The escape velocity of Earth is $v=\sqrt{\frac {2GM}{R}}$, where $M$ is the mass of the Earth and $R$ it's radius (approximating it as a sphere), and is much less than light speed $c$.
But I want to know the escape velocity of black holes. Is it much more than light speed?
Best Answer
In General relativity, energy formula of a body thrown straight up to the infinity is
$\large {E=\frac{mc^2}{\sqrt{1-R_S/R}}}$
As we know relativistic energy formula is
$\large {E=\frac{mc^2}{\sqrt{1-v^2/c^2}}}$
So
$\large {\frac{mc^2}{\sqrt{1-v_e^2/c^2}}=\frac{mc^2}{\sqrt{1-R_S/R}}}$
hence escape velocity equation in General relativity is
${\large {v_e^2=c^2\frac{R_S}{R}}}$
where $R_S=2GM/c^2$ - Schwarzschild radius of a black hole, and $R>R_S$
It's easy to derive that
${\large {v_e=c\sqrt{\frac{R_S}{R}}}=\sqrt{\frac {2GM}{R}}}$
So escape velocity formula in General relativity and Newton gravity is the same.