The reason given in most places about why one cannot escape out from an event horizon is the fact that the escape velocity at the event horizon is equal to the speed of light, and no one can go faster than speed of light.

But, you don't really need to reach the escape velocity to get away from a massive object like a planet. For example, a rocket leaving earth doesn't have escape velocity at launch, but it still can get away from earth since it has propulsion.

So, if a rocket is just inside the event horizon of a black hole, it doesn't need to have the escape velocity to get out, and it should at least be able to come out of the event horizon through propulsion. Also, if the black hole is sufficiently large, the gravitational force near the event horizon will be weaker, so a normal rocket should be able to get out easily.

Is this really theoretically possible? If it was just the escape velocity being too high was the problem of getting out, I don't see any reason why a rocket cannot get out.

This is a similar question, but my question is not about a ship with Alcubierre drive.

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    $\begingroup$ Closely related physics.stackexchange.com/q/557947 $\endgroup$
    – ProfRob
    Commented Jun 24, 2021 at 10:20
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    $\begingroup$ Something at the horizon and travelling in the directly outwards (radial) direction and travelling at the speed of light (e.g. a lightwave) will not simply fail to get a long way away from the horizon. It will fail to make any progress at all! It will remain permanently at the horizon. The horizon itself is not strictly speaking a "place" but you can define a place just outside and near to it; the light at the horizon will never arrive at such a place. $\endgroup$ Commented Jun 24, 2021 at 11:32
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    $\begingroup$ A heuristic consistent with the analogy: in a Newtonian situation, escape velocity is a function of your position. If you apply rockets to escape, your speed must eventually surpass the local escape velocity. But light trajectories are unique in that they always travel at the speed of light, so if escape velocity is the speed of light somewhere, it's also the speed of light everywhere on the escaping light ray. If you attempt to "follow" the light out with rockets, you can never surpass the local escape velocity because accelerating to the speed of light requires infinite kinetic energy. $\endgroup$
    – jawheele
    Commented Jun 24, 2021 at 16:40
  • $\begingroup$ @jawheele I didn't get it. Do you mean If there is a black hole somewhere in the universe, escape velocity from that black hole at the current position of the earth is also the speed of light, so we cannot escape from it? $\endgroup$ Commented Jun 25, 2021 at 1:26
  • $\begingroup$ @LahiruChandima Well, no, because even the light on a trajectory exactly where the "escape velocity" becomes $c$ wouldn't make it to the Earth (or anywhere outside the event horizon), so the bit about "everywhere on the escaping light ray" says nothing about the Earth. To rephrase, the observation is essentially that the light ray doesn't escape, and you can't surpass the light ray even with rockets because you can never match its speed, so you can't escape either. Compare to the Newtonian case, where you'd have to eventually pass any less-than-escape-velocity trajectory as you boosted away. $\endgroup$
    – jawheele
    Commented Jun 25, 2021 at 2:26

7 Answers 7


It is often said that the escape velocity at the event horizon is the speed of light, but while this is true in a sense it is not very useful. The problem is that the speed is an observer dependent quantity. An observer far from the black hole would say the escape velocity at the event horizon was zero, which is obviously nonsensical and proves only that speed is not a useful quantity to describe the motion near an event horizon.

There is more on this in the question Does light really travel more slowly near a massive body? though this may be excessively technical.

A better way to understand what is going on is to ask how powerful a rocket motor would you need to hover at a fixed distance from the black hole. For example to hover at the Earth's surface your rocket motor needs to be able to generate an acceleration of $g$ i.e. a force $mg$ where $m$ is the mass of the rocket. If your rocket motor is more powerful than this you will accelerate upwards away from the Earth and if it is less powerful you will fall downwards towards the Earth.

In Newtonian gravity the acceleration required to hover at a distance $r$ from a mass $M$ is given by the well known equation for Newtonian gravity:

$$ a = \frac{GM}{r^2} \tag{1} $$

The event horizon is at $r = 2GM/c^2$ so if Newtonian gravity applied we could substitute this into equation (1) to give:

$$ a = \frac{c^4}{2GM} \tag{2} $$

which is a large number, but some future physicist might be able to build a rocket that powerful. The problem is that when we move to general relativity equation (1) is no longer valid. The GR equivalent is derived in twistor59's answer to What is the weight equation through general relativity? The details are a little involved, but in GR the equation becomes:

$$ a = \frac{GM}{r^2} \frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} \tag{3} $$

If you now substitute $r = 2GM/c^2$ into this equation you find that the acceleration required is infinite i.e. no matter how powerful a rocket motor you build you cannot hover at the event horizon. Once at the horizon you are doomed to fall in.

And this explains why you cannot start at the event horizon and move away from it slowly using your rocket motor. You would need an infinitely powerful rocket!

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    $\begingroup$ Good answer, but could you comment how you derived (3)? I am curious how this relativistic factor pops up. $\endgroup$
    – Koschi
    Commented Jun 24, 2021 at 8:01
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    $\begingroup$ @Koschi I have added a link to the derivation of equation (3). $\endgroup$ Commented Jun 24, 2021 at 8:45
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    $\begingroup$ Thank you for taking the time to craft an explanation that appears (perhaps deceivingly ;-) ) accessible. One question: If I read (3) correctly, there is no solution in $\mathbb{R}$ for radii smaller than the event horizon because the term under the root becomes negative, right? $\endgroup$ Commented Jun 25, 2021 at 7:35
  • $\begingroup$ @Peter-ReinstateMonica Correct. The calculation of the proper acceleration in equation (3) assumes the spaceship is hovering at fixed $r$ but this is not physically possible for $r \le r_s$ so equation (3) returns a physically meaningless result. $\endgroup$ Commented Jun 25, 2021 at 10:22
  • $\begingroup$ This is a nice answer for force applied in the radial direction. It just makes me wonder if there is an equally simple explanation for what happens with a force in tangential direction. With Newtonian physics, as long as the rocket keeps firing, the tangential speed would slowly increase and the orbit would rise above the event horizon. $\endgroup$
    – jpa
    Commented Jun 27, 2021 at 7:23

That's why "escape velocity > speed of light" is not a good way to describe the event horizon of a black hole. It's convenient for understanding, but it's not precise. You simply cannot leave a black hole once you're within the horizon. That's because all possible trajectories point inwards.

See Wiki for an even more precise version of what I wrote above:

One of the best-known examples of an event horizon derives from general relativity's description of a black hole, a celestial object so dense that no nearby matter or radiation can escape its gravitational field. Often, this is described as the boundary within which the black hole's escape velocity is greater than the speed of light. However, a more detailed description is that within this horizon, all lightlike paths (paths that light could take) and hence all paths in the forward light cones of particles within the horizon, are warped so as to fall farther into the hole. Once a particle is inside the horizon, moving into the hole is as inevitable as moving forward in time - no matter what direction the particle is traveling, and can actually be thought of as equivalent to doing so, depending on the spacetime coordinate system used.

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    $\begingroup$ Or as its sometimes described, all futures are inward. Doesn't matter what you do, what direction you go, how powerful or fast, how close to light speed. None of that matters. The black holes gravity has so distorted spacetime, that EVERY direction at ANY speed leads only further inward. Trying to escape that is a bit like trying to not move into the future as time passes, or trying to escape from inside a sphere by finding the edge. Its actually not possible. $\endgroup$
    – Stilez
    Commented Jun 24, 2021 at 22:12
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    $\begingroup$ @Stilez: If we get the warp drive working all bets are off. $\endgroup$
    – Joshua
    Commented Jun 25, 2021 at 19:16

The reason given in most places about why one cannot escape out from an event horizon is the fact that the escape velocity at the event horizon is equal to the speed of light, and no one can go faster than speed of light.

This is misleading.

In general relativity, in simple terms, there is no gravitational force that would pull you anywhere, and the seeming pull is only artefact of the fact that spacetime is curved. Observer in free fall, if he is small enough so that tidal effects can be neglected, experiences no strange things near him and feels like he is in spacetime without any gravity at all.

If there is no pull, why then we cannot escape black hole? Just direct your engines away from the black hole, and go! The problem is, spacetime is curved so much, there is no direction away from the black hole. In fact, the black hole (singularity to be precise) is not at certain place to make you direct your spaceship away from it, it is in your future, and you cannot escape the future no matter which directions will you take.

But then, if the black hole is in the future and not in some point in space, why do we observe black holes to be somewhere? The answer is, because we are far away from them, and we are observing only event horizon and this we see as area of space. But once we get under the event horizon, it becomes our past, while singularity will become our future.

It is probably hard to understand if you are not familiar with the math, but the reason why nothing can escape event horizon is that spacetime is curved so much, there is no direction which would lead out. All possible directions of travel lead to central singularity (assuming the simplest, Schwarzschild black hole). You can imagine it like space itself is getting collapsed into singularity and this collapse of space drags you inward.

Here is a nice video about it.

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    $\begingroup$ "You cannot escape the black hole because it is in your future, and you cannot escape the future" -- you very successfully transformed GR math in prose there. Although it sounds like the ultimate abusive relationship, doesn't it ... $\endgroup$ Commented Jun 25, 2021 at 7:45

What you say about the concept of escape velocity is true, the problem with the event Horizon is that you need infinite proper acceleration just to stay still at the horizon, that is, if you compute the acceleration needed to stay still at the horizon you get a divergent quantity.

Instead, if you are inside the event horizon you simply cannot stand still. The way you cannot stand still or go outside of it it's similar to the way you cannot stand still or go backwards in time in this very moment. You will necessarily end up in the singularity


Absolutely not, mass can not exceed the speed of light, so if a rocket(made of mass) goes into the black holes event horizon it is too late, in fact at around 2r no object with mass can escape. Also, any acceleration would just make you go deeper into the black hole faster, not get you out.

This means you cannot use a rocket to escape from the event horizon once you have went past the horizon, as any acceleration would lead towards the black hole no matter the amount of acceleration(unless you exceed the speed of light before you hit the singularity, but this is not possible). In fact, to even just not fall in and stay still you would have to use infinite energy on the edge of the horizon, let alone actually inside the black hole.

So, in summary, you can not do as you say because you need to exceed the speed of light to even go away from the singularity once inside(not possible), you can not get further by slowly leaving as this slow, gradual escape makes you go towards the singularity faster, it does not bring you away.

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    $\begingroup$ Please add that last comment to your full answer ;) $\endgroup$ Commented May 20 at 12:16
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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented May 20 at 12:53

It should be nearly impossible to do. You have to be roughly the speed of light to escape. But If your position is far from the black hole, it looks like you can still escape.


The force of gravity is small. But it's a small force on a thing that contains zero energy. So the force gravity is infinite in a sense.

The rocket lost all of its energy when it was lowered to the event horizon.

It may be worth mentioning that the rocket motor burns fuel at infinitely slow rate. That has some effect on how much force the rocket generates.

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    $\begingroup$ I do not understand why you talk about zero energy, and infinitely small burning rate, and your conclusion "gravity is small, but also small on something containing zero energy, which makes gravity infinite in a sense". $\endgroup$
    – Koschi
    Commented Jun 24, 2021 at 7:57
  • $\begingroup$ @Koschi Lowering an object with local mass m to the horizon and then dropping it causes zero mass increase of the black hole. Lowering an object with local mass m to the horizon causes mass m to appear to the place where the original mass was originally, as braking energy. Person at the lower end of the rope used to do the lowering says that local rope segment transmits many units of momentum per local second. Person at the upper end says that local rope segment transmits not so many units of momentum per local second. Low person's second is upper persons year. $\endgroup$
    – stuffu
    Commented Jun 24, 2021 at 11:19

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