If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by:

$A$ particle of mass $m_0$ is projected from the midpoint of the line joining two fixed particles,each of mass $m$. If the separation between the fixed particles is $l$,the minimum velocity of projection of the particle so as to escape is equal to:

The velocity with which a projectile must be fired so that it escapes Earth's gravitation does not depend on

Explain the escape energy of a body of mass $m$ lying on the surface of the Earth.

$A$ meteor of mass $m$ having a speed $v$ at infinity reaches the surface of the earth with a speed of (where $v_e$ is the escape speed from the earth's surface).

If the Earth has a mass nine times and a radius twice that of a planet $P$. Then $\frac{v_e}{3} \sqrt{x} \; ms^{-1}$ will be the minimum velocity required by a rocket to escape the gravitational force of $P$,where $v_e$ is the escape velocity on Earth. The value of $x$ is