The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and the same mass density is:

If $g$ is the acceleration due to gravity at the earth's surface and $r$ is the radius of the earth,the escape velocity for a body to escape out of the earth's gravitational field is

$A$ satellite is moving with a constant speed '$V$' in a circular orbit about the Earth. An object of mass '$m$' is ejected from the satellite such that it just escapes from the gravitational pull of the Earth. At the time of its ejection,the kinetic energy of the object is

The escape velocity for a planet whose mass is six times the mass of Earth and whose radius is twice the radius of Earth will be (where $V_{e}$ is the escape velocity from the 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 for that planet. The maximum height attained by the body is

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: