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 escape velocity from the surface of Earth of mass $M$ and radius $R$ is $V_{e}$. The escape velocity from the surface of a planet whose mass and radius are $3$ times that of the Earth will be:

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

If the escape velocity of a body of mass $1\,kg$ from the surface of the Earth is $11.2\,km/s$,then what is the escape velocity for a body of mass $10\,kg$?

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: The kinetic energy needed to project a body of mass $m$ from the Earth's surface to infinity is $\frac{1}{2} mgR$,where $R$ is the radius of the Earth.
Reason $R$: The maximum potential energy of a body is zero when it is projected to infinity from the Earth's surface.
In the light of the above statements,choose the correct answer from the options given below.

$A$ body is projected vertically upwards from the surface of the Earth with a velocity equal to one-third of the escape velocity. The maximum height attained by the body will be $...... \ km$. (Take radius of Earth $R = 6400 \ km$ and $g = 10 \ m/s^2$)