If the radius of a planet is four times that of Earth and the value of $g$ is the same for both,the escape velocity on the planet will be ......... $km/s$.

The escape velocity from the earth is about $11 \, km/s$. The escape velocity from a planet having twice the radius and the same mean density as the earth is ......... $km/s$.

$Assertion$: The escape speed does not depend on the direction in which the projectile is fired.
$Reason$: Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the Earth rotates about its axis.

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is

$A$ spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V$. Due to the rotation of the planet about its axis,the acceleration due to gravity $g$ at the equator is $1/2$ of $g$ at the poles. Find the escape velocity of a particle on the planet in terms of $V$.

The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is four times and density is nine times that of the earth,is (in $,v_e$)