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 mutual separation a.The time period of the circular motion is $2π \sqrt{\frac{a^{3}}{xGm}}$ then find the $x$ value

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answer is 3.

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Detailed Solution

Resultant Gravitational force acting on each particle is:

$F = \sqrt{{(\frac{{Gm}^{2}}{a^{2}})}^{2} + {(\frac{{Gm}^{2}}{a^{2}})}^{2} + 2 (\frac{{Gm}^{2}}{a^{2}}) (\frac{{Gm}^{2}}{a^{2}}) \cos 60} = \sqrt{3} \frac{{Gm}^{2}}{a^{2}}$

The radius of the circle that particles are moving will be $r = \frac{a}{\sqrt{3}}$

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

$\begin{array}{l} \frac{{mv}^{2}}{r} = \sqrt{3} \frac{{Gm}^{2}}{a^{2}} \\ \frac{\sqrt{3} {mv}^{2}}{a} = \sqrt{3} \frac{{Gm}^{2}}{a^{2}} \\ v = \sqrt{\frac{Gm}{a}} \end{array}$