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:

$A$ body is thrown from the surface of the earth with velocity $V \ m/s$. The maximum height above the earth's surface up to which it will reach is ($R =$ radius of earth,$g =$ acceleration due to gravity).

The energy required to take a body from the surface of the earth to a height equal to the radius of the earth is $W$. The energy required to take this body from the surface of the earth to a height equal to twice the radius of the earth is:

$A$ 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 $g$)

$A$ 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 (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).