Question

Three particles, each of mass $M$, are situated at the vertices of an equilateral triangle of side '$a^{\prime }$. Theonly forces acting on the particles are their mutualgravitational forces. It is desired that each particlemoves in a circle while maintaining their original separation '$a$'. The initial velocity that should be given to each particle and time period of circular motion are respectively.

• A. $\sqrt{\frac{2GM}{a}},2\Pi \sqrt{\frac{a^3}{3GM}}$
• B. $\sqrt{\frac{GM}{a}},\frac{1}{2\Pi }\sqrt{\frac{3GM}{a^3}}$
• C. $\sqrt{\frac{GM}{a}},2\Pi \sqrt{\frac{a^3}{3GM}}$
• D. $\sqrt{\frac{GM}{3a}},2\Pi \sqrt{\frac{a^3}{3GM}}$

Answer 1

Answer: (C)

$\sqrt{\frac{GM}{a}},2\Pi \sqrt{\frac{a^3}{3GM}}$

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 $T=\frac{2\pi r}{v}$

$T=\frac{2\pi a}{\sqrt{3}}\sqrt{\frac{a}{Gm}}$

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