$A$ body of mass $m$ is taken from the earth's surface to a height equal to twice the radius $(R)$ of the earth. The change in potential energy of the body will be

$A$ rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V$. Assuming the radius of the earth to be $R$ and negligible air resistance,the maximum height attained by the rocket above the surface of the earth is

$A$ body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h = 3R$ ($R$ is the radius of the earth). The change in gravitational potential energy of the body is

Two planets have masses $M$ and $16\, M$ and their radii are $a$ and $2\, a$ respectively. The separation between the centres of the planets is $10\, a.$ $A$ body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach the surface of the smaller planet,the minimum firing speed needed 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 $g$)

$A$ 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).