$A$ mass '$m$' on the surface of the Earth is shifted to a height equal to the radius of the Earth. If '$R$' is the radius and '$M$' is the mass of the Earth,then the work done in this process is:

$A$ 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]

$A$ body of mass $m$ falls from a height $R$ above the surface of the earth,where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)

$A$ 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)

If the gravitational acceleration at the surface of the Earth is $g$,then the increase in potential energy in lifting an object of mass $m$ to a height equal to half of the radius of the Earth from the surface will be:

Two hypothetical planets of masses $m_1$ and $m_2$ are at rest when they are at an infinite distance apart. Because of the gravitational force,they move towards each other along the line joining their centres. What is their speed when their separation is $d$? (Speed of $m_1$ is $v_1$ and that of $m_2$ is $v_2$)