Consider a nebula in the form of a ring of radius $R$ and mass $M$. A star of mass $m(m<<M)$ is located at distance $R$ from the centre of the ring on its axis, initially at rest. The speed with which it crosses the centre of the ring is $v=\sqrt{(p-\sqrt{q})\frac{GM}{R}}.$ Find $(p+q).$

Answer 1

Answer: 4

In the figure,
Total energy at $P=$ Total energy at $O$

$-\frac{GmM}{\sqrt{2}R}=\frac{1}{2}m{v}^2-\frac{GmM}{R}$
$\therefore v=\sqrt{(2-\sqrt{2})\frac{GM}{R}}$