Four particles of equal masses $M$ move along a circle of radius $R$ under the action of their mutua

Question

Four particles of equal masses $M$ move along a circle of radius $R$ under the action of their mutual gravitational attraction. Find the speed of each particle.

Professor PC Thomas Classes · Answer

Net force towards centre $F_{net} = \frac{GM^2}{(2R)^2} + \sqrt{2}\frac{GM^2}{(\sqrt{2}R)^2} = \frac{GM^2}{R^2}\left(\frac{1}{4} + \frac{\sqrt{2}}{2}\right) = \frac{GM^2}{R^2}\left(\frac{1+2\sqrt{2}}{4}\right)$. Equating to centripetal force $\frac{Mv^2}{R}$, we get $v = \sqrt{\frac{GM}{R}\left(\frac{2\sqrt{2}+1}{4}\right)}$.