Question

Four particles, each of mass $M$, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is
(Assume that each particle is diameterically opposite to other in pair.)

• A. $\frac{GM}{R}$
• B. $\sqrt{2\sqrt{2}\frac{GM}{R}}$
• C. $\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}$
• D. $$\sqrt{\frac{GM}{R}(\frac{2\sqrt{2}+1}{4})}$$

Answer 1

The correct option is D $$\sqrt{\frac{GM}{R}(\frac{2\sqrt{2}+1}{4})}$$

Consider particle $1$ which is in circular motion about the center under the influence of gravitational forces by particles $2$, $3$ and $4$.


Distance between the adjacent particles $=2R\cos {45}^{\circ }=\sqrt{2}R$

From, universal law of gravitation,

$F_{21}=F_{41}=\frac{G{M}^2}{(\sqrt{2}R{)}^2}=\frac{G{M}^2}{2R^2}$

And, $F_{31}=\frac{G{M}^2}{(2R{)}^2}=\frac{G{M}^2}{4R^2}$

Components of $F_{21}$ & $F_{41}$ along the center of circle will be

$F_{21}\cos {45}^{\circ }=\frac{G{M}^2}{2R^2}\times \frac{1}{\sqrt{2}}=\frac{G{M}^2}{2\sqrt{2}{R}^2}$

Let force towards the center (centripetal force) is $F_c$

$F_c=F_{21}\cos {45}^{\circ }+F_{41}\cos {45}^{\circ }+F_{31}$

$\Rightarrow F_c=\frac{G{M}^2}{2\sqrt{2}{R}^2}+\frac{G{M}^2}{2\sqrt{2}{R}^2}+\frac{G{M}^2}{4R^2}=\left(\frac{2\sqrt{2}+1}{4}\right)\frac{G{M}^2}{R^2}$

Let $v$ be the circular speed of the particle.

Centripetal force $=\frac{M{v}^2}{R}$
[R is the radius of circular orbit]

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

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

Hence, option $\left(d\right)$ is correct.

Why this question The particles in circular motion under the influence of mutual gravitation is a commonly asked question. This concept can be extended to any number of particles, $2,3,4$ or many.