Three identical particles each of mass ‘m’ are arranged at the corner of an equilateral triangle of side “a”. If they are to be in equilibrium, the speed with which they must revolve under the influence of one another’s gravity in a circular orbit circumscribing the triangle is

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$\sqrt{\frac{3 G M}{a}}$

$\sqrt{\frac{3 G M}{a^{2}}}$

$\sqrt{\frac{G M}{3 a}}$

$\sqrt{\frac{G M}{a}}$

answer is B.

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Detailed Solution

$W h e n F_{1} = F_{2} t h e n F_{n e t} = 2 F \cos \frac{θ}{2} = 2 F \cos \frac{60}{2} = 2 F \cos 30 = \frac{2 F \sqrt{3}}{2} = F \sqrt{3} = \frac{\sqrt{3} G m m}{a^{2}} F_{n e t} p r o v i d e s n e c e s s a r y c e n t r i p e t a l f o r c e \Rightarrow F_{n e t} = \frac{m v^{2}}{r} \Rightarrow \frac{\sqrt{3} G m m}{a^{2}} = \frac{m v^{2}}{\frac{a}{\sqrt{3}}} (∵ r = \frac{a}{\sqrt{3}}) \Rightarrow \frac{\sqrt{3} G m}{a} = \sqrt{3} v^{2} \Rightarrow v = \sqrt{\frac{G m}{a}}$