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:

$A$ particle is projected vertically up with velocity $v = \sqrt{\frac{4 g R_e}{3}}$ from the Earth's surface. The velocity of the particle at a height equal to half of the maximum height reached by it is .........

$A$ particle falls towards the Earth from infinity. Its velocity on reaching the Earth would be:

If the escape velocity on Earth is $11.2 \text{ km/s}$, what is its value for a planet having double the radius and $8$ times the mass of Earth (in $\text{ km/s}$)?

The mass and radius of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is:

$A$ body of mass $m$ is dropped from a height $h = \frac{R}{2}$ above the surface of the Earth,where $R$ is the radius of the Earth. Find its speed when it hits the Earth's surface. (Given: $v_e$ is the escape velocity from the Earth's surface).