The escape velocity of a body depends upon its mass as:

$A$ body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity of that planet. Then,the maximum height attained by the body is

$A$ 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 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 $g = 9.8\text{ m/s}^2$)

$A$ planet has twice the radius but the mean density is $\frac{1}{4}^{th}$ as compared to Earth. What is the ratio of the escape velocity from Earth to that from the planet?

The escape velocity of a body from the earth's surface is $v_e$. The escape velocity of the same body from a height equal to $R$ from the earth's surface will be