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

Answer 1

Answer: (D)

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