To project a body of mass $m$ from Earth's surface to infinity,the required kinetic energy is (assume,the radius of Earth is $R_E$,$g =$ acceleration due to gravity on the surface of Earth):

$A$ body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity of that planet. Then,the maximum height attained by the body is

If $V, R$ and $g$ denote respectively the escape velocity from the surface of the earth,the radius of the earth,and the acceleration due to gravity,then the correct equation is:

$A$ rocket is launched with velocity $10 \ km/s$. If the radius of the Earth is $R$,then the maximum height attained by it will be: (in $R$)

Consider a spherical planet which is rotating about its axis such that the speed of a point on its equator is $v$ and the effective acceleration due to gravity on the equator is $\frac{1}{3}$ of its value at the poles. What is the escape velocity for a particle at the pole of this planet?

If Earth has a mass nine times and radius twice that of planet $P$, then $v_e / (3 \sqrt{x}) \text{ ms}^{-1}$ is the minimum velocity required by a rocket to escape the gravitational force of planet $P$, where $v_e$ is the escape velocity on Earth. The value of $x$ is: