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 ......... $km/s$.

$A$ rocket is launched straight up from the surface of the earth. When its altitude is $\frac{1}{3}$ of the radius of the earth,its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth,the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is $11.2 \ km/s$.) (in $km/s$)

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

The mass of a planet is half that of the Earth and the radius of the planet is one-fourth that of the Earth. If we plan to send an artificial satellite from the planet,the escape velocity will be (escape velocity on Earth $v_e = 11 \ km \ s^{-1}$): (in $km \ s^{-1}$)

$A$ 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

$A$ body of mass $m$ is projected with velocity $\lambda v_{e}$ in a vertically upward direction from the surface of the earth into space. It is given that $v_{e}$ is the escape velocity and $\lambda < 1$. If air resistance is considered to be negligible,then the maximum height from the center of the earth,to which the body can go,will be: ($R$: radius of earth)