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?

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    $\begingroup$ The escape velocity formula is $v_e=\sqrt{\frac {2GM}{R}}$ $\endgroup$
    – voix
    Commented Aug 10, 2012 at 21:05
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    $\begingroup$ 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. $\endgroup$
    – user10851
    Commented Aug 29, 2012 at 4:42

2 Answers 2


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

$\large {E=\frac{mc^2}{\sqrt{1-v^2/c^2}}}$


$\large {\frac{mc^2}{\sqrt{1-v_e^2/c^2}}=\frac{mc^2}{\sqrt{1-R_S/R}}}$

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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 11, 2012 at 12:36
  • $\begingroup$ @AlanSE - Yes, for a point-like object. $\endgroup$
    – voix
    Commented Aug 11, 2012 at 20:24
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    $\begingroup$ 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 $v_e=\sqrt{2GM/R - (2GM/R)^2}$. $\endgroup$
    – jinawee
    Commented Aug 15, 2017 at 21:29
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    $\begingroup$ 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. $\endgroup$
    – jinawee
    Commented Sep 14, 2019 at 9:59
  • $\begingroup$ Is this only true for radial motion? $\endgroup$
    – ProfRob
    Commented May 20, 2022 at 14:52

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 $c$.

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    $\begingroup$ Where the "surface" is the event horizon. There isn't necessarily any material surface there. $\endgroup$ Commented Aug 10, 2012 at 21:38
  • $\begingroup$ Yes, that means event horizon. $\endgroup$
    – Anixx
    Commented Aug 10, 2012 at 21:38
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    $\begingroup$ But if the escape velocity is $c$, why can't light escape? $\endgroup$
    – seriousdev
    Commented Aug 11, 2012 at 14:18
  • $\begingroup$ It can if it comes from a place just above the horizon. But it looses much of its energy and becomes redshifted. $\endgroup$
    – Anixx
    Commented Aug 11, 2012 at 22:20
  • $\begingroup$ There is no "the escape velocty". In GR, the escape velocity depends on the direction of launch. $\endgroup$
    – ProfRob
    Commented Feb 15 at 13:19