$A$ body of mass $m$ starts falling from a point $2R$ above the Earth's surface. What is its kinetic energy when it has fallen to a point $R$ above the Earth's surface? [$R$ = Radius of Earth,$M$ = Mass of Earth,$G$ = Gravitational Constant]

Two uniform solid spheres of equal radii $R$,but mass $M$ and $4M$ have a centre-to-centre separation $6R$,as shown in the figure. The two spheres are held fixed. $A$ projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the centre of the second sphere. Obtain an expression for the minimum speed $v$ of the projectile so that it reaches the surface of the second sphere.

$A$ 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)

From the pole of the earth,a body of mass $m$ is imparted a velocity $v_0$ directed vertically up. If $M$ is the mass of the earth,$R$ its radius and $g$ is the free-fall acceleration on its surface,then the height $h$ to which the body will ascend is (neglect air resistance).

$A$ body (mass $m$) starts its motion from rest from a point distant $R_0$ $(R_0 > R)$ from the centre of the Earth. The velocity acquired by the body when it reaches the surface of the Earth will be ($G =$ universal constant of gravitation,$M =$ mass of Earth,$R =$ radius of Earth).

$A$ missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is