The escape velocity for the Earth is $11 \ km/s$. The escape velocity for a planet having twice the radius and the same density as the Earth is .......... $km/s$.

If a body of mass $1\text{ kg}$ falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are . . . . . . . (Take radius of earth to be $6400\text{ km}$ and $g = 9.8\text{ m/s}^2$)

The mass of a planet is half that of the Earth and the radius of the planet is one-fourth that of the Earth. If we plan to send an artificial satellite from the planet,the escape velocity will be (escape velocity on Earth $v_e = 11 \ km \ s^{-1}$): (in $km \ s^{-1}$)

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

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $V_{e}$,then the escape velocity from the planet is:

$A$ body is projected from the earth's surface with a speed $\sqrt{5}$ times the escape speed $(V_{e})$. The speed of the body when it escapes from the gravitational influence of the earth is