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The masses and radii of the earth and moon are $M_{1}$ , $R_{1}$ and $M_{2}$ , $R_{2}$ respectively. Their centres are at distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centres so that it escapes to infinity is:
$2 \sqrt{\frac{G m}{d} (M_{1} + M_{2})}$
$0$
$2 \sqrt{\frac{G}{d} (M_{1} + M_{2})}$
$2 \sqrt{\frac{G}{m d} (M_{1} + M_{2})}$

A

0

B

2 \sqrt{\frac{G}{m d} (M_{1} + M_{2})}

C

2 \sqrt{\frac{G m}{d} (M_{1} + M_{2})}

D

2 \sqrt{\frac{G}{d} (M_{1} + M_{2})}

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Solution

Verified by Toppr

The sum total of kinetic energy and potential energy of system must be conserved. For minimum velocity, this must be $0$ at infinity.

Hence,

$\frac{1}{2} m v^{2} - \frac{G M_{1} m}{\frac{d}{2}} - \frac{G M_{2} m}{\frac{d}{2}} = 0$

$\frac{1}{2} m v^{2} = 2 \frac{G m}{d} (M_{1} + M_{2})$

$v = 2 \sqrt{\frac{G}{d} (M_{1} + M_{2})}$

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