# Escaping Black Holes

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Orbital velocities are slower than escape velocities, so you could theoretically orbit a blackhole inside the event horizon(assume its so massive that there isn't a whole lot of sheer) without breaking any laws of physics(i think). To escape you could change your apoapsis till it is above the event horizon and still be going below c. when you get out side the event horizon you just bring your periapsis out of the event horizon by accelerating again and you could then accelerate to the escape velocity at that altitude. any flaws here?

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What´s an "apoapsis"? And as a follow-up question: What´s a "periapsis" ?

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its very simple buddy you can never destroy any "information" so anything that goes into black hole comes out of it. it can not permanently be there in. it will take 100 billions years or more i dont know but it will come out of it. "YOU CAN NEVER DESTROY ANY INFORMATION" on the edge of event horizon you can not match the speed of light but after years of decay it is possible. if i believe other theory than it is possible that at the end there are only blackholes in the universe and than they pull each other and whole universe will again be point. and than again for a moment few nanoseconds all laws of physics vanish. And again a "BIGBANG" back to the square one.

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Orbital velocities are slower than escape velocities' date=' so you could theoretically orbit a blackhole inside the event horizon(assume its so massive that there isn't a whole lot of sheer) without breaking any laws of physics(i think). To escape you could change your apoapsis till it is above the event horizon and still be going below c. when you get out side the event horizon you just bring your periapsis out of the event horizon by accelerating again and you could then accelerate to the escape velocity at that altitude. any flaws here?[/quote']

If this is the correct definition of "event horizon" then I think this should work (theoretically of course, don't try it at home )

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Orbital velocities are slower than escape velocities' date=' so you could theoretically orbit a blackhole inside the event horizon(assume its so massive that there isn't a whole lot of sheer) without breaking any laws of physics(i think). To escape you could change your apoapsis till it is above the event horizon and still be going below c. when you get out side the event horizon you just bring your periapsis out of the event horizon by accelerating again and you could then accelerate to the escape velocity at that altitude. any flaws here?[/quote']Outside of the Event Horizon there is a Photon Sphere where light is able to maintain orbit.

Below the Photon Sphere You would need to go faster than light to even maintain an orbit.

Mathematically speaking, the photon sphere occurs at 3/2 the Schwarzschild Radius. It's the only place where light rays can have (very) unstable orbits around the black hole.
http://www.gothosenterprises.com/black_holes/static_black_holes.html
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oops got my event horizons mixed up. apoapsis is the highest point in an orbit and periapsis is the lowest point in an orbit.

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Outside of the Event Horizon there is a Photon Sphere where light is able to maintain orbit.

Below the Photon Sphere You would need to go faster than light to even maintain an orbit. http://www.gothosenterprises.com/black_holes/static_black_holes.html

Is there another horizon outside the event horizon and photon sphere that corresponds to the escape velocity of massive particles that approach light speed? I think that may be the one you can (in theory) "climb" or accelerate out of as Insane Alien describes.

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hang on i've been looking at the site and i'm wondering about this photon sphere. how can it be orbit outside the event horizon at c? if its going at c and the escape velocity at that distance from the black hole is less than c why doesn't it just escape?

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Outside of the Event Horizon, light or even a spaceship can escape, but if "escaping" then it's no longer in orbit. (Edit: Escaping inside the Photon Sphere requires a non orbital path.)

Light rays in the Photon Sphere have just the right direction of travel to maintain their orbit.

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yes but how can light orbit the blackhole if the escape velocity at that distance is less than c? is there some weird relativistic effect here? if something is going faster than escape velocity then it WILL escape(or should do at least) the area where the orbital velocity is c should be within the event horizon if the event horizon is the distance from the singularity where the escape velocity is c.

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yes but how can light orbit the blackhole if the escape velocity at that distance is less than c? is there some weird relativistic effect here? if something is going faster than escape velocity then it WILL escape(or should do at least) the area where the orbital velocity is c should be within the event horizon if the event horizon is the distance from the singularity where the escape velocity is c.

The event horizon is the point (surface) from which nothing (quantum effects aside) can escape. It is not near the point (surface) at which the escape velocity of any mass would approach c. This point (surface) is outside both the event horizon and photon sphere.

When you first posted your argument made sense to me, but it was because I was confusing the two points (surfaces again), so I had to think about it.

Does this other surface have a name?

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surfaces? right i'm confused now. On the website linked aboe it said that the event horizon was where the escape velocity was c. it then said that the photon sphere was outsie this and since orbital velocity is slower than escape velocity this should be the other way around. I think this is what i'm meaning i don't know i've confused myself.

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surfaces? right i'm confused now. On the website linked aboe it said that the event horizon was where the escape velocity was c. it then said that the photon sphere was outsie this and since orbital velocity is slower than escape velocity this should be the other way around[/b']. I think this is what i'm meaning i don't know i've confused myself.

I think it is for mass...but not for light!

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To escape you could change your apoapsis till it is above the event horizon ...
Apoapsis is the highest point in an orbit and periapsis is the lowest point in an orbit.
Any flaws here?

Well, you say is that you escape the inside of the event horizont by escaping the inside of the event horizont. That´s flawless argumentation, at least. Whether a statement of "i can do something that´s not possible by doing it" is correct or wrong is for the philosophers to decide. For the physics part: Once you are behind the event horizont, there´s no non-spacelike path out - not even for a short time as part of a stationary orbit.

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but it does not require going faster than c as long as you are above the distance from the singularity where the orbital velocity is c.

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but it does not require going faster than c as long as you are above the distance from the singularity where the orbital velocity is c.

This is correct but also outside the event horizon. (50% further out according to the link as Spyman pointed out)

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sorry, but to orbit at c, you have to be 1.5 times as far away as the event horizon. anything closer requires ftl.

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sorry, but to orbit at c, you have to be 1.5 times as far away as the event horizon. anything closer requires ftl.

Yes, 50% further out.

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but it does not require going faster than c as long as you are above the distance from the singularity where the orbital velocity is c.

No, it doesn´t require to go faster than c - it just requires to get out of the event horizont. And the effect that you can escape a black hole when you are outside of its event horizont is a rather trivial statement (every spherical symmetric mass distribution has the same metric as a black hole outside the mass distribution, even earth).

I´m too drunk to give a definite statement about the "light has a stable orbit at 1.5 times the schwarzschild radius" statement presented above. But from my understanding of the schwarzschild metric it seems simply wrong (should be exactly the ss-radius). Maybe I´ll look into it tomorrow.

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No' date=' it doesn´t require to go faster than c - it just requires to get out of the event horizont. And the effect that you can escape a black hole when you are outside of its event horizont is a rather trivial statement (every spherical symmetric mass distribution has the same metric as a black hole outside the mass distribution, even earth).

I´m too drunk to give a definite statement about the "light has a stable orbit at 1.5 times the schwarzschild radius" statement presented above. But from my understanding of the schwarzschild metric it seems simply wrong (should be exactly the ss-radius). Maybe I´ll look into it tomorrow.[/quote']

We will wake you up early!

Black What?

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its very simple buddy you can never destroy any "information" so anything that goes into black hole comes out of it. it can not permanently be there in. it will take 100 billions years or more i dont know but it will come out of it. "YOU CAN NEVER DESTROY ANY INFORMATION" on the edge of event horizon you can not match the speed of light but after years of decay it is possible. if i believe other theory than it is possible that at the end there are only blackholes in the universe and than they pull each other and whole universe will again be point. and than again for a moment few nanoseconds all laws of physics vanish. And again a "BIGBANG" back to the square one.

I recently read in new scientist that "scientists" are comeign up with white hole therorys that say there are white holes where what entered a black hole exits, either in the same universe or in another universe.

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yes but how can light orbit the blackhole if the escape velocity at that distance is less than c? is there some weird relativistic effect here? if something is going faster than escape velocity then it WILL escape(or should do at least) the area where the orbital velocity is c should be within the event horizon if the event horizon is the distance from the singularity where the escape velocity is c.
Light rays in the Photon Sphere have just the right direction of travel[/b'] to maintain their orbit.
There is no weird effects here:

- if the light rays are moving straigh up then they will escape from the EH and out' date='

- if the light rays are moving straigh down then they will get caught from infinity and in, (eventually),

- if the light rays are moving in an orbital direction at correct distance then they will orbit.

Likewise with a spaceship, moving away, (with higher than escape velocity), from Earth or a BH, will allow it escape, (if above EH), moving against will allow it to land or make it crash, but moving with correct speed depending on radius with a direction around 90 degrees from against/away will let it orbit.

Escape velocity is calculated for the direction straight up.

If You where to stand on the Moon and point a flashlight towards Earth, will the light ray "escape" Earths gravity field or will the photons reach the surface ?

(Can You see the Moon and stars at night ?)

The event horizon is the point (surface) from which nothing (quantum effects aside) can escape. It is not near the point (surface) at which the escape velocity of any mass would approach c. This point (surface) is outside both the event horizon and photon sphere.

Does this other surface have a name?

From my understanding a spaceship can, (with very high speed), pass the Photon Sphere, almost touch the Event Horizon and still escape.

Any object with any mass should be able to orbit just above the Photon Sphere, with very high speed.

So, I am at a loss what You mean here ?

I´m too drunk to give a definite statement about the "light has a stable orbit at 1.5 times the schwarzschild radius" statement presented above. But from my understanding of the schwarzschild metric it seems simply wrong (should be exactly the ss-radius).
Direction is the keyword, (of the light rays).

(And from my quote: "(very) unstable orbits".)

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• 2 weeks later...

Are you thinking of the Roche Limit? I.e. give it no angular momentum on here and it will orbit indefinitly, do, and it will escape or fall in depending on which side you push it.

Inside the roche limit, anything that tries not to escape, will not.

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