Two particles of equal mass m move in a circl | vedclass

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(C) The centripetal force required for circular motion is provided by the gravitational force of attraction between the two particles.The distance bet

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$v = \frac{1}{2}\sqrt {\frac{{Gm}}{R}} $

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(C) The centripetal force required for circular motion is provided by the gravitational force of attraction between the two particles.The distance between the two particles is $2R$.The gravitational force between them is $F_g = \frac{G \cdot m \cdot m}{(2R)^2} = \frac{Gm^2}{4R^2}$.The centripetal force required for a particle of mass $m$ moving with speed $v$ in a circle of radius $R$ 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^2}{4R^2} \cdot \frac{R}{m}$$v^2 = \frac{Gm}{4R}$$v = \sqrt{\frac{Gm}{4R}} = \frac{1}{2}\sqrt{\frac{Gm}{R}}$

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

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