$A$ planet having mass $9 M_e$ and radius $4 R_e$,where $M_e$ and $R_e$ are the mass and radius of the Earth respectively,has an escape velocity in $km/s$ given by: (Given escape velocity on Earth $V_e = 11.2 \times 10^3 \, m/s$)

The escape velocity of a body from the surface of the earth is $11.2 \,km/s$. The escape velocity of a body from a planet having the same mean density as the earth but twice the radius of earth is: (in $\,km/s$)

The initial velocity $v_{i}$ required to project a body vertically upward from the surface of the earth to reach a height of $10 R$,where $R$ is the radius of the earth,may be described in terms of escape velocity $v_{e}$ such that $v_{i} = \sqrt{\frac{x}{y}} \times v_{e}$. The value of $x$ will be ...... .

If the radius of a planet is $R$ and its density is $\rho$,the escape velocity from its surface will be

$A$ body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text{rd}}$ of escape velocity. The maximum height reached by the body is ($R=$ radius of earth).

If the escape velocity on Earth is $11.2 \text{ km/s}$, what is its value for a planet having double the radius and $8$ times the mass of Earth (in $\text{ km/s}$)?