If we say that the escape velocity from a planet is say 10 km/s we think that a slower body will move away from that planet but will be eventually forced to fall back on the planet. In simple words we don't say the body won't move at all but it couldn't leave for ever the planet. What is confusing for me is the escape velocity at the black hole event horizon. If it is the speed of light does it mean that a slower body would leave the horizon but fall down again or that is impossible for that body to make a path at all even 1mm away from the event horizon?

  • $\begingroup$ You can't throw a mass up in the first place. $\endgroup$ Commented May 27, 2022 at 20:41

2 Answers 2


Your arguments make sense in pre-relativistic Physics, but they don't apply to actual black holes, which are described by General Relativity. In this answer, I'll try to briefly sketch the difference.

"Black Stars" in Newtonian Gravity

In Newtonian gravity, it is not difficult to compute the escape velocity at a distance $d$ from the center of a spherical mass to be $$v = \sqrt{\frac{2GM}{d}},$$ which is even quoted at Wikipedia. In this framework of Classical Mechanics, we can then ask ourselves: what if a massive body was so compact that the escape velocity at its surface would be $c$? With some algebra, we figure out that the distance $d$ needed for this to happen is $$d = \frac{2 G M}{c^2}.$$

In this framework, your arguments are correct. Suppose you have a spherical body of mass $M$ and radius, say, $R = 0.99 d$. Then one could shoot a rock upwards at speeds below that of light, it would be able to surpass the position $r = d$ and would eventually fall back. The limitation would be simply that objects with speeds below that of light would need to fall back eventually.

However, this argumentation is based on pre-relativistic Physics. In particular, it employs Newtonian gravity, which is not an appropriate description of strong gravitational fields and is inconsistent with the Special Theory of Relativity.

Black Holes in General Relativity

In General Relativity, the setup is completely different. $c$ is now an incredibly important parameter in the theory that not only describes the speed of light. It describes the speed of causality itself. What happens this time is that one tries to describe the gravitational field of a spherical object with mass $M$ and finds that is it done by means of a curved spacetime with metric given by $$\mathrm{d}s^2 = -\left(1 - \frac{2G M}{c^2 r}\right)c^2\mathrm{d}t^2 + \left(1 - \frac{2G M}{c^2 r}\right)^{-1}\mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2.$$

In this setup, what happens at $r = \frac{2GM}{c^2}$ is far different. Previously, in pre-relativistic Physics, this equation determined a region of space. In GR, this equation determines a region of spacetime that does not behave as a spatial region, but rather it is what is called a null surface. In layperson's terms, it behaves a bit more as an instant in time than as a surface in space. In this framework, crossing this region is not like stepping through your doorstep, but rather like literally walking backwards or forwards in time. Nothing can escape the event horizon without literally moving backwards in time.

The previous concepts of escape velocity or even of force don't really make sense anymore. If you attempt to define the force needed to hover over $r = \frac{2 G M}{c^2}$ (i.e., to stay fixed at this radius without falling in), you'll find out that this force is infinite, which is greatly different from the Newtonian prediction. The reason is simply that the Newtonian theory fails to describe these strong fields, and one must instead resort to a different theory (namely, General Relativity).

In Short

No, after falling in, it is impossible for any body to move even a millimeter past the event horizon. To get out of the event horizon, as little as it might be, would actually be equivalent to travelling to the past. The apparent difficulty in reconciling this with the notion of escape velocity is that escape velocity is a notion appropriate to pre-relativistic Physics, while black holes and the event horizon are described in a different framework—namely, General Relativity.

  • $\begingroup$ Is the metric you wrote called something? it looks very much like the Schwarzschild Metric (although, a difference(at least to my knowledge, correct me if I am wrong) is the $c^2$'s) $\endgroup$ Commented Jan 7, 2023 at 17:17
  • 1
    $\begingroup$ @WilliamMartens It is the Schwarzschild metric, but I wrote it without setting $c=1$. When writing it, I probably thought it would be more pedagogical to keep the factors of $c$ in this answer. Most books will set $c=1$ and disappear with them (which is indeed way easier when you know what you are doing) $\endgroup$ Commented Jan 7, 2023 at 18:44
  • $\begingroup$ I see, yes! it did strike me later, but thanks; (I do remember the book by Hamilton I read, explains this with Setting c = 1) so; yes and thanks for the amazingly fast reply! $\endgroup$ Commented Jan 7, 2023 at 18:50

The people who first postulated the existence of an object like a star or planet massive enough that the escape velocity would be equal to that of light did apparently think in those terms (light struggling to leave the object but inevitably being drawn back in, or light beams hovering motionlessly in space right next to the object).

Then hundreds of years later when general relativity was understood to predict the existence of black holes and they (later) became subjects of study by mathematical physicists, those ways of imagining the workings of a black hole had to be abandoned and replaced with something entirely different.

My understanding of this is that once any object falls through the event horizon of an uncharged, nonspinning black hole its world line points in only one direction- towards the singularity- and so it gets carried in that direction in the same way that it gets carried forward in time. Since no world lines can bend outward to cross the EH on an exit path, it makes no sense to think of a photon for instance on the inside of a black hole trying to head out, slowing down, and falling inwards again.

This is a frightfully complicated business and I hope someone here can give you a more complete answer than this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.