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 $\text{ km/s}$)?

$A$ body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text{rd}}$ of escape velocity. The maximum height reached by the body is ($R=$ radius of earth).

If $g$ is the acceleration due to gravity at the earth's surface and $r$ is the radius of the earth,the escape velocity for a body to escape out of the earth's gravitational field is

$A$ particle is projected from the surface of the earth with a velocity equal to twice the escape velocity. When the particle is very far from the earth,its speed would be

$A$ particle is kept at rest at a distance $R$ from the surface of the Earth (of radius $R$). The minimum speed with which it should be projected so that it does not return is

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