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

Question

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

Toppr Admin · Accepted Answer

The particle m is at a distance d-2 from the centre of the earth -mass M1- as well as from the centre of the moon -mass M2- Therefore- the total gravitational potential energy of the particle due to the earth and the moon is-x2212-GM1md-2-x2212-GM2md-2-x2212-2G-M1-M2-mdClearly- the work required to move the particle from the point midway between the earth and the moon to infinity would be 2G-M1-M2-md- Thus- the minimum velocity of projection for the particle m to escape is given by12mve2-2G-M1-M2-mdor-xA0-  ve-2-x221A-G-M1-M2-mdve-x221A-4G-M1-M2-md