$M_{1}, R_{1}$ 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 $n \sqrt{\frac{G (M_{1} + M_{2})}{d}}$ . The value of n is :

2

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6

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3

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4

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Solution

The correct option is A $2$
From the law of energy conservation
$\frac{1}{2} m v^{2} - \frac{G M_{1} m}{d / 2} - \frac{G M_{2} m}{d / 2} = 0$
$\frac{v^{2}}{2} = \frac{2 G (M_{1} + M_{2})}{d}$
$v = 2 \frac{2 G (M_{1} + M_{2})}{d}$
Answer is $2 \frac{2 G (M_{1} + M_{2})}{d}$
Therefore $n = 2$

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Q. The masses and radii of the earth and the Moon are

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R_{1}

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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