The escape velocity of a body from the earth's surface is $v_e$. The escape velocity of the same body from a height equal to $R$ from the earth's surface will be

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

The mass of the moon is $1/144$ times the mass of a planet and its diameter is $1/16$ times the diameter of a planet. If the escape velocity on the planet is $v$,the escape velocity on the moon will be:

There is no atmosphere on the moon because

$A$ 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 $km/s$)

Two spherical stars $A$ and $B$ have densities $\rho_A$ and $\rho_B$,respectively. $A$ and $B$ have the same radius,and their masses $M_A$ and $M_B$ are related by $M_B = 2M_A$. Due to an interaction process,star $A$ loses some of its mass,so that its radius is halved,while its spherical shape is retained,and its density remains $\rho_A$. The entire mass lost by $A$ is deposited as a thick spherical shell on $B$ with the density of the shell being $\rho_A$. If $v_A$ and $v_B$ are the escape velocities from $A$ and $B$ after the interaction process,the ratio $\frac{v_B}{v_A} = \sqrt{\frac{10n}{15^{1/3}}}$. The value of $n$ is. . . . .