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

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 escape velocity from the surface of the Earth is $V_e$. The escape velocity from the surface of a planet whose mass and radius are $3$ times those of the Earth will be:

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 $6.4 \times 10^6 \ m$,$g = 9.8 \ m/s^2$).

Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B,$ then the ratio of the escape velocity $\frac{V_A}{V_B}$ is

For a satellite,the escape velocity is $11 \ km/s$. If the satellite is launched at an angle of $60^\circ$ with the vertical,then the escape velocity will be ........... $km/s$.