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

$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

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:

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and the same mass density is:

$A$ body is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half of the escape velocity. The maximum height reached by the body is