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:

$A$ 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 $m$ starts falling from a distance $3R$ above the Earth's surface. When it reaches a distance $R$ above the surface of the Earth (radius $R$,mass $M$),its kinetic energy is:

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).

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$ is

$A$ stationary object is released from a point $P$ at a distance $3R$ from the centre of the moon,which has radius $R$ and mass $M$. Which of the following gives the speed of the object on hitting the moon?