The value of escape velocity on a certain planet is $2 \, km/s$. Then the value of orbital speed for a satellite orbiting close to its surface is

What is the escape energy of an object of mass $m$ at a height of $4R_e$ from the surface of the Earth? ($R_e$ = radius of the Earth)

$A$ body is projected from the earth's surface with a speed $\sqrt{5}$ times the escape speed $(V_{e})$. The speed of the body when it escapes from the gravitational influence of the earth is

Two stars of masses $3\times10^{31} \ kg$ each,and at distance $2\times10^{11} \ m$ rotate in a plane about their common centre of mass $O$. $A$ meteorite passes through $O$ moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star,the minimum speed that the meteorite should have at $O$ is: (Take Gravitational constant $G = 6.67\times10^{-11} \ Nm^2 \ kg^{-2}$)

$A$ geostationary satellite is revolving around the $Earth$. To make it escape from the gravitational field of the $Earth$,its velocity must be increased by ........ $\%$

Why is there no atmosphere on the Moon? Explain it in terms of escape velocity.