The escape velocity for a rocket from Earth is $11.2 \ km/s$. Its value on a planet where the acceleration due to gravity is double that on the Earth and the diameter of the planet is twice that of Earth will be in $km/s$:

If the radius of a planet is $R$ and density is $\rho$,then the escape velocity $v_{e}$ of any body from its surface will be proportional to:

Two spherical stars $A$ and $B$ have densities $\rho_A$ and $\rho_B$,respectively. $A$ and $B$ have the same radius,and their masses $M_A$ and $M_B$ are related by $M_B = 2M_A$. Due to an interaction process,star $A$ loses some of its mass,so that its radius is halved,while its spherical shape is retained,and its density remains $\rho_A$. The entire mass lost by $A$ is deposited as a thick spherical shell on $B$ with the density of the shell being $\rho_A$. If $v_A$ and $v_B$ are the escape velocities from $A$ and $B$ after the interaction process,the ratio $\frac{v_B}{v_A} = \sqrt{\frac{10n}{15^{1/3}}}$. The value of $n$ is. . . . .

When the speed of a satellite is increased by $x$ percentage,it will escape from its orbit,where the value of $x$ is ....... $\%$

The least velocity required to throw a body away from the surface of a planet so that it may not return is (radius of the planet is $6.4 \times 10^6 \ m$,$g = 9.8 \ m/s^2$).

What is the escape velocity for the surface of a black hole?