When the speed of a satellite is increased by $x$ percentage,it will escape from its orbit,where the value of $x$ is ....... $\%$

$A$ 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

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 value of the escape velocity $v_e$ for a stationary object on the surface of a planet is directly proportional to the mass and radius of the planet." Is this statement correct? If not,correct it.

Consider a spherical planet which is rotating about its axis such that the speed of a point on its equator is $v$ and the effective acceleration due to gravity on the equator is $\frac{1}{3}$ of its value at the poles. What is the escape velocity for a particle at the pole of this planet?

$A$ rocket is launched straight up from the surface of the earth. When its altitude is one-fourth of the radius of the earth,its fuel runs out and it coasts. What is the minimum velocity the rocket must have when it starts to coast if it is to escape from the gravitational pull of the earth? (Escape velocity on the surface of the earth is $11.2 \ km/s$)