If the acceleration due to gravity $g$ doubles and the radius of earth becomes half that of the present value,then the value of escape velocity is (Assume,$g=10 \ m/s^2$ and radius of earth,$R=6400 \ km$)

$A$ satellite is orbiting close to the Earth and has a kinetic energy $K$. The minimum extra kinetic energy required by it to just overcome the gravitational pull of the Earth is

If the radius and acceleration due to gravity both are doubled,the escape velocity of the Earth will become ......... $km/s$.

The escape velocity for a rocket from Earth is $11.2 \ km/s$. Its value on a planet where the acceleration due to gravity is double that on the Earth and the diameter of the planet is twice that of Earth will be in $km/s$:

$A$ satellite is moving with a constant speed $v$ in a circular orbit around 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 ejection,the kinetic energy of the object is

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet is