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:

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

$A$ particle falls towards the Earth from infinity. Its velocity on reaching the Earth would be:

The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is

For a satellite,the escape velocity is $11 \ km/s$. If the satellite is launched at an angle of $60^\circ$ with the vertical,then the escape velocity will be ........... $km/s$.

An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$. The maximum height to which it rises from the center of the earth is (radius of earth is $R$):