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

There is no atmosphere on the moon because

If Earth has a mass nine times and radius twice that of planet $P$, then $v_e / (3 \sqrt{x}) \text{ ms}^{-1}$ is the minimum velocity required by a rocket to escape the gravitational force of planet $P$, where $v_e$ is the escape velocity on Earth. The value of $x$ is:

The masses and radii of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centers are at a distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centers so that it escapes to infinity is:

The radius in kilometer to which the present radius of earth $(R = 6400 \ km)$ must be compressed so that the escape velocity is increased $10$ times is ............ $km$.

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is: