The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is four times and density is nine times that of the earth,is (in $,v_e$)

$A$ body is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half of the escape velocity. The maximum height reached by the body is

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. . . . .

$A$ particle is projected from the surface of the earth with a velocity equal to twice the escape velocity. When the particle is very far from the earth,its speed would be

$A$ ball $A$ of mass $m$ falls to the surface of the earth from infinity. Another ball $B$ of mass $2m$ falls to the earth from a height equal to six times the radius of the earth. The ratio of the velocities of $A$ and $B$ on reaching the earth is:

$A$ spaceship moves from the Earth to the Moon and back. The greatest energy required for the spaceship is to overcome the difficulty in: