Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is :

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:

Earth has mass $8$ times and radius $2$ times that of a planet. If the escape velocity from the Earth is $11.2 \ km/s$,the escape velocity in $km/s$ from the planet will be:

The magnitude of potential energy per unit mass of an object at the surface of the Earth is $E$. Then,the escape velocity of the object is ..........

$A$ particle of mass $m$ is kept at rest at a height $3R$ from the surface of the Earth,where $R$ is the radius of the Earth and $M$ is the mass of the Earth. The minimum speed with which it should be projected upward so that it does not return back is ($g$ = acceleration due to gravity on the Earth's surface).

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: