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Three particles each of mass $m$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

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$\sqrt{\frac{G m}{\sqrt{3} R}}$

$\sqrt{\frac{2 G m}{3 R}}$

$\sqrt{\frac{3 G m}{2 R}}$

$\sqrt{\frac{G m}{3 R}}$

answer is A.

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Detailed Solution

$\frac{m V^{2}}{R} = 2 \frac{G m m}{{(I)}^{2}} \cos 30 °$

$\frac{V^{2}}{R} = 2 \frac{G m}{I^{2}} \frac{\sqrt{3}}{2} = \sqrt{3} \frac{g m}{I^{2}}$

$R \cos 30 ° = \frac{I}{2}$

$R \frac{\sqrt{3}}{2} = \frac{I}{2} \Rightarrow I = \sqrt{3} R$

$\frac{V^{2}}{R} = \frac{\sqrt{3} G m}{{(\sqrt{3} R)}^{2}}$

$\frac{V^{2}}{R} = \frac{\sqrt{3} G m}{3 R^{2}}$

$V = \sqrt{\frac{G m}{\sqrt{3} R}}$

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