If the Earth has a mass nine times and a radius twice that of a planet $P$. Then $\frac{v_e}{3} \sqrt{x} \; ms^{-1}$ will be the minimum velocity required by a rocket to escape the gravitational force of $P$,where $v_e$ is the escape velocity on Earth. The value of $x$ is

The escape velocity from the earth is about $11 \, km/s$. The escape velocity from a planet having twice the radius and the same mean density as the earth is ......... $km/s$.

$A$ particle is projected from the surface of the earth with a velocity equal to twice the escape velocity. When the particle is very far from the earth,its speed would be

The mass of the moon is $1/144$ times the mass of a planet and its diameter is $1/16$ times the diameter of a planet. If the escape velocity on the planet is $v$,the escape velocity on the moon will be:

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 $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

Given below are two statements:
Statement $I:$ For a planet,if the ratio of mass of the planet to its radius increases,the escape velocity from the planet also increases.
Statement $II:$ Escape velocity is independent of the radius of the planet.
In the light of the above statements,choose the most appropriate answer from the options given below: