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Escape velocity is the ballistic speed required to escape from a gravitational field to infinity, ignoring any third body dynamics. The operative word here being ballistic, meaning unpowered.

In response to prior question removal:   
    
The below reference to a Fermi Drive is a tongue-in-cheek handy 
Gedanken construct for a non-ballistic mechanism. 
It is not unpublished personal theory.

There is no claim of a mechanism to escape a Black Hole.

Now let's suppose I have a Fermi spacecraft with effectively unlimited fuel. This could be via some as yet undefined reactionless drive or a highly efficient ion drive of some kind.

Now the escape velocity from Earth is around 25000 mph, but this is an unpowered ballistic velocity. However I could hop in my Fermi spacecraft and travel at a sedate 50 mph continuously straight up for as long as I like, until I eventually land on the Moon or go elsewhere. Escape velocity is not a factor here because this is a powered ascent not a ballistic one.

Now let's take this same concept to a Black Hole event horizon.

I can sedately and continuosly travel for as long as I like. I don't have to deal with super luminal escape velocities.

So where does this leave me? Can I escape from a Black Hole, or am I running up against Lorentz contractions in time or some such rather than escape velocity issues?

Fermi Drive - Patent Pending

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The problem is that you assume that an event horizon is simply an area within which the escape velocity becomes greater than $c$. This isn't really the full picture, otherwise we could just model black holes using Newtonian gravity; rather, it's just a heuristic that we use to calculate the Schwarzschild radius. It gives the "right answer for the wrong reasons", in a sense.

What really happens, according to general relativity, is that when you're within the event horizon, spacetime is contorted in such a way that there is literally no direction you can travel that will lead you outside the event horizon, regardless of how much kinetic energy you give yourself.

It's not like escaping Earth, where you can identify a particular direction (namely, upwards) that, if you keep moving along it, will eventually allow you to leave Earth's sphere of influence. Inside the event horizon, there is no such thing as "upwards" anymore. No matter which direction you move, you will end up closer to the singularity.

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  • $\begingroup$ I am fairly confident that this is correct, but I'm still hoping for more clarity. $\endgroup$ – user10216038 Sep 26 '20 at 21:11
  • $\begingroup$ @user10216038 What are you looking for in terms of clarity? Which specific points are unclear? $\endgroup$ – probably_someone Sep 27 '20 at 2:13
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Now the escape velocity from Earth is around 25000 mph, but this is an unpowered ballistic velocity. However I could hop in my Fermi spacecraft and travel at a sedate 50 mph continuously straight up for as long as I like, until I eventually land on the Moon or go elsewhere. Escape velocity is not a factor here because this is a powered ascent not a ballistic one.

This is a very interesting take. As you note, for continuous thrust the escape velocity is not relevant. However, that is not what defines the event horizon, it is more of an interesting coincidence. However, there is another interesting property that is more directly relevant to your question:

Notice that your Fermi drive has no minimum escape speed, but it does have a minimum escape acceleration. If the Fermi drive does not produce more than 1 g of proper acceleration at ground level then you will not escape earth by this method.

So what is the minimum proper acceleration to escape a black hole? It turns out to become infinite at the horizon. So regardless of what acceleration your Fermi drive can produce, it will be insufficient to go outward at any velocity at the horizon.

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  • $\begingroup$ That doesn't seem to hang together in my mind. As you pointed out, on Earth it would be a 1G+ acceleration at ground level, but it would actually drop with altitude to maintain constant velocity. Your logic seems to suggest infinite gravity at the event horizon, but isn't that the definition at the singularity? It would have to be less at the horizon wouldn't it? I suspect it's more of a closed space-time horizon as suggested by @probably_someone but I lack the ability to visualize more than 3 dimensions of space. $\endgroup$ – user10216038 Sep 30 '20 at 0:24
  • $\begingroup$ No, the proper acceleration for a hovering object becomes infinite at the event horizon, not the singularity. The tidal gravity is what becomes infinite at the singularity $\endgroup$ – Dale Sep 30 '20 at 0:44
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Imagine yourself on your infinite propellant drive competing against a photon. The photon travels with fixed velocity $c$, you travel with a much smaller velocity initially, but your acceleration is constant and nonzero. Question: will you eventually overtake the photon?

The answer to this question is no.

This is because the kinematical relation $$ x(t) = \frac{a t^2}{2} $$ is Newtonian. It gets relativistic corrections when your velocity approaches the speed of light.

The relativistic correction to this formula makes it very clear that you will never reach the speed of light or exceed it. Hence, the photon will always win in this race: $$ x(t) = \frac{c^2}{a} \left( \sqrt{1 + \left( \frac{a t}{c}\right)^2} - 1 \right) \lt c t. $$

When we say that the inner region of the event horizon is defined such that even light moving radially out of the black hole can't escape it, it means nothing else can escape. You and your infinite propellant drive will always stay behind the photon that is in free motion, and the photon will always stay behind the event horizon, hence you will stay behind the event horizon, too.

This argument, though it appears heuristic, is actually accurate within General Relativity. This is because of the equivalence principle: a small region of space in the immediate vicinity of the event horizon is actually almost flat (especially for supermassive black holes), hence Special Relativity applies.

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  • $\begingroup$ I understand the Special Relativity correction, but that's not what I was asking. I'm not attempting to exceed or even approach light speed. $\endgroup$ – user10216038 Sep 23 '20 at 21:51
  • $\begingroup$ @user10216038 you’re asking if you can escape the event horizon of a black hole with a rocket engine. My answer explains why you can’t. $\endgroup$ – Prof. Legolasov Sep 23 '20 at 21:53

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