$v_e$ and $v_p$ denote the escape velocity from the Earth and another planet having twice the radius and the same mean density as the Earth. Then:

$A$ body of mass $m$ is thrown vertically upward with speed $\sqrt{3} v_e$, where $v_e$ is the escape velocity of a body from the Earth's surface. The final velocity of the body is

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:

The escape velocity from the earth is about $11 \, km/s$. The escape velocity from a planet having twice the radius and the same mean density as the earth is ......... $km/s$.

The escape velocity of a body from the surface of the Earth:

$A$ 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).