$A$ planet has twice the radius but the mean density is $\frac{1}{4}^{th}$ as compared to Earth. What is the ratio of the escape velocity from Earth to that from the planet?

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ........... $km/s$.

$A$ 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 of a rocket launched from the surface of the earth:

$A$ 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