Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I often hear that nothing can escape a black hole because it's "escape velocity" is greater than c. If that is accurate, what about the following. I know that the following has a lot of most likely impossible assumptions, but I'm trying to understand whether the escape velocity explanation makes sense.

Suppose you've got some sort of space ship that's inside the event horizon, and it hasn't been destroyed by the gravity or tidal forces. Also suppose that you've got a whole lot of reaction mass and a fantastic power supply.

Could this space ship, through applying enough constant force escape the event horizon?

share|improve this question
Mitchell, you might want to accept the answer as given. It answers your question very well. – dotancohen Jul 28 '14 at 15:03
up vote 14 down vote accepted

One of the clearest way to see what's going on is to look at a Penrose diagram of a Schwarzschild black hole. A Penrose diagram is like a map of the spacetime drawn in such a way as to preserve angles and put the light rays diagonally at $45^\circ$ angles, forming the light cones.

Since we're mapping all of an infinite space(time) into a finite drawing, distances are necessarily distorted, but that's a small price to pay.

Schwarzschild causal diagram, original by Andrew Hamilton

Time goes up on the diagram, and a typical infalling trajectory is in blue.

Because every massive object must be locally slower than light, it must stay within those light cones. Hence, no matter how you accelerate, at every point of your trajectory you must go in a direction that stays within those $45^\circ$ diagonals at that point. But once you're inside the horizon, every direction that stays within the light cones leads to the singularity.

Accelerating this way or that just means that you get to choose where you hit the singularity--a little on the left on the diagram or a little to the right. Trying to escape using high acceleration brings you closer to the light's $45^\circ$ lines, which because of time dilation will actually shorten your lifespan. You'll actually get to the singularity sooner if you struggle in that manner.

This particular image came from Prof. Andrew Hamilton. Note that it pictures an eternal Schwarzschild black hole, i.e. one that has always and will exist. An actual black hole is formed by stellar collapse and will eventually evaporate, so its diagram will be slightly different (in particular, there will be no "antihorizon"). However, in all respects relevant here, it's the same situation.

share|improve this answer

This is more of a physics question than it is astronomy, but nevertheless, think of it this way (most of what I say requires verification as I am not qualified to give a 100% correct in details answer): Light can't escape a black hole, and no propulsion system can accelerate an object to the speed of light. The faster you go, the more you weigh, the more force you have to input to maintain acceleration, the more energy you waste over time. If drawn on a graph, the energy required to increase speed by dX when at speed X, the graph has an asymptote at the speed of light. Meaning that it won't reach it no matter how much energy is input.

Take a look at LHC, where hadrons are accelerated to such a point when not their speed is measured anymore, because it hardly changes, but rather the energy of the particles, which increases a lot when approaching the speed of light. This will give you perspective about how unreachable it is.

So, if you can't accelerate to the speed of light then you, like light, are trapped inside the event horizon.

Also, AFAIK, light doesn't exit not because photons have weight, but because time stops. So light doesn't travel, nothing happens because there is no time.

Again, I'm not qualified. I just like to read about this stuff.

share|improve this answer
My question revolves around my understanding that an escape velocity is an initial velocity, that without any additional acceleration is sufficient to escape the gravitational pull of a planet. So, for example, if we had a good enough power source and reaction mass, we should be able to get arbitrarily far from the earth, while moving at a constant slow speed. Correspondingly could this work with a black hole, or is there something about the curvature of space that makes escape impossible. – Mitchell Kaplan Nov 25 '13 at 12:20
Like I've explained, you can't even reach the speed of light, and not even light can escape. There is nothing you could to to reach the speed of light. You need an infinite amount of energy to do this, which means in other words that it's impossible. Also, movement is described as a change in position in a time limit, and is defined by speed (t^-1), acceleration (t^-2) and higher level equations (t^-n), which are all dependent on time (t). Well beyond the event horizon there is no time, so "movement" doesn't make any sense as a concept. – AlexanderMP Nov 25 '13 at 18:10
Hmm, photons follow null geodesics of space-time. Why do you say that time stops? Also, what do you mean by this? Do you mean time as seen by an observer outside of the event horizon, or do you mean the local coordinate time of somebody who has fallen in. I think that part of your answer should be omitted. – astromax Jan 5 '14 at 17:02

I am only trying to explain using classical physics without reference to theory of relativity (which becomes relevant at such masses and speeds approaching that of light). For any movement away from singularity within event horizon, the expression 1/2 mv2 -GMm/r has to be +ve where m is the mass of the so called space ship, r distance between the ship and singularity and M is mass of the black hole. For a planet like earth even if v of a rocket with a means of self propulsion is less than the escape velocity i.e. 11.2 Km/Second relative to earth, the expression 1/2 mv2 can be > GMm/r and hence a +ve value (which at infinite distance becomes 0 for an object without propulsion but with an initial velocity>escape velocity) is possible. But for a space ship within event horizon the first part of the expression i.e. 1/2mv2 does not become large enough to overcome - GMm/r even if v becomes c i.e. speed of light. Hence no positive movement away from singularity i.e. no escape. (In fact a large -ve cumulative value indicates any effort to escape at less or even at speed of light will result in further fall towards singularity)

share|improve this answer
-1 does not answer the question: in non-relativistic physics, a rocket can go arbitrarily far away from a gravitating body while stays below the escape velocity, by use of continuous thrust. – Stan Liou Jun 21 '14 at 11:03
@Stan Liou I totally with you, but why it doesn't work like that under the event horizon - how relativistic effects make it impossible? – Mithoron Jan 9 '15 at 21:28

Although I partially agree with @Alexander here are a few more things.

According to Einstein's special theory of relativity, particles cannot travel at the speed of light (c). It would take infinite energy as stated above. However, the special theory of relativity states that an object can travel slower or faster than the speed of light but not at the speed of c. We may refer to the objects that travel faster than the speed of light as tachyons. There are celestial objects which are accelerated faster than the speed of light.

A common thing that everyone says is that nothing can escape a blackhole. But it is totally wrong. Blackholes do emit particles and radiation called as Hawking radiation. Even stephen hawking has changed his mind about blackholes as the general concept that blackholes absorb everything can be disproved by quantum mechanics theories.

So, If there exists a possibility that we can indeed escape a blackhole.

At CERN, reports says that ghostly subatomic particles called neutrinos were observed to travel faster than light. Researches are still trying to confirm it.

There is also another possible theory existing around the scientific community is the concept of wormholes. If a ship were trying to escape a blackhole, opening a wormhole to escape the blackhole and reaching another spot is a proposed theory.

There are also researches that say that objects are emanated from blackhole.

share|improve this answer
I think they've discovered an experimental error at CERN, something about the timing of the 2 clocks. So they no longer think that the neutrinos may have traveled faster than light. I think the things emanating from black holes are 1/2 of the virtual particle pairs that spontaneously appear in empty space, when a particular pair appears right outside the event horizon. I'm wondering if you need to be moving anywhere near c to escape a black hole as long as you can apply continuous acceleration. @AlexanderMP's comment about "time" may be my answer. But I also wonder about curvature. – Mitchell Kaplan Nov 26 '13 at 12:26
Although neutrino's have been proven wrong, the concept of tachyons still exist. One thing is for sure that there exists no element on earth to build a ship that can escape a blackhole even if we were able to harness the energy to do so. @MitchellKaplan The final results were that neutrino speed were equal to light. The margin of error was 15ns. – Mahe Nov 26 '13 at 12:51
@Mahe tachyons are not particles, they are instabilities, and the Hawking radiation arises not from the inside of the black hole but from the event horizon or slightly above. – Dilaton Nov 27 '13 at 1:32
Your comment about nothing escaping a blackhole is also misguided - see @Dilaton's response to it. Black holes are said to evaporate, but that doesn't mean that the things that were going in are now coming out (see Hawking Radiation - – astromax Jan 5 '14 at 16:57
There's so much wrong with the answer, I don't even know where to begin. Please don't speculate, this is science, not a layman's guessing game. – Florin Andrei Jun 20 '14 at 18:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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