The masses and radii of the earth and the 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 teh two centres so as to escape to infinity is $n \sqrt{\frac{G (M_{1} + M_{2})}{d}}$ . The value of $n$ is:

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Solution

From the law of energy conservation,

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

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

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

Answer is $2 \sqrt{\frac{G (M_{1} + M_{2})}{d}}$ .

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The masses and radii of the earth and the Moon are M1, R1 and M2,R2 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 teh two centres so as to escape to infinity is n√G(M1+M2)d. The value of n is:

From the law of energy conservation,12mv2−GM1md/2−GM2md/2=0v22=2G(M1+M2)dv=2√G(M1+M2)dAnswer is 2√G(M1+M2)d.

From the law of energy conservation,

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

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

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

Answer is $2 \sqrt{\frac{G (M_{1} + M_{2})}{d}}$ .