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:

An object is kept at rest at a distance of $3R$ above the earth's surface,where $R$ is the earth's radius. The minimum speed with which it must be projected so that it does not return to earth is (Assume $M =$ mass of earth,$G =$ Universal gravitational constant).

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?

Maximum height reached by an object projected perpendicular to the surface of the earth with a speed equal to $50\%$ of the escape velocity from the earth's surface is - ($R =$ Radius of Earth)

The ratio of the radius of a planet $A$ to that of planet $B$ is $r$. The ratio of acceleration due to gravity on the planets is $x$. The ratio of the escape velocities from the two planets is

$v_e$ and $v_p$ denote the escape velocity from the Earth and another planet having twice the radius and the same mean density as the Earth. Then: