Escape velocity at the surface of the Earth is $11.2 \, km/s$. If the radius of a planet is double that of the Earth but its mean density is the same as that of the Earth,then the escape velocity will be ........ $km/s$.

Explain escape energy and provide its definition. Also, explain escape speed.

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:

The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is

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

$A$ geostationary satellite is revolving around the $Earth$. To make it escape from the gravitational field of the $Earth$,its velocity must be increased by ........ $\%$