The escape velocity of a body from the earth's surface is $v_e$. The escape velocity of the same body from a height equal to $R$ from the earth's surface will be

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:

The angular velocity of rotation of a star (of mass $M$ and radius $R$) at which the matter starts to escape from its equator will be:

The mass and radius of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is:

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

The escape velocity of a body on the surface of the earth is $11.2 \, km/s$. If the earth's mass increases to twice its present value and the radius of the earth becomes half,the escape velocity would become ......... $km/s$.