$A$ rocket is projected in the vertically upwards direction with a velocity $kv_e$,where $v_e$ is the escape velocity and $k < 1$. The distance from the centre of the Earth up to which the rocket will reach is:

The escape velocity of a body projected vertically from the surface of the Earth is $11 \, km/s$. If the body is projected at an angle of $60^\circ$ with the vertical,the escape velocity will be ......... $km/s$.

$A$ 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)

The escape velocity of a body from a planet whose mass is $6$ times the mass of Earth and radius is $2$ times the radius of Earth will be (where $V_{e}$ is the escape velocity of a body from the Earth's surface).

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 escape velocity of a body depends upon its mass as: