What is the escape energy of an object of mass $m$ at a height of $4R_e$ from the surface of the Earth? ($R_e$ = radius of the Earth)

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:

For an object projected from the Earth,on which two factors does the escape velocity $NOT$ depend?

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is

$A$ body is projected from the earth's surface with thrice the escape velocity. What will be its velocity when it escapes the gravitational pull?