The minimum speed with which a body is projected from the surface of the earth so that it reaches to the surface of the moon is v while the minimum speed with which a body is projected from the surface of the moon so that it reaches to the surface of the earth is v'. Then, disregarding the spinning and orbital motions of the earth and the moon, and using the fact that $\frac{M_{E}}{R_{E}} > \frac{M_{M}}{R_{M}}$ where $M_{E}, M_{M}, R_{E}$ and $R_{M}$ are mass of earth, mass of moon, radius of earth and radius of the moon respectively, we can say that :

Question

The minimum speed with which a body is projected from the surface of the earth so that it reaches to the surface of the moon is v while the minimum speed with which a body is projected from the surface of the moon so that it reaches to the surface of the earth is v'. Then, disregarding the spinning and orbital motions of the earth and the moon, and using the fact that

\frac{M_{E}}{R_{E}} > \frac{M_{M}}{R_{M}}

where

M_{E}, M_{M}, R_{E}

and

R_{M}

are mass of earth, mass of moon, radius of earth and radius of the moon respectively, we can say that :

Toppr Admin · Accepted Answer

Let the potential of the point where net gravity is zero be-xA0- -V-Then from the energy conservation- we have-x2212-GMEmRE-12mv2-x2212-GMMmRM-12mv-x2032-2-x2212-mV-0

Let the potential of the point where net gravity is zero be -V.Then from the energy conservation, we have,−GMEmRE+12mv2=−GMMmRM+12mv′2=−mV+0

Let the potential of the point where net gravity is zero be -V.
Then from the energy conservation, we have,
$\frac{- G M_{E} m}{R_{E}} + \frac{1}{2} m v^{2} = \frac{- G M_{M} m}{R_{M}} + \frac{1}{2} m v^{' 2} = - m V + 0$