First slide

Gravitational potential and potential energy

Question

A point P is on the axis of a fixed ring of mass M and radius R, at a distance 2R from the centre O. A small particle starts from P and reaches O under the gravitational attraction only. Its speed at O will be

Moderate

A

zero
B

$\sqrt{\frac{2 GM}{R}}$
C

$\sqrt{\frac{2 GM}{R} (\sqrt{5} - 1)}$
D

$\sqrt{\frac{2 GM}{R} (1 - \frac{1}{\sqrt{5}})}$

Solution

Gravitational potential at ‘P’,
$v_{p} = \frac{- GM}{\sqrt{5} R}$
Gravitational potential at ‘O’,
$v_{o} = - \frac{GM}{R}$
Using work energy theorem,
$\begin{array}{l} W = ΔK \Rightarrow m [v_{P} - v_{O}] = \frac{1}{2} {mv}^{2} \\ m [\frac{GM}{R} - \frac{GM}{\sqrt{5} R}] = \frac{1}{2} {mv}^{2} \\ \sqrt{\frac{2 GM}{R} [1 - \frac{1}{\sqrt{5}}]} = v \end{array}$

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