Question

A small particle of mass m lies on the axis of a ring of mass $M$ and radius a at a distance a from the centre. The particle reaches the centre under gravitational attraction only. Its speed at the centre will be

• A. $\sqrt{\frac{2GM}{a}}$
• B. $\sqrt{\frac{2GM}{a}(\sqrt{2}-1)}$
• C. $\sqrt{\frac{2GM}{a}(1-\frac{1}{\sqrt{2}})}$
• D. $0$

Answer 1

Answer: (C)

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

We know that, Potential energy due to a ring of radius 'a' at distance x from the center on the axis is given by

$V=\frac{-GMm}{\sqrt{a^2+x^2}}$

applying energy conservation :

Loss in potential Energy = gain in kinetic Energy

Loss in $PE=P{E}_i-P{E}_f=\frac{GMm}{a}(1-\frac{1}{\sqrt{2}})$

Gain in $KE=\frac{1}{2}m{v}^2$

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

Option $C$ is right answer.