First slide

Gravitaional potential energy

Question

Two thin rings having masses $m_{1}$ and $m_{2}$ , each of radius R, are coaxially placed at a distance R. Then the work done in moving a particle of mass m from the centre of one ring to that of the other is

Moderate

A

Zero
B

$\frac{Gm (m_{1} - m_{2}) (\sqrt{2} - 1)}{\sqrt{2} R}$
C

$\frac{\sqrt{2} Gm (m_{1} + m_{2})}{R}$
D

$\frac{Gm (\sqrt{2} m_{1} + m_{2})}{R}$

Solution

$V_{1} = - \frac{{Gm}_{1}}{R} - \frac{{Gm}_{2}}{\sqrt{2} R} V_{2} = - \frac{{Gm}_{2}}{R} - \frac{{Gm}_{1}}{\sqrt{2} R} ∴ ΔV = V_{2} - V_{1} = G (m_{1} - m_{2}) (\frac{1}{R} - \frac{1}{\sqrt{2} R}) Hence, W = m . ΔV = \frac{G . m (m_{1} - m_{2}) (\sqrt{2} - 1)}{\sqrt{2} R}$

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