The work done to raise a mass $m$ from the surface of the earth to a height $h$,which is equal to the radius of the earth,is

$A$ rocket is fired vertically with a speed of $5 \; km/s$ from the Earth's surface. How far from the Earth's surface does the rocket go before returning? (Mass of the Earth $M_e = 6.0 \times 10^{24} \; kg$,mean radius of the Earth $R_e = 6.4 \times 10^{6} \; m$,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$)

$A$ body which is initially at rest at a height $R$ above the surface of the earth of radius $R$,falls freely towards the earth. Its velocity on reaching the surface of the earth is:

An object of mass $1 \, kg$ is taken to a height from the surface of Earth which is equal to three times the radius of Earth. The gain in potential energy of the object will be $.... \, MJ$ [Given: $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$].

Two stars each of one solar mass $\left(=2 \times 10^{30} \; kg\right)$ are approaching each other for a head-on collision. When they are at a distance of $10^{9} \; km$,their speeds are negligible. What is the speed with which they collide? The radius of each star is $10^{4} \; km$. Assume the stars remain undistorted until they collide. (Use $G = 6.67 \times 10^{-11} \; N \cdot m^{2}/kg^{2}$)

$A$ body is released from a height equal to the radius $(R)$ of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given $g$ = acceleration due to gravity on the earth.)