Question

A point P lies on the axis of a fixed ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. If speed at C is $\sqrt{\frac{2GM}{a}(1-\frac{1}{\sqrt{x}})}$. The value of $x$ is:

Answer 1

Initially, each point mass on the ring is at a distance of $\sqrt{2}a$ from m and finally it is at a distance $a$. Hence, conserving total energy of the system, we get

$-\frac{GMm}{\sqrt{2}a}=-\frac{GMm}{a}+\frac{1}{2}m{v}^2$

$v=\sqrt{\frac{2GM}{a}(1-\frac{1}{\sqrt{2}})}$

Therefore, $x=2$