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.

Answer 1

Assume that three particles are at points A, B and C on the circumference of a circle.
BC = CD = $\sqrt{2}a$



The force on the particle at C due to gravitational attraction of the particle at B is ${\overrightarrow{F}}_{\mathrm{CB}}=\frac{\mathrm{G}{M}^2}{2{\mathrm{R}}^2}\hat{j}$.
The force on the particle at C due to gravitational attraction of the particle at D is ${\overrightarrow{F}}_{\mathrm{CD}}=-\frac{G{M}^2}{2{\mathrm{R}}^2}\hat{i}$.
Now, force on the particle at C due to gravitational attraction of the particle at A is given by

$$\begin{gathered} {\overrightarrow{F}}_{\mathrm{CA}}=-\frac{\mathrm{G}{M}^2}{4R^2}\cos 45\hat{i}+\frac{\mathrm{G}{M}^2}{4R^2}\sin 45\hat{j} \\ \therefore {\overrightarrow{F}}_{\mathrm{C}}={\overrightarrow{F}}_{\mathrm{CA}}+{\overrightarrow{F}}_{\mathrm{CB}}+{\overrightarrow{F}}_{\mathrm{CD}} \\ =\frac{-\mathrm{G}{M}^2}{4R^2}\left(2+\frac{1}{\sqrt{2}}\right)\hat{i}+\frac{\mathrm{G}{M}^2}{4R^2}\left(2+\frac{1}{\sqrt{2}}\right)\hat{j} \end{gathered}$$

So, the resultant gravitational force on C is ${\mathrm{F}}_{\mathrm{C}}=\frac{\mathrm{G}{m}^2}{4{\mathrm{R}}^2}\sqrt{2\sqrt{2}+1}$.

Let v be the velocity with which the particle is moving.
Centripetal force on the particle is given by
$$\begin{gathered} F=\frac{m{v}^2}{R} \\ \Rightarrow v\mathit{=}\sqrt{\frac{\mathit{G}\mathit{M}}{\mathit{R}}\left(\frac{\mathit{2}\sqrt{\mathit{2}}\mathit{+}\mathit{1}}{\mathit{4}}\right)} \end{gathered}$$