Four particles, each of mass M and lying on the vertices of square, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :

Engineering

Physics

Gravitational Force and Field

Question

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

$\sqrt{2 \sqrt{2} \frac{GM}{R}}$

$\frac{1}{4} \sqrt{\frac{GM}{R} (1 + 2 \sqrt{2})}$

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

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$\frac{{mv}^{2}}{R} = \frac{{Gm}^{2}}{{(\sqrt{2} R)}^{2}} \cos 45 ° \times 2 + \frac{{Gm}^{2}}{{(2 R)}^{2}}$

$= \frac{Gm}{\sqrt{2} R} + \frac{Gm}{4 R}$

$\frac{Gm}{4 R} [1 + 2 \sqrt{2}]$