The radius of a planet is $\frac{1}{4}$ of earth's radius and its acceleration due to gravity is double that of earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to its value on earth's surface?

$A$ body is projected vertically from the Earth's surface of radius $R$ with a velocity equal to half the escape velocity. The maximum height reached by the body is

The ratio of the radii of two planets is $r$ and the ratio of accelerations due to gravity on the planets is $x$. Then the ratio of the escape velocities from the planets is

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:

$A$ 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)

$A$ rocket is fired vertically with a speed of $4 \ km/s$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth (in $km$)? (Take radius of Earth $R = 6.4 \times 10^6 \ m$ and $g = 10 \ m/s^2$)