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Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Velocity of a particle is given by

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a

$\sqrt[3]{\frac{Gm}{r}}$

b

${[\frac{1}{2} \frac{Gm}{r} (\frac{1 + 2 \sqrt{2}}{2})]}^{\frac{1}{2}}$

c

$\sqrt{\frac{Gm}{r} (1 + 2 \sqrt{2})}$

d

${[\frac{Gm}{r} (\frac{1 + 2 \sqrt{2}}{4})]}^{\frac{1}{2}}$

answer is A.

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Detailed Solution

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Centripetal force = net gravitational force

$\begin{array}{l} \Rightarrow \frac{{mv}_{0}^{2}}{r} = 2 Fcos 45^{\circ} + F_{1} = \frac{2 {Gm}^{2}}{(\sqrt{2} r)^{2}} \frac{1}{\sqrt{2}} + \frac{{Gm}^{2}}{4 r^{2}} \\ \frac{{mv}_{0}^{2}}{r} = \frac{{Gm}^{2}}{4 r^{2}} [2 \sqrt{2} + 1] \\ \Rightarrow v_{0} = {[\frac{Gm (2 \sqrt{2} + 1)}{4 r}]}^{\frac{1}{2}} \end{array}$

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