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A stone being thrown from the surface must have a velocity of $11.2$ km/sec if it wants to escape Earth's gravity. However, a stone that has a constant force being applied on it need not have an escape velocity. A stone with a thruster with infinite fuel can travel as slow as 1 m/s and still escape Earth.

So now imagine a black hole. Let's name it “Black Hole $001$”. With just enough gravitational pull that it prevents light from escaping.

Photons once generated from the source have an inbuilt energy much like the stone and thus can't escape a black hole's pull. But shouldn't our previous 'stone with a thruster' be able to escape any classical (= non-quantum) black hole as long as it generates a force slightly more than the gravitational force of Black Hole $001$ on the stone?

Is this possible?

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    $\begingroup$ Escape velocity has nothing to do with why light cannot escape from a black hole. $\endgroup$ – StephenG Sep 13 '19 at 15:57
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    $\begingroup$ @Brick: if you define black hole through the event horizon then we can only have tautological statements: if object crosses event horizon it cannot escape, if it can escape it was not in a black hole. But by that logic we cannot call the result of large star collapse a black hole, because we have not yet excluded a possibility that a wormhole would open inside it at some future date, disgorging parts of its interior. $\endgroup$ – A.V.S. Sep 13 '19 at 16:50
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    $\begingroup$ @Brick: I see. Your formalization of the problem is the motion of test object on a fixed geometry. But if our object did escape, then this would be the change in causal structure. Which means backreaction must be taken into account. And there are plenty of model spacetimes with apparent horizon but without event horizons. Could the “stone with infinite fuel thruster” produce such a spacetime is an open question. $\endgroup$ – A.V.S. Sep 13 '19 at 17:21
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    $\begingroup$ @safesphere The light cone tips over more, in a sense that could be made precise. In the context of a question that has a stone with an infinite source of fuel (and finite mass!), it seemed appropriate to try to relate the point to the terms used by the OP, even if imprecise. A precise answer to this infinite fuel vehicle cannot be had anyway. $\endgroup$ – Brick Sep 13 '19 at 19:37
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    $\begingroup$ What's a "classical black hole"? $\endgroup$ – probably_someone Sep 14 '19 at 10:49
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While we can look at Newtonian physics and see that an object with escape velocity greater than $c$ would give rise to many of the properties of what we now know as blacks holes, the behavior of black holes is more complicated than what is visible from a Newtonian perspective. The fact that objects can't escape from a black hole is much more than just "everything is travelling slower than the escape velocity".

Even if it weren't from the general relativity effects, special relativity effects such as objects gaining mass-energy as they accelerate, and mass-energy approaching infinity as an object approaches $c$, would prevent an object from escaping. Again, this doesn't take into account the actual general relativity effects, but since you presumably aren't familiar with that level of physics, the following addresses why even without them, your idea of a rocket escaping a black hole doesn't work.

A stone with a thruster with infinite fuel can travel as slow as 1 m/s and still escape Earth.

Infinite fuel is not possible. So I'll interpret this as "sufficient fuel". How much is sufficient? Well, if it were already travelling at escape velocity, then a certain percentage of its mass-energy would be made up of its kinetic energy component. Call that percentage $p$. For a rocket starting at zero velocity to escape Earth, the percentage of its mass that is fuel must be at least (in reality, due to inefficiencies, much more than) $p$.

As an object approaches the speed of light, the percentage of its mass-energy that is kinetic energy approaches 100%. So if the escape velocity of a body is $c$, then a rocket starting at zero velocity at its surface would have to be made up of nothing but fuel and have a way of converting that fuel to kinetic energy with perfect efficiency. Once the escape velocity exceeds $c$, more than 100% of the rocket would have to fuel.

This shows that this sort of analysis breaks down when you have an object massive enough to be a black hole. To understand exactly how it breaks down, you need to understand general relatively, but at the very least this shows that what you propose doesn't work.

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  • $\begingroup$ This answer is misleading, as it implies that the restriction is technical and so it is possible to at least try to escape by accelerating in some direction that, if not for the technical limitation with the fuel, would supposedly lead outside. There is however no direction pointing outside, even if the engine were driven by a divine power. The restriction has nothing to do with the force or the speed. There simply is no route to escape. $\endgroup$ – safesphere Sep 14 '19 at 0:48
  • $\begingroup$ Actually I don't require to travel at the speed of light as stated in the question. All I want to do is to apply a force greater than the pull towards the center of the black hole. Please explain why it's not possible to exert such a force? $\endgroup$ – SamwellTarly Sep 14 '19 at 14:03
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    $\begingroup$ @SamwellTarly "the pull towards the center of the black hole" - Despite a wide spread misconception, a black hole does not have a center. A singularity is not a point-like object in space that attracts staff. This is an urban legend perpetuated by the ignorant press. A Schwarzschild singularity is an infinitely long line located in the future of the inner space a black hole. So there is no pull to resist or overcome. Imagine that time now is 11:59 am. In which direction would you apply a force to prevent yourself from reaching noon in one minute? $\endgroup$ – safesphere Sep 14 '19 at 18:58
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    $\begingroup$ @SamwellTarly I addressed your question on a level of energy. The amount of energy it takes to get an object to the escape velocity is the amount of energy it takes to get an object out of the gravity well. It doesn't matter whether the object reaches the escape velocity or not: if it leaves the gravity well, it must have been given at least as much energy as if it had been going at the escape velocity. This breaks down with black holes, because objects with mass can't go the speed of light, and so they can't be given as much energy as if they were traveling at the speed of light. $\endgroup$ – Acccumulation Sep 16 '19 at 14:32
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As long as it never crosses the event horizon, sure. Once it crosses the event horizon, it's gone forever. Gravity isn't really a force. It's a side effect of the curvature of space. Once you cross the event horizon, space becomes so curved there is simply no path out anymore. All worldlines point towards the singularity at the center of the black hole.

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    $\begingroup$ "Once you cross the event horizon, space becomes so curved there" - How would you rconsile this statement with the often quoted fact that, according to the equivalence principle, spacetime is locally flat everywhere, including in the vicinity of the event horizon? $\endgroup$ – safesphere Sep 13 '19 at 19:36
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    $\begingroup$ Is the question rhetorical, @safesphere? The local flatness that you mention only has to hold in a neighborhood of a reference point. As you move to regions that are more and more curved (as measured by the Riemann tensor), you become limited to smaller and smaller neighborhoods if you want to see local flatness to a fixed order. $\endgroup$ – Brick Sep 13 '19 at 19:42
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    $\begingroup$ @safesphere That's a bit like asking "If every polynomial is differentiable, then how can they be curved?" A function having a derivative means that the slope of its secants line converge, hence it is "locally straight". That doesn't mean it's globally straight. $\endgroup$ – Acccumulation Sep 13 '19 at 21:21
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    $\begingroup$ @Brick I am talking about a small neighborhood at the event horizon. The curvature within this neighborhood does not change to make this point the point of no return, as this answer claims. Thus the answer is incorrect along with your comment. $\endgroup$ – safesphere Sep 13 '19 at 23:26
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    $\begingroup$ @Acccumulation Apparently you did not understand my comment, so your objection is irrelevant. Please see my response to Brick above. The curvature is differentiable at the horizon and also can be arbitrary small for very large black holes. Therefore the curvature at the horizon cannot be the reason for the horizon to be exactly there. There is a different reason for the horizon to be where it is, but this reason is not the curvature at that point. $\endgroup$ – safesphere Sep 13 '19 at 23:36
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Here are two thoughts that might help your intuition.

  1. In special relativity, if you have a body subject to a constant force then it has a constant proper acceleration. It accelerates and accelerates. But its velocity still only tends to $c$ (relative to anything else).

  2. In the case of a black hole, once a body is inside the horizon it is carried to the singularity just as surely as you and I are carried into next week. It is not so much a question of slow and fast, as future and past. For events inside the horizon the whole future lies within the horizon. Once you have crossed it from outside, the horizon lies in your past. So rather than 'once inside the horizon' it might be better to say 'after crossing the horizon'.

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  • $\begingroup$ The second point is what I wish you elaborated. The question I would like to clarify is "What if after crossing the event horizon, the body exerts a force greater than the black hole's pull towards the singularity?" Why is such a strong opposing force not a possibility? Again to clarify, the body is not going to travel at the speed of light. It's going to travel much slower. Let's funnily assume it has a cruise control set at 0.3c $\endgroup$ – SamwellTarly Sep 14 '19 at 14:13
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    $\begingroup$ @SamwellTarly : Because, in effect, any and every path that a "force" can, no matter how strong, take you on, ends up at the black hole singularity. $\endgroup$ – The_Sympathizer Sep 16 '19 at 8:14
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I'm +1 on the answer by Ryan_L because I think that represents a very reasonable interpretation of the intent of your question with an answer that is correct within that interpretation. After some of the comments on the question, though, here's an answer taking a different interpretation of your question.

The premise of the question is that your stone can have "infinite fuel" and also (implicitly) have finite mass. (If it had infinite mass, then by $F=ma$ no finite force would be great enough to accelerate it, not to mention that the center of mass of the Earth-stone system would be at the stone, so you'd be thinking in that case of the stone moving the Earth. I suppose with "infinite" fuel you could also hypothesize an "infinite" force generated, but I won't go there because by then you've left mainstream physics.)

Even in the Newtonian case, that's not possible, of course. But if we somehow accept it, in the Newtonian case, perhaps as an approximation to a fuel with an extreme energy density, then the presence of this fuel does not change the dynamics. Newton's laws, we assume because we're doing physics in the Newtonian framework, apply and you can use $F=ma$ to move your stone, maybe, if you're tracking everything, using the rocket equation to account for the change in mass as fuel is spent.

In the case of general relativity, however, we care about the total mass-energy density for determining the spacetime in addition to the dynamics of your stone within the spacetime. So your stone with "infinite fuel" (whatever that means) is presumably going to curve spacetime itself by quite a bit. Your asserted amount of fuel now either creates a consistency issue or changes the problem drastically from what you originally setout to describe. The latter possibility is in contrast to the Newtonian case where the assumption was suspect but did not change the equations of motion. Now the more you try to add fuel to escape, the more you change the spacetime structure, which in turn increases the amount of fuel that you need. Even as a limiting process, I suspect that diverges or otherwise does "nasty" things that are beyond what you intended to probe with your question.

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  • $\begingroup$ The question specifies "infinite" fuel, @safesphere, which I understand the OP to believe gets them around some of the other limitations. My understanding of the question is that the OP also intends for this fuel to be burned for an infinite amount of time and thinks there is (or may be) some limiting process associated with that to get out. It is true that a force doesn't necessarily require a lot of fuel but that seems to be a different question. If you interpret the question to have a fixed background (as I did originally), then the other answer is right, as noted. $\endgroup$ – Brick Sep 13 '19 at 19:59
  • $\begingroup$ If you consider the stone/fuel system to be more than a test mass, then you don't have Schwarzschild anymore, which is what I was getting at here in the limit of it the stone/fuel becoming very significant. Your stone with its fuel may have the bigger effect on curvature if you go to that extreme. I think that if we agreed on the question, we'd agree on the answer, @safesphere. $\endgroup$ – Brick Sep 13 '19 at 20:01
  • $\begingroup$ Agreed. A caveat for the OP still is that, no matter how powerful the engine is, there is no itection to point to that would lead to outside. It's like escaping from today back to yesterday. $\endgroup$ – safesphere Sep 13 '19 at 23:52
  • $\begingroup$ @safesphere , What about a body generated inside the event horizon. For simplicity lets assume it's a hypothetical heavy atom constantly converting its mass into energy and moving out with a speed much less than the speed of light. $\endgroup$ – SamwellTarly Sep 14 '19 at 14:06
  • $\begingroup$ @SamwellTarly Notwithstanding the multiple reasons that your particle won't escape a black hole one inside the horizon and the potential to interpret your question more than one way, there is no way that something already inside a black hole is getting out, at least not according to accepted theories of classical physics. $\endgroup$ – Brick Sep 16 '19 at 12:37
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A rocket above a black hole burns during one millisecond (time at infinity) one ton (proper mass) of fuel, aiming the exhaust to the black hole. The mass of the black hole increases by 500 kg (as measured at infinity).

Another rocket at lower position than the previous one burns during two milliseconds (time at infinity) one ton (proper mass) of fuel aiming the exhaust to the black hole. The mass of the black hole increases by 250 kg (as measured at infinity).

Those two rockets are supposed to be identical. They burn the same proper mass of fuel during the same proper time.

Observes inside those two rockets feel the same force. So we say that the two rockets produce the same proper force.

If the lower rocket does not ascend but the upper rocket does, we can see, observing from a far away position, various reasons for that:

1) exhaust has less mass 2) fuel burns slower 3) force of gravity is larger

The third one is less important than the two other ones. Gravitational force changes by a small amount when the distance to the center of the black hole changes by a small amount, while time dilation approaches infinity as the distance to the event horizon approaches zero.

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