Possible to get above a black hole's event horizon from below it?

[2552]

Below the event horizon of a black hole, the escape velocity is more than the speed of light (which means, from below the event horizon, establishing an escape trajectory from the black hole's SOI is impossible). But, is it hypothetically possible to, from below the event horizon, change your orbit to take you to a point just above the event horizon, then circularize your orbit from there, then escape from a point (outside the horizon) where the escape velocity is below c?

Put another way, assuming you had a circular orbit of the black hole just outside the event horizon, and lowered your periapsis to just below the event horizon, would your orbit take you just below the horizon and back? Or would tidal forces lower your whole orbit beneath the horizon and spaghettify you?

 

[Jarvitä]

I know this is an uneducated assumption and likely wrong, but couldn't you use a solar-sail like drive and escape with the help of Hawking radiation?

 

[TSPenguin]

Hawking radiation is beeing emitted at the event horizon, so that would not work.
Besides that, to escape the event horizon, you will need to go past the speed of light.
So if you fall in, even then you could not get back out.
The space around the singularity is just too distortet to even go in on a normal parrabel.


-----Posted Added-----


Keep in mind, that I am not nearly informed enough to give detailed (or even 100% correct) answers.
However, once you get into the event horizon, there is no way to get out by means of our physics.

 

[David]

is it hypothetically possible to, from below the event horizon, change your orbit to take you to a point just above the event horizon, then circularize your orbit from there, then escape from a point (outside the horizon) where the escape velocity is below c?

That actually seems feasible, to me. As far as I know, a black hole, conceptually, is simply a gravity well (with an "event horizon" outside the "surface" of the central body). Escape velocity is a function of radial distance, so that inside a black hole's event horizon is that radial distance at which the escape velocity > c, but I cannot think of a theoretical reason why one could not orbit a black hole, elliptically, with a periapsis within the event horizon and an apoapsis outside the event horizon. And then, at apoapsis, the escape velocity would be < c, so that one could accelerate to escape from there.

I think that the behaviors of black holes, are generally considered wrt objects upon which only gravity is acting (inanimate materials, surrounding the black hole). With a spacecraft capable of producing additional forces (thrust), the consideration would be different. Of course, the extent to which your proposition is technologically feasible (a fuel supply, sufficient to enable acceleration to escape velocity, from a point outside but near to the event horizon) would be something else to consider; the more elliptical the orbit, the less of a problem, this would be.

Another consideration, somewhat suggested by TSPenguin's remark about spatial distortion, would be the extent to which one's orbit would be gravitationally affected by other materials within the event horizon. Perhaps tidal forces would also be significant; this would depend upon the dimensions of one's vessel and the dimensions of the black hole (its gravity gradient).

I am not aware that black holes are conceptually defined by energy considerations, such that objects within the event horizon, would necessarily be characterized as having insufficient energy to escape, or something like that; presumably, the energy contained in fuel, would not be theoretically constrained from participating in a process that would enable escape.

 

[Linguofreak]

Below the event horizon of a black hole, the escape velocity is more than the speed of light (which means, from below the event horizon, establishing an escape trajectory from the black hole's SOI is impossible). But, is it hypothetically possible to, from below the event horizon, change your orbit to take you to a point just above the event horizon, then circularize your orbit from there, then escape from a point (outside the horizon) where the escape velocity is below c?

Put another way, assuming you had a circular orbit of the black hole just outside the event horizon, and lowered your periapsis to just below the event horizon, would your orbit take you just below the horizon and back?

Under Newtonian physics, what you have described would work in theory.

Under General Relativity, the basic problem is that space itself can be seen as moving inwards towards a gravitating body at escape velocity. Once escape velocity exceeds c, there simply is no way to move outwards, only inwards.

Or would tidal forces lower your whole orbit beneath the horizon and spaghettify you?

Spaghettification and tidal forces are a completely different issue from the behavior of the event horizon. Depending on your size, structural strength, and the size of the black hole, you may get spaghettified well before you reach the horizon, or a long time after you pass through it. Tidal forces and spaghettification are a feature of Newtonian gravity (In fact, Saturn's rings result from it having spaghettified a moon). Event horizons and the inescapability of black holes are features of gravity under General Relativity.

 

[tgep]

Since Black holes do evaporate ( over an incredibly looooooooong period of time ) there's a good chance that at least of of your original sub-atomic particles will get spit out as a particle pair at some point but there's no garruntee that your part of the pair wont be the one that gets re-absorbed.
:beach:

As for a complete object, the only feasible chance would be if the black hole in question were super-massive because the tidal forces are ironicly less severe the larger a black hole is.

I'd go dig up some math on that but quadratic equations make my head hurt.:sorry:

I suggest Prof. Hawkings " Black Holes and Baby Universes and Other Essays " for a good read on the subject.

Bantam Books
1993
ISBN 0-553-09523-4

 

[Linguofreak]

Since Black holes do evaporate ( over an incredibly looooooooong period of time ) there's a good chance that at least of of your original sub-atomic particles will get spit out as a particle pair at some point but there's no garruntee that your part of the pair wont be the one that gets re-absorbed.
:beach:

Not really how it works. The particles that get radiated out aren't really copies of the ones that went in. It's just that, over the time it takes the hole to evaporate the mass, charge, and angular momentum totals of the particles that get radiated end up matching the totals for the matter that went in in the first place.

As for a complete object, the only feasible chance would be if the black hole in question were super-massive because the tidal forces are ironicly less severe the larger a black hole is.

Nope. You will never get anything back from inside the event horizon of a black hole. Just because the object manages to get inside the event horizon intact (as it would with a supermassive black hole), doesn't mean that it will ever come out again.

Even Hawking Radiation doesn't actually extract anything from the black hole. The process occurs just above the horizon, and can really be described more as canceling out what's already gone in than as anything having escaped in any way.

 

[TSPenguin]

You would have to reach c to escape, but you can not do that because you have mass. Nothing with mass can ever reach c, only come infinitely close to it.

(I just thought this was missing from the thread)

 

[ryan]

And also we have even really confirmed the existence of black holes.

 

[cjp]

You don't need to have escape velocity to escape a body. If you manage to maintain a continuous vertical velocity of 1 m/s for a long time, then you will escape earth sooner or later. Of course, doing so with rocket engines is terribly inefficient.

So, an interesting question would be whether the amount of fuel would be a problem. If m is the rest mass of your fuel, then the maximum amount of energy you can get from it is E = m*c^2. But, besides fuel energy, you also have potential and kinetic energy. I don't know how this works out in the relativistic case, but in Newtonian physics, an object approaching in a hyperbolic orbit will automatically have enough energy to leave the system again. Now, conservation of energy remains valid, so the only question is whether there is a way to avoid losing energy (and gaining energy during the slingshot would be a fun thing too).

Finally, there is the question of time. How long will it take to finish the maneuvre? How many ship-years will have passed, and what will be the year when you finally get out of the black hole?

 

[Eagle]

You don't need to have escape velocity to escape a body. If you manage to maintain a continuous vertical velocity of 1 m/s for a long time, then you will escape earth sooner or later. Of course, doing so with rocket engines is terribly inefficient.

Yeah, that's not really much different from reaching escape velocity. Technically, when you're on the other side of the Universe and are still moving 1 m/s away from the body, you're going faster than escape velocity (which is defined as your relative speed approaching zero as the distance to the body becomes infinite.)

The problem here is to pass the event horizon the spacecraft's acceleration still has to be greater than c.

I've always wondered if a hyperbolic trajectory (just barely into the event horizon) would work. Classically, inside the event horizon the craft would move faster than c. General relativity doesn't really allow for this though. So my guess is you get an interesting distribution of x-rays.

 

[cjp]

The problem here is to pass the event horizon the spacecraft's acceleration still has to be greater than c.

:facts::graduate: c is a velocity, not an acceleration. Can you be more precise in what you mean?

I've always wondered if a hyperbolic trajectory (just barely into the event horizon) would work. Classically, inside the event horizon the craft would move faster than c. General relativity doesn't really allow for this though. So my guess is you get an interesting distribution of x-rays.

Arggh... X-rays like in a cyclotron? Coming from your ship? That would mean you're losing energy!!! But wait... in a cyclotron, the X-rays only appear because the particles are charged, so you only need to make sure your ship doesn't have any net electrical charge. Now you only need to worry about gravitons I think (X-ray = photons = electromagnetic field particles = related to charge; gravitons = gravitation field particles = related to mass)

 

[Eagle]

:facts::graduate: c is a velocity, not an acceleration. Can you be more precise in what you mean?

X-rays like in a cyclotron?...

I forget if its c/s or c/(infinitesimally small unit of time). I have a feeling its the second, not so discrete number.

Maybe, I don't know. Its either that or nothing. For the net charge thing, depending upon when atoms break down, we may see radiation from the electrons and protons themselves. The reasons I believe we should see 'some' radiation is that the space around black holes spew gamma, x and other rays (especially at the poles), I figure if our ship is big enough we might see SOME change.

Meh, the experts don't know, and I know even less.

 

[tblaxland]

And also we have even really confirmed the existence of black holes.

There is general scientific consensus that they exist.

 

[Urwumpe]

When you move fast through the gravity field of a black hole, you should at least produce gravity waves, according to the general theory of relativity, which also take energy from you. Also, you can NOT become faster than c, as c already means infinite kinetic energy. Regardless where you start from, you should always reach the event horizon with a lower speed as the escape velocity - you can only pass if you keep enough distance to the black hole and have enough excess velocity left.

 

[Linguofreak]

You don't need to have escape velocity to escape a body. If you manage to maintain a continuous vertical velocity of 1 m/s for a long time, then you will escape earth sooner or later. Of course, doing so with rocket engines is terribly inefficient.

Once again, space itself is moving inwards at c at the event horizon, and faster than c inside of it. At the event horizon, any vertical velocity greater than zero with respect to an outside observer is faster than light with respect to local space. Just below the event horizon, even a vertical velocity of zero as seen from the outside is FTL as seen locally, only negative vertical velocities correspond to local velocities of c or slower, and the further you go below the event horizon, the more negative your vertical velocity as seen from the outside has to be to remain locally slower than light.

Furthermore, once you pass the event horizon, the way to maximize your time to live is not to apply any thrust at all. Due to the way relativity works, any thrust, even outwards, will make you hit the singularity more quickly.


-----Posted Added-----

Also, you can NOT become faster than c, as c already means infinite kinetic energy. Regardless where you start from, you should always reach the event horizon with a lower speed as the escape velocity - you can only pass if you keep enough distance to the black hole and have enough excess velocity left.

You cannot exceed c with respect to local space, but local space at the event horizon does reach c with respect to a stationary observer at infinity (which is exactly why you can't get back out). Below the event horizon it exceeds c.

If you fall in from infinity, you will reach c at the horizon with respect to a stationary observer at infinity.

 

[TSPenguin]

Furthermore, once you pass the event horizon, the way to maximize your time to live is not to apply any thrust at all. Due to the way relativity works, any thrust, even outwards, will make you hit the singularity more quickly.

This is interesting, could you explain this in more detail?

 

[Urwumpe]

You cannot exceed c with respect to local space, but local space at the event horizon does reach c with respect to a stationary observer at infinity (which is exactly why you can't get back out). Below the event horizon it exceeds c.

If you fall in from infinity, you will reach c at the horizon with respect to a stationary observer at infinity.

Yes, but what does that mean for you? The expected behavior is, that an invincible observer would disappear inside the black hole, so conservation of energy does seem to have a problem inside it, which makes sure you can not leave. If you reach c, right at the point where the escape velocity is c, you should be able to escape from a black hole on a parabola or a long elliptic orbit. So, conservation of energy alone can not explain it.

 

[Kyle]

Ya'll hear of Energy = mass constant (Sq)? We'll, put that in the equation. I believe the words were that nothing can go faster than the speed of light?

 

[Linguofreak]

I've drawn up a (very) rough illustration of how space and time work in the vicinity of a black hole. "t" is the time axis and "x" is the space axis. Velocity is the angle between t and an object's path, with a 45 degree angle being c and a 90 degree angle (path parallel to x) being infinite speed. An angle to t greater than 90 corresponds to backwards time travel. The lines at 45 degree angles to the x and t axes form the light cone, and are the paths that light will take. The blue region on the inside of the lightcone is the future. For objects inside the event horizon, the singularity is not a place in space but a moment in time, since all paths that move into the future hit the singularity (in other words, t points more towards the singularity than towards the top of the diagram).

This diagram dives into general relativity, and so for people who don't have a good grasp of special relativity it might be a bit hard to understand (General relativity is basically the answer to the question "How do we fit gravity into special relativity," and so if you don't know special relativity, an explanation of how gravity fits in will go straight over your head). TSPenguin's request for more info on why thrusting reduces your time to live once you're past the event horizon also requires some background from special relativity. General relativity tells us why we can't avoid hitting the singularity, but we have to have background on special relativity to know why freefall gives you the longest time before you hit it. (Basically the answer is that under special relativity, getting from one place and time to another place and time takes less time if you accelerate to get there than if you coast. At some point I'll have to explain special relativity so you know exactly what that means).

As for Urwumpe's question, I can't answer exactly, but the basic answer is that Newtonian gravity is the limit of General Relativity for weak fields. Under Newtonian gravity you would indeed come back out on a parabola (space is flat and gravity drops off with the inverse square of distance). Under General Relativity things are different (as observed in the precession of Mercury's orbit), and when escape velocity approaches c, things become very different (space is curved, and gravity drops off with something radically different from the inverse square of distance). So you don't come out on a parabola, and all your kinetic energy, as well as linear and angular momentum, gets transfered to the black hole in what is effectively a completely inelastic collsion.

http://www.orbiter-forum.com/gallery/data/541/Black_hole_lightcone_illustration.jpg