Escape velocity at the surface of the Earth is $11.2 \, km/s$. If the radius of a planet is double that of the Earth but its mean density is the same as that of the Earth,then the escape velocity will be ........ $km/s$.

Maximum height reached by an object projected perpendicular to the surface of the earth with a speed equal to $50\%$ of the escape velocity from the earth's surface is - ($R =$ Radius of Earth)

The radius of a planet is $\frac{1}{4}$ of earth's radius and its acceleration due to gravity is double that of earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to its value on earth's surface?

If the radius and acceleration due to gravity both are doubled,the escape velocity of the Earth will become ......... $km/s$.

$A$ 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).

Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is :