$A$ body is projected vertically upwards 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 from the surface of the earth is

Explain the escape energy of a body of mass $m$ lying on the surface of the Earth.

The escape velocity of a body from a planet whose mass is $6$ times the mass of Earth and radius is $2$ times the radius of Earth will be (where $V_{e}$ is the escape velocity of a body from the Earth's surface).

$A$ 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 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

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g$,where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \ km/s$,the escape speed on the surface of the planet in $km/s$ will be: