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(B) The gravitational force between the two particles provides the necessary centripetal force for circular motion.The distance between the two partic

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$\frac{4\pi r^{3/2}}{\sqrt{Gm}}$

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(B) The gravitational force between the two particles provides the necessary centripetal force for circular motion.The distance between the two particles is $2r$.The gravitational force is $F = \frac{G \cdot m \cdot m}{(2r)^2} = \frac{Gm^2}{4r^2}$.The centripetal force required for a particle of mass $m$ moving in a circle of radius $r$ with velocity $v$ is $F_c = \frac{mv^2}{r}$.Equating the two forces: $\frac{mv^2}{r} = \frac{Gm^2}{4r^2}$.Solving for $v$: $v^2 = \frac{Gm}{4r} \implies v = \frac{1}{2} \sqrt{\frac{Gm}{r}}$.The time period $T$ is given by $T = \frac{2\pi r}{v}$.Substituting the value of $v$: $T = \frac{2\pi r}{\frac{1}{2} \sqrt{\frac{Gm}{r}}} = 4\pi r \sqrt{\frac{r}{Gm}} = \frac{4\pi r^{3/2}}{\sqrt{Gm}}$.

Two particles of equal mass $m$ each move in a circle of radius $r$ under the action of their mutual gravitational attraction. The time period of each particle is

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