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 moves 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. $\sqrt{\frac{3GM}{a}},2\pi \sqrt{\frac{a^3}{3GM}}$
• 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

The correct option is C $\sqrt{\frac{GM}{a}},2\pi \sqrt{\frac{a^3}{3GM}}$

The resultant force on each particle due to other two particles is

$F_R=\sqrt{F_1^2+F_2^2+2F_1{F}_{2}cos\theta }=\sqrt{3}{F}_1=\sqrt{3}\frac{G{M}^2}{a^2}...........\left(i\right)$
If each particle is given a tangential velocity v, so that F acts as the centripetal force, they will move in a circle of radius
$r=\left(\frac{2}{3}\right)asin{60}^{\circ }=\frac{a}{\sqrt{3}}$
Now $F_R=\frac{M{v}^2}{r}=\sqrt{3}\frac{M{v}^2}{a}.......\left(ii\right)$
From (i) and (ii),
$\sqrt{3}\frac{M{v}^2}{a}=\frac{G{M}^2\sqrt{3}}{a^2}orv=\sqrt{GM/a}$
Time period.
$T=\frac{2\pi r}{v}=\frac{2\pi a}{\sqrt{3}}\sqrt{\frac{a}{GM}}=2\pi \sqrt{\frac{a^3}{3GM}}$