$A$ 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.)

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).

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:

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).

Two masses $m_1$ and $m_2$ are at rest at infinity. Find their relative velocity of approach due to gravitational attraction when their separation is $d$.

Write the change in gravitational potential energy of a body lifted at height $h$ from the surface of Earth.