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

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Detailed Solution

Net force on any one particle $= \frac{G M^{2}}{{(2 R)}^{2}} + \frac{G M^{3}}{{(R \sqrt{2})}^{2}} \cos 45^{0} + \frac{G M^{2}}{{(R \sqrt{2})}^{2}} \cos 45^{0} = \frac{G M^{2}}{R^{2}} [\frac{1}{4} + \frac{1}{\sqrt{2}}]$

This force will be equal to centripetal force so $\frac{M u^{2}}{R} = \frac{G M^{2}}{R^{2}} [\frac{1 + 2 \sqrt{2}}{4}]$

$u = \sqrt{\frac{G M}{4 R} [1 + 2 \sqrt{2}]} = \frac{1}{2} \sqrt{\frac{G M}{R} (2 \sqrt{2} + 1)}$

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