Q: Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction, the speed of each particle is

(a) $\large \sqrt{\frac{G M}{R}}$

(b) $\large \sqrt{2\sqrt{2}\frac{G M}{R}}$

(c) $\large \sqrt{(1+2\sqrt{2})\frac{G M}{R}}$

(d) $\large \frac{1}{2}\sqrt{(1+2\sqrt{2})\frac{G M}{R}}$

Ans: (d)

Sol: Net force acting on any one of the particle is

$\large F = \frac{GM^2}{(2R)^2} + \frac{GM^2}{(\sqrt{2}R)^2} cos45^o + \frac{GM^2}{(\sqrt{2}R)^2} cos45^o$

$\large F = \frac{GM^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}}) $

This force will provide required centripetal force .

$\large \frac{M v^2}{R} = \frac{GM^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}}) $