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

$A$ particle of mass $m$ is thrown upwards from the surface of the earth with a velocity $u$. The mass and the radius of the earth are $M$ and $R$, respectively. $G$ is the gravitational constant and $g$ is the acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth is:

The escape velocity of a sphere of mass $m$ is given by ($G =$ Universal gravitational constant; $M_e =$ Mass of the earth and $R_e =$ Radius of the earth).

If a body of mass $1\text{ kg}$ falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are . . . . . . . (Take radius of earth to be $6400\text{ km}$ and $g = 9.8\text{ m/s}^2$)

$A$ body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body is:

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: