Work done in raising a body of mass $m$ from the surface of the earth to a height $\frac{R}{3}$ is ......... (where,$R$ is radius of earth,$M$ is mass of the earth and $G$ is gravitational constant)

$A$ 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 $R_e = 6.4 \times 10^{6} \; m$,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$)

Two planets have masses $M$ and $16\, M$ and their radii are $a$ and $2\, a$ respectively. The separation between the centres of the planets is $10\, a.$ $A$ body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach the surface of the smaller planet,the minimum firing speed needed is

The radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth,then the change in potential energy will be:

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 of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h = 3R$ ($R$ is the radius of the earth). The change in gravitational potential energy of the body is