The escape velocity of a satellite from the surface of the Earth does $NOT$ depend on

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

The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

$A$ particle of mass $m$ is kept at rest at a height $3R$ from the surface of the Earth,where $R$ is the radius of the Earth and $M$ is the mass of the Earth. The minimum speed with which it should be projected upward so that it does not return back is ($g$ = acceleration due to gravity on the Earth's surface).

For a satellite orbiting very close to the surface of the Earth,what is the relationship between its escape velocity $(v_{e})$ and its orbital velocity $(v_{0})$?

$A$ mass of $6 \times 10^{24} \,kg$ is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4 \,ms^{-1}$. The radius of the sphere is (Universal gravitational constant $G = 6.66 \times 10^{-11} \,N \,m^2 \,kg^{-2}$) (in $\,km$)