$A$ 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 $g = 10 \ m/s^2$)

$A$ particle of mass $m$ is kept at rest at a height $3R$ from the surface of the Earth,where $R$ is the radius of the Earth and $M$ is the mass of the Earth. The minimum speed with which it should be projected upward so that it does not return back is ($g$ = acceleration due to gravity on the Earth's surface).

The least velocity required to throw a body away from the surface of a planet so that it may not return is (radius of the planet is $6.4 \times 10^6 \ m$,$g = 9.8 \ m/s^2$).

Planet $A$ has mass $M$ and radius $R$. Planet $B$ has half the mass and half the radius of Planet $A$. If the escape velocities from the Planets $A$ and $B$ are $v_{A}$ and $v_{B}$ respectively,then $\frac{v_{A}}{v_{B}}=\frac{n}{4}$. The value of $n$ is

$A$ 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?

If the acceleration due to gravity on the surface of a planet is two times that on the surface of the Earth and its radius is double that of the Earth,then the escape velocity from the surface of that planet in comparison to the Earth will be: