On which two factors does the escape velocity of an object projected from the Earth $NOT$ depend?

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

$A$ 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 Earth has a mass nine times and radius twice that of planet $P$, then $v_e / (3 \sqrt{x}) \text{ ms}^{-1}$ is the minimum velocity required by a rocket to escape the gravitational force of planet $P$, where $v_e$ is the escape velocity on Earth. The value of $x$ is:

The escape velocity for a planet is $v_e$. $A$ tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet,its speed will be

There is no atmosphere on the moon because