Question

Two particles of equal masses go round a circle of radius R under the action of their mutual gravitational attraction. What would be the speed of each particle? [UK PMT 2015]

Moderate

Solution

The gravitational force between the particles will give the required centripetal force.
So, $\frac{G M^{2}}{(2 R)^{2}} = \frac{M v^{2}}{R} \Rightarrow \frac{G M^{2}}{4 R^{2}} = \frac{M v^{2}}{R} \Rightarrow v = \sqrt{\frac{G M}{4 R}}$

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