$A$ satellite revolving around a planet has an orbital velocity of $10 \ km/s$. The additional velocity required for the satellite to escape from the gravitational field of the planet is: (in $km/s$)

$A$ body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body is:

The mass of the Earth is $81$ times that of the Moon and the radius of the Earth is $3.5$ times that of the Moon. The ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon will be:

For a satellite,the escape velocity is $11 \ km/s$. If the satellite is launched at an angle of $60^\circ$ with the vertical,then the escape velocity will be ........... $km/s$.

$A$ 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 $\,km$)

$A$ projectile is projected with velocity $k{v_e}$ in the vertically upward direction from the ground into space. (${v_e}$ is the escape velocity and $k < 1$). If air resistance is considered to be negligible,then the maximum height from the center of the Earth to which it can go will be: ($R$ = radius of Earth)