Q: 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
(a) $\large \sqrt{\frac{G M}{R}}$
(b) $\large \sqrt{2\sqrt{2}\frac{G M}{R}}$
(c) $\large \sqrt{(1+2\sqrt{2})\frac{G M}{R}}$
(d) $\large \frac{1}{2}\sqrt{(1+2\sqrt{2})\frac{G M}{R}}$
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Ans: (d)
Sol: Net force acting on any one of the particle is
$\large F = \frac{GM^2}{(2R)^2} + \frac{GM^2}{(\sqrt{2}R)^2} cos45^o + \frac{GM^2}{(\sqrt{2}R)^2} cos45^o$
$\large F = \frac{GM^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}}) $
This force will provide required centripetal force .
$\large \frac{M v^2}{R} = \frac{GM^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}}) $