Can you exit the event horizon with a rocket?

Question

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.

John Rennie · Accepted Answer

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!

Allure · Answer

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.

stuffu · Answer

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.

AstroFedale · Answer

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

Umaxo · Answer

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.

Erwin Pratama · Answer

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.

Can you exit the event horizon with a rocket?

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