The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is

$v_e$ and $v_p$ denote the escape velocity from the Earth and another planet having twice the radius and the same mean density as the Earth. Then:

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 $\text{ km/s}$)?

$A$ 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)

$A$ body is projected vertically upwards from the surface of the earth with a speed of $k{v_e}$,where $k < 1$ and ${v_e}$ is the escape velocity of the earth. What is the maximum height from the center of the earth that the body will reach? (Given: $R$ is the radius of the earth)

$A$ 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