Question

A point $P$ lies on the axis of a ring of mass $M$ and radius $a$ at a distance $a$ from its centre $O$ . A small particle of mass $m$ starts from $P$ and reaches $O$ under gravitational attraction only. Its speed when it reaches $O$ is

• A. $\sqrt{\frac{2 G M}{a}}$
• B. $\sqrt{\frac{G M}{a}}$
• C. $\sqrt{\frac{2 G M}{a} \left(\sqrt{2} - 1\right)}$
• D. $\sqrt{\frac{2 G M}{a} \left(1 - \frac{1}{\sqrt{2}}\right)}$

Answer 1

Answer: (D)

The gravitational potential at P, $V_{P}=-\frac{G M}{A P}$
$V_{P}=-\frac{G M}{\sqrt{a^{2} + a^{2}}}=-\frac{G M}{\sqrt{2} a}$ and gravitational potential at O, $V_{O}=-\frac{G M}{a}$
Let v be the velocity of the particle when it reaches O.
$\therefore $ K.E. at $O=\frac{1}{2}mv^{2}$
Gravitational P.E. = Gravitational potential × mass
$\therefore $ P.E. at $P=-\frac{G M m}{\sqrt{2} a}$ and P.E. at $O=-\frac{G M m}{a}$
By the principle of conservation of mechanical energy
$\frac{1}{2}mv^{2}-\frac{G M m}{a}=-\frac{G M m}{\sqrt{2} a}$
$\therefore v^{2}=2\left[\frac{G M m}{a} - \frac{G M m}{\sqrt{2} a}\right]=\frac{2 G M}{a}\left[1 - \frac{1}{\sqrt{2}}\right]$
$\therefore v=\sqrt{\frac{2 G M}{a} \left(1 - \frac{1}{\sqrt{2}}\right)}$