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With what minimum speed should m be projected from point C in presence of two fixed masses M each at A and B as shown in the figure such that mass m should escape the gravitational attraction of A and B

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

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

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

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

answer is B.

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Detailed Solution

$G .P .E = K .E$

$- \frac{GMm}{\sqrt{2} R} - \frac{GMm}{\sqrt{2} R} = \frac{1}{2} {mv}^{2}$

$- \frac{2GMm}{\sqrt{2} R} = \frac{1}{2} {mv}^{2}$

$- \frac{4GM}{\sqrt{2} R} = v^{2} \Rightarrow v^{2} = \frac{2 \sqrt{2} GM}{R}$

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

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