The escape velocity from a spherical planet $A$ is $10 \ km/s$. The escape velocity from another planet $B$ whose density and radius are $10\%$ of those of planet $A$, is . . . . . . $m/s$.

Escape velocity at the surface of the Earth is $11.2 \, km/s$. If the radius of a planet is double that of the Earth but its mean density is the same as that of the Earth,then the escape velocity will be ........ $km/s$.

The escape velocity for a planet whose radius is $1.7 \times 10^6 \ m$ and acceleration due to gravity is $1.7 \ m s^{-2}$ is

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:

If the acceleration due to gravity $g$ doubles and the radius of earth becomes half that of the present value,then the value of escape velocity is (Assume,$g=10 \ m/s^2$ and radius of earth,$R=6400 \ km$)

$A$ 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)