Question

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

Answer 1

The gravitational force on each mass due to other $3$ masses provides the necessary centripetal force
Consider the particle $A$. Net force on particle $A$ is given by
${\overrightarrow{F}}_A={\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{AD}+{\overrightarrow{F}}_{AC}$
$\Rightarrow F_{AY}=\frac{G{M}^2}{(AB{)}^2}\cos {45}^{\circ }+\frac{G{M}^2}{(AD{)}^2}\cos {45}^{\circ }+\frac{G{M}^2}{(AC{)}^2}$
$=\frac{G{M}^2}{(\sqrt{2}R{)}^2}\frac{1}{\sqrt{2}}+\frac{G{M}^2}{(\sqrt{2}R{)}^2}\frac{1}{\sqrt{2}}+\frac{G{M}^2}{(2R{)}^2}=\frac{2G{M}^2}{2\sqrt{2}{R}^2}+\frac{G{M}^2}{4R^2}=\frac{G{M}^2}{R^2}[\frac{1}{\sqrt{2}}+\frac{1}{4}]$ along $AO$
This must provide the necessary centripetal force
Hence, $\frac{G{M}^2}{R^2}[\frac{1}{\sqrt{2}}+\frac{1}{4}]=\frac{m{v}^2}{R}$
$\Rightarrow v=\sqrt{\frac{GM}{R}\left[\frac{2\sqrt{2}+1}{4}\right]}$.