Two particles of equal mass 'm' move in a cir | vedclass

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Question

(D) 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

$\sqrt{\frac{ Gm }{4 r }}$

Vedclass · Answer

(D) The gravitational force between the two particles acts as the centripetal force required 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 speed '$v$' is $F_c = \frac{mv^2}{r}$.Equating the two forces: $\frac{Gm^2}{4r^2} = \frac{mv^2}{r}$.Solving for '$v$': $v^2 = \frac{Gm^2}{4r^2} \cdot \frac{r}{m} = \frac{Gm}{4r}$.Therefore,$v = \sqrt{\frac{Gm}{4r}}$.

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 will be:

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