Consider four point masses each of mass m are fixed at the vertex of a Rhombus of side $l$ . The escape speed from the centre of the Rhombus is (G is universal gravitational constant)

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$\sqrt{\frac{8 G m}{l} (1 + \frac{1}{\sqrt{5}})}$

$\sqrt{4 \sqrt{2} \frac{G m}{l}}$

$\sqrt{\frac{8 G m}{l} (1 + \frac{1}{\sqrt{3}})}$

$\sqrt{\frac{8 G m}{l}}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} m_{0} v^{2} = \frac{2 G m m_{0}}{(\frac{l}{2})} + \frac{2 G m m_{0}}{\sqrt{3} \frac{l}{2}} \\ = \sqrt{\frac{8 G m}{l} + \frac{8 G m}{\sqrt{3} l}} = \sqrt{\frac{8 G m}{l} (1 + \frac{1}{\sqrt{3}})} \end{array}$

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