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 $,v_e$)

Maximum height reached by an object projected perpendicular to the surface of the earth with a speed equal to $50\%$ of the escape velocity from the earth's surface is - ($R =$ Radius of Earth)

The radius of a planet is $\frac{1}{4}$ of earth's radius and its acceleration due to gravity is double that of earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to its value on earth's surface?

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. . . . .

$A$ body is thrown with a velocity equal to $n$ times the escape velocity $(v_e)$. What will be the velocity of the body at a large distance away?

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