Given that the mass of the Earth is $M$ and its radius is $R$. $A$ body is dropped from a height equal to the radius of the Earth above the surface of the Earth. When it reaches the ground,its velocity will be:

If $g$ is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass $m$ raised from the earth's surface to a height equal to the radius $R$ of the earth is

Three solid spheres each of mass $1 \ kg$ and radius $2 \ m$ are arranged at the three corners of an equilateral triangle of side $10 \ m$,such that the centers of the spheres coincide with the corners of the triangle. When they are released from that position,the speed of any one sphere at the time of collision would be ($G$ is the universal gravitational constant).

The change in the gravitational potential energy when a body of mass $m$ is raised to a height $h = nR$ above the surface of the Earth is (where $R$ is the radius of the Earth).

$A$ 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 of mass $m$ is released from a height of $2R$ above the Earth's surface. What will be its kinetic energy at a height of $R$ from the Earth's surface? (Where $R$ is the radius of the Earth)