$A$ satellite is orbiting close to the Earth and has a kinetic energy $K$. The minimum extra kinetic energy required by it to just overcome the gravitational pull of the Earth is

$A$ spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V$. Due to the rotation of the planet about its axis,the acceleration due to gravity $g$ at the equator is $1/2$ of $g$ at the poles. Find the escape velocity of a particle on the planet in terms of $V$.

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. . . . .

The escape velocity of a body on an imaginary planet,which has thrice the radius of the Earth and double the mass of the Earth,is (where $v_e$ is the escape velocity of Earth):

There are two planets. The ratio of the radii of the two planets is $K$,and the ratio of the acceleration due to gravity of both planets is $g$. What will be the ratio of their escape velocities?

$A$ body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity of that planet. Then,the maximum height attained by the body is