$A$ rocket is launched straight up from the surface of the earth. When its altitude is one-fourth of the radius of the earth,its fuel runs out and it coasts. What is the minimum velocity the rocket must have when it starts to coast if it is to escape from the gravitational pull of the earth? (Escape velocity on the surface of the earth is $11.2 \ km/s$)

The radius and mean density of a planet are four times that of the Earth. The ratio of the escape velocity on the Earth to the escape velocity on the planet is:

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:

The escape velocity of a body on a planet $A$ is $12 \, km/s$. The escape velocity of the body on another planet $B$,whose density is four times and radius is half of the planet $A$,is ................... $km/s$.

$A$ rocket is launched normal to the surface of the Earth,away from the Sun,along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2.5 \times 10^4$ times larger than the radius of the Earth. The escape velocity from the Earth's gravitational field is $v_e = 11.2 \text{ km s}^{-1}$. The minimum initial velocity $(v_s)$ required for the rocket to be able to leave the Sun-Earth system is closest to:
(Ignore the rotation and revolution of the Earth and the presence of any other planet)

If the escape velocity of a body of mass $1\,kg$ from the surface of the Earth is $11.2\,km/s$,then what is the escape velocity for a body of mass $10\,kg$?