Q: Four particles of equal masses m move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle.

(a) $ \displaystyle \frac{1}{2}\sqrt{(2\sqrt2 +1)\frac{Gm}{R}} $

(b) $ \displaystyle \sqrt{(\sqrt2 +1)\frac{Gm}{R}} $

(c) $ \displaystyle \sqrt{(2\sqrt2 +1)\frac{Gm}{R}} $

(d) $ \displaystyle \frac{1}{2}\sqrt{(\sqrt2 +1)\frac{Gm}{R}} $

Ans: (a)