Question

Three particles each of mass m, are situated at the vertices of an equilateral triangle of side length 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 the original separation a. Find the initial velocity that should be given to each particle and also the time required for the circular motion

• A. $v=\sqrt{2\frac{Gm}{a}},T=2\pi (\frac{a^3}{3Gm}{)}^{\frac{1}{2}}$.
• B. $v=\sqrt{\frac{Gm}{a}},T=2\pi (\frac{a^3}{3Gm}{)}^{\frac{1}{2}}$.
• C. $v=\sqrt{5\frac{Gm}{a}},T=2\pi (2\frac{a^3}{3Gm}{)}^{\frac{1}{2}}$.
• D. $v=\sqrt{\frac{Gm}{3a}},T=2\pi (\frac{a^3}{5Gm}{)}^{\frac{1}{2}}$.

Answer 1

The correct option is B

$v=\sqrt{\frac{Gm}{a}},T=2\pi (\frac{a^3}{3Gm}{)}^{\frac{1}{2}}$.

Figure shows three particles located at vertices A, B and C of an equilateral triangle of sides AB = BC = CA = a. These particles move in a circle with 0 as the centre and radius
r = OA = OB - OC

$\frac{BD}{cos30}=\frac{\frac{a}{2}}{\sqrt{\frac{3}{2}}}=\frac{a}{\sqrt{3}}$
Let us find the net gravitational force acting on one particle, say at A, due to particles at B and C.
Particle at A is attracted.

Towards B with a force, $F_1=\frac{Gmm}{a^2}$
And towards C with a force, $F_2=\frac{Gmm}{a^2}$
Notice that $F_1=F_2=F\left(say\right)=\frac{G{m}^2}{a^2}$
The angle between these equal forces is $\theta ={60}^{\circ }$.
The resultant force on the particle at A is

$F_r=(F^2+F^2+2F^{2}cos{60}^{\circ }{)}^{\frac{1}{2}}\Rightarrow F_r=\sqrt{3}F$

$F_r=\sqrt{3}\frac{G{m}^2}{a^2}$ directed along AO.

Thus the net force on particle at A is radial.Similarly, the net force on particle at B and at C is $F_r$,each directed towards center 0.

This force provides the necessary centripetal force.If v is the required initial velocity of each particle,

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

or $v^2=\sqrt{3}\frac{Gm}{\sqrt{3a}}=\frac{Gm}{a}$ $(r=\frac{a}{\sqrt{3}})$ or $v=\sqrt{\frac{Gm}{a}}$.

Time period $T=\frac{2\pi r}{v}=\frac{2\pi \times \frac{a}{\sqrt{3}}}{\sqrt{\frac{Gm}{a}}}=2\pi (\frac{a^3}{3Gm}{)}^{\frac{1}{2}}$