The escape velocity of a body from a planet whose mass is $6$ times the mass of Earth and radius is $2$ times the radius of Earth will be (where $V_{e}$ is the escape velocity of a body from the Earth's surface).

The escape velocity on Earth is $11.2 \, km/s$. On another planet having twice the radius and $8$ times the mass of the Earth,the escape velocity will be ......... $km/s$.

$A$ 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 a body of mass $1\text{ kg}$ falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are . . . . . . . (Take radius of earth to be $6400\text{ km}$ and $g = 9.8\text{ m/s}^2$)

Two stars of masses $3\times10^{31} \ kg$ each,and at distance $2\times10^{11} \ m$ rotate in a plane about their common centre of mass $O$. $A$ meteorite passes through $O$ moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star,the minimum speed that the meteorite should have at $O$ is: (Take Gravitational constant $G = 6.67\times10^{-11} \ Nm^2 \ kg^{-2}$)

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