Four 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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$\sqrt{\frac{G M}{2 R} (1 + 2 \sqrt{2})}$

$\frac{1}{2} \sqrt{\frac{G M}{R} (1 + \sqrt{2})}$

$\sqrt{\frac{G M}{R} (1 + 2 \sqrt{2})}$

$\frac{1}{2} \sqrt{\frac{G M}{R} (1 + 2 \sqrt{2})}$

answer is D.

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Detailed Solution

Let a be the distance between two particles.

The resultant gravitational force on any one of the particle is given by

$F_{R} = \frac{G M^{2}}{a^{2}} (\sqrt{2} + \frac{1}{2}) .$
Which provides necessary centripetal force for motion of mass M in circle, so
$\begin{array}{l} (\sqrt{2} + \frac{1}{2}) (\frac{G M^{2}}{a^{2}}) = \frac{M V^{2}}{\frac{a}{\sqrt{2}}} \\ V^{2} (\frac{2 \sqrt{2} + 1}{2 \sqrt{2}}) \frac{G M}{a} = (\frac{2 \sqrt{2} + 1}{2 \sqrt{2}}) \frac{G M}{\sqrt{2} R} \\ \Rightarrow V = \frac{1}{2} \sqrt{\frac{G M}{R} (1 + 2 \sqrt{2})} \end{array}$