The escape velocity of a body from the surface of the Earth:

The escape velocity of a satellite from the surface of the Earth does $NOT$ depend on

$A$ 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 $km/s$)

$A$ planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 \ km/s$,the escape velocity from the surface of the planet would be ........ $km/s$.

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

The ratio of the orbital velocity of a body near the surface of a planet to the escape velocity of a body from the surface of the same planet is: