Question

Three particles, each of mass $m$ are situated at the vertices of an equilateral triangle of side $a$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle should move in a circle while maintaining the original mutual separation $a$. Then their time period of revolution is :

• A. $2\pi \sqrt{\frac{a^2}{3Gm}}$
• B. $2\pi \sqrt{\frac{a^3}{3Gm}}$
• C. $2\pi \sqrt{\frac{3a^4}{Gm}}$
• D. $2\pi \sqrt{\frac{a^4}{Gm}}$

Answer 1

Answer: (B)

$2\pi \sqrt{\frac{a^3}{3Gm}}$

As per the figure on the left hand side.
Resultant Gravitational force acting on each particle is:
$F=\sqrt{(\frac{G{m}^2}{a^2}{)}^2+(\frac{G{m}^2}{a^2}{)}^2+2\left(\frac{G{m}^2}{a^2}\right)\left(\frac{G{m}^2}{a^2}\right)\cos 60}=\sqrt{3}\frac{G{m}^2}{a^2}$

As per figure on right side, as the particles are expected to move in a circle of radius $r$ (say) while maintaining original separation $a$.
$\frac{a/2}{r}=\cos 30$

$r=\frac{a}{\sqrt{3}}$

Because of circular motion, the centripetal force is balanced by resultant gravitational force.

If $v$ is the velocity:
$\frac{m{v}^2}{r}=\sqrt{3}\frac{G{m}^2}{a^2}$

$\frac{\sqrt{3}m{v}^2}{a}=\sqrt{3}\frac{G{m}^2}{a^2}$

$v=\sqrt{\frac{Gm}{a}}$

Time period, $T=\frac{2\pi r}{v}$$=\frac{2\pi a}{\sqrt{3}}\sqrt{\frac{a}{Gm}}$$=2\pi \sqrt{\frac{a^3}{3Gm}}$