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(A) The two particles of mass $m$ are moving in a circle of radius $r$. The distance between them is the diameter of the circle,which is $2r$.The grav

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$\sqrt{\frac{G m}{4 r}}$

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(A) The two particles of mass $m$ are moving in a circle of radius $r$. The distance between them is the diameter of the circle,which is $2r$.The gravitational force between the two particles acts as the centripetal force required for circular motion.The gravitational force is given by $F_g = \frac{G m m}{(2r)^2} = \frac{G m^2}{4r^2}$.The centripetal force required for a particle of mass $m$ moving with speed $v$ in a circle of radius $r$ is $F_c = \frac{m v^2}{r}$.Equating the two forces: $\frac{m v^2}{r} = \frac{G m^2}{4r^2}$.Solving for $v^2$: $v^2 = \frac{G m}{4r}$.Taking the square root,we get $v = \sqrt{\frac{G m}{4r}}$.

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 ($G=$ Universal gravitational constant).

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