The escape velocity for the earth is $11.2 \ km/s$. The mass of another planet is $100$ times that of the earth and its radius is $4$ times that of the earth. The escape velocity for this planet will be ......... $km/s$.

$A$ 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 $km/s$)

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 planets $P$ and $Q$ have the same uniform density $\rho$,masses $M_P$ and $M_Q$,and surface areas $A$ and $4A$,respectively. $A$ spherical planet $R$ also has uniform density $\rho$ and its mass is $(M_P + M_Q)$. The escape velocities from the planets $P, Q,$ and $R$ are $V_P, V_Q,$ and $V_R$ respectively. Then:

$A$ body is projected vertically upwards from the Earth's surface with velocity $2 V_e$,where $V_e$ is the escape velocity from the Earth's surface. The velocity when the body escapes the gravitational pull is

Let the escape velocity of a body kept at the surface of a planet be $u$. If it is projected at a speed of $200 \%$ more than the escape speed,then its speed in interstellar space will be ...........