The masses and radii of the earth and the Moon are $M_{1}$ , $R_{1}$ and $M_{2}$ , $R_{2}$ , respectively. Their centres are at distance $d$ apart. The minimum speed with which a particle of $m$ should be projected from a point midway the two centres so as to escape to infinity is:

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

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\sqrt{\frac{4 G (M_{1} + M_{2})}{d}}

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\sqrt{\frac{2 G M_{1} M_{2}}{d}}

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

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Solution

The correct option is C $\sqrt{\frac{4 G (M_{1} + M_{2})}{d}}$
The particle m is at a distance $d / 2$ from the centre of the earth (mass $M_{1}$ ) as well as from the centre of the moon (mass $M_{2}$ ). Therefore, the total gravitational potential energy of the particle due to the earth and the moon is
$- \frac{G M_{1} m}{d / 2} - \frac{G M_{2} m}{d / 2} = - \frac{2 G (M_{1} + M_{2}) m}{d}$
Clearly, the work required to move the particle from the point midway between the earth and the moon to infinity would be $\frac{2 G (M_{1} + M_{2}) m}{d}$ .
Thus, the minimum velocity of projection for the particle m to escape is given by
$\frac{1}{2} m {v_{e}}^{2} = \frac{2 G (M_{1} + M_{2}) m}{d}$
or $v_{e} = 2 \sqrt{\frac{G (M_{1} + M_{2}) m}{d}}$
$v_{e} = \sqrt{\frac{4 G (M_{1} + M_{2}) m}{d}}$

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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 a distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity is given by