Two identical particles each of mass m go rou | vedclass

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(C) The gravitational force between the two particles acts as the centripetal force required for circular motion.The distance between the two particle

Vedclass · Accepted Answer

$\sqrt{\frac{G m}{4 a^3}}$

Vedclass · Answer

(C) The gravitational force between the two particles acts as the centripetal force required for circular motion.The distance between the two particles is $r = 2a$.The gravitational force is given by $F = \frac{G m m}{(2a)^2} = \frac{G m^2}{4a^2}$.The centripetal force required for a particle of mass $m$ to move in a circle of radius $a$ with angular speed $\omega$ is $F_c = m \omega^2 a$.Equating the two forces: $\frac{G m^2}{4a^2} = m \omega^2 a$.$\omega^2 = \frac{G m}{4 a^3}$.Therefore,the angular speed is $\omega = \sqrt{\frac{G m}{4 a^3}}$.

Two identical particles each of mass $m$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be

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