The escape velocity of an atmospheric particle which is $1000 \, km$ above the Earth's surface is .......... $km/s$ (radius of Earth is $6400 \, km$ and $g = 9.8 \, m/s^2$).

If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by:

What is the value of escape velocity of a body lying on the surface of Earth and Moon?

The ratio of escape velocity at earth $(v_e)$ to the escape velocity at a planet $(v_p)$ whose radius and mean density are twice as that of earth is:

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by: