$A$ body is projected up with a velocity equal to $\frac{3}{4}$ of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth $= R$)

The escape speed of a projectile on the earth's surface is $11.2 \; km/s$. $A$ body is projected out with thrice this speed. What is the speed (in $km/s$) of the body far away from the earth? Ignore the presence of the sun and other planets.

The escape velocity of a body from the earth's surface is $v_e$. The escape velocity of the same body from a height equal to $R$ from the earth's surface will be

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?

$A$ 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 $km/s$)

$A$ rocket is projected in the vertically upwards direction with a velocity $kv_e$,where $v_e$ is the escape velocity and $k < 1$. The distance from the centre of the Earth up to which the rocket will reach is: