What is the escape velocity of a Black Hole?

Question

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?

You might be interested in my response to this related question, as it points out that escape velocities from black holes are not quite the same as Newtonian escape velocities. — , Aug 29 '12 at 04:42

voix · Answer 1 · 2012-08-10T22:21:09.353

12

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.

edited Aug 10 '12 at 22:21

answered Aug 10 '12 at 22:08

voix

2,084

So the escape velocity at the event horizon would go to the speed of light as the ratio in the radical goes to 1? Am I reading that right? That would still quite simply imply infinite energy, so it seems consistent. – Alan Rominger Aug 11 '12 at 12:36
@AlanSE - Yes, for a point-like object. – voix Aug 11 '12 at 20:24
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I don't buy your derivation. The energy of a particle is not given by the second formula you've written since it neglects the gravitational field, you are considering a free particle. The correct energy is $E=\gamma mc^2 \sqrt{1-2GM/R}$, which leads to $v_e=\sqrt{2GM/R - (2GM/R)^2}$. – jinawee Aug 15 '17 at 21:29
After coming back to this question, your answer is valid for a Schwarzschild BH when the escaping particle travels radially and you measuring the speed from the starting point. The formula I wrote assumes you measure the distance from a point at infinity, which is not the same since speed is coordinate dependent. – jinawee Sep 14 '19 at 09:59
Is this only true for radial motion? – ProfRob May 20 '22 at 14:52

score 7 · Answer 2 · edited Aug 10 '12 at 21:44

7

The escape velocity from the surface (i.e., the event horizon) of a Black Hole is exactly $c$, the speed of light.

Actually the very prediction of the existence of black holes was based on the idea that there could be objects with escape velocity equal to $c$.

edited Aug 10 '12 at 21:44

Keith Thompson

7,643

answered Aug 10 '12 at 21:34

Anixx

11,159

1

Where the "surface" is the event horizon. There isn't necessarily any material surface there. – Keith Thompson Aug 10 '12 at 21:38
Yes, that means event horizon. – Anixx Aug 10 '12 at 21:38
1

But if the escape velocity is $c$, why can't light escape? – seriousdev Aug 11 '12 at 14:18
It can if it comes from a place just above the horizon. But it looses much of its energy and becomes redshifted. – Anixx Aug 11 '12 at 22:20
There is no "the escape velocty". In GR, the escape velocity depends on the direction of launch. – ProfRob Feb 15 '24 at 13:19

What is the escape velocity of a Black Hole?

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