Q. The masses and radii of the earth and moon are $M_{1},R_{1}$ and $M_{2},R_{2}$ respectively. Their centres are distance $d$ apart. The minimum velocity with which a particle of mass $m$ shoule be projected from a point midway between their centres so that it escapes to infinity is
A) $2\sqrt{\frac{G}{d}(M_{1} + M_{2})}$
B) $2\sqrt{\frac{2G}{d}(M_{1} + M_{2})}$
C) $2\sqrt{\frac{Gm}{d}(M_{1} + M_{2})}$
D) $2\sqrt{\frac{Gm(M_{1} + M_{2})}{d(R_{1} + R_{2})}}$
Explanation
(a) Gravitational potential at mid point
$V = \frac{- GM_{1}}{d/2} + \frac{- GM_{2}}{d/2}$
Now, $PE = m \times V = \frac{- 2Gm}{d}(M_{1} + M_{2})$
$(m =$ mass of particle)
So, for projecting particle from mid point to infinity
$KE = |PE|$
$\Rightarrow \frac{1}{2}mv^{2} = \frac{2\ Gm}{d}\left( M_{1} + M_{2} \right) \Rightarrow v = 2\sqrt{\frac{G(M_{1} + M_{2})}{d}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.