$A$ planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 \ km/s$,the escape velocity from the surface of the planet would be ........ $km/s$.

$A$ planet has twice the radius but the mean density is $\frac{1}{4}^{th}$ as compared to Earth. What is the ratio of the escape velocity from Earth to that from the planet?

Given the mass of the moon is $1/81$ of the mass of the earth and its radius is $1/4$ of the earth's radius. If the escape velocity on the earth's surface is $11.2 \, km/s$,the value of the escape velocity on the surface of the moon is ......... $km/s$.

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,$R=6400 \ km$)

The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centers are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

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 $G = 6.67\times10^{-11} \ Nm^2 \ kg^{-2}$)