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Escaping from a Black Hole?

  1. Jul 1, 2012 #1
    Let's use the classical example of a black hole: A massive planet that has shrunk tremendously, until the escape velocity at its surface is greater than the speed of light.

    If you were sitting on the surface of said planet, would it still be possible for you to escape?


    Let's say you had a jetpack. The thrust of the jetpack just needs to overcome your weight.

    As you are propelled upwards, the gravitational field strength decreases.

    If you can sustain jet propulsion, could you eventually escape this "black hole"?
     
  2. jcsd
  3. Jul 1, 2012 #2

    phinds

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    I think you misunderstand the concept of black hole. There is a thing called the "event horizon" which would be farther out than any of the mass that created the black hole and anywhere inside it you cannot escape. In any event, you would have to be a REALLY long way away from a black hole for anything like a jet pack to put you out of the gravity well
     
  4. Jul 1, 2012 #3

    jcsd

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    I think it was John Tyndall who first proposed the idea of star whose surface escape velocity was greater than the speed of light, he called it a 'dark star'.

    The (about to be be) qualified answer to your question is yes, it would be possible to escape a dark star and you needn't ever go faster than the speed of light to do so either. However black holes and dark stars are not synoymous, they live in two very different regimes. The former in Newton's theory of gravity, the latter in Einstein's general theory of relativity. Whilst there's a few superficial simlairities between a dark star and a black hole, the two concepts are actually two very different beasts

    For general relativity it is not possible to escape a black hole in general relativity, in fact black holes are usually defined by the impossibilty of escape.
     
  5. Jul 1, 2012 #4

    jcsd

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    let's be clear:

    A dark star is an object in Newtonian physics whose surface escape velocity is greater than the speed of light.

    A black hole is an 'object' in general relativity whose escape velocity is gretaer than c within its event horizon (actually this is a bit of a gross simplification).

    A dark star is not synonymous with a black hole.
     
  6. Jul 1, 2012 #5
    And to the OP: At first glance these appear to be the same thing but the different mathematics between the two theories of gravity lead to very different physical entities.
     
  7. Jul 1, 2012 #6
    Is time dilation also infinite at the surface of a dark star?
     
  8. Jul 1, 2012 #7

    jcsd

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    No because you don't have time dilation in Newtonian physics. Assuming general relativity is correct there are no such things as dark stars, instead you have black holes.
     
  9. Jul 1, 2012 #8
    Got it now.

    If a black hole is a singularity, why should it have a specific event horizon, and how do you calculate this event horizon based on the black holes mass?
     
  10. Jul 2, 2012 #9
  11. Jul 2, 2012 #10
    But why should a black hole have such an event horizon in the first place? What is the derivation for the Schwarzschild radius?
     
  12. Jul 2, 2012 #11

    Nugatory

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    It comes from the GR solution to the Einstein field equations for a spherically symmetric and constant in time gravitational field. Google will tell you more.
     
  13. Jul 2, 2012 #12
    John Michell actually: http://en.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics)
    As above linked article explains, there is only a slim and indirect way of anything escaping a notional dark star. Not that it matters.
     
  14. Jul 2, 2012 #13

    jcsd

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    I should'v e remembered, I was reading about it a few days ago.

    No it doesn't say that, it's talking about the em radiaition that might be seen from a dark star. What I said is perfectly correct.
     
  15. Jul 2, 2012 #14

    PAllen

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    One additional point:

    Supposing you have a supermassive collapsing ball of matter (e.g. a billion solar masses), with a well defined surface (ignore how it came to exist). Suppose it has a well defined surface and you are standing on it. When it has collapsed enough so that it is inside its event horizon, the density and tidal forces for such a super-massive ball are still quite modest. You could absolutely take off from the surface with a jet pack. However, the massive ball is collapsing steadily, and there is no possibility of stopping its collapse. Further, no matter how powerful your jet pack (even though you can pull away from the surface) you will be drawn in with the collapse and (pretty quickly) be dragged into the forming singularity. If you could magically see the event horizon, you would see it moving away from from the surface as fast as outgoing light, and since you cannot overtake light, you would be ever further inside the event horizon, no matter how you fired your jets.
     
  16. Jul 2, 2012 #15
    OK then. Mind explaining how? It had occurred to me after reading that linked Wiki article that rather than a random boost from some stray exterior particle, a conceivable method of escape might be via a form of 'multi-stage rocket' propulsion. Not any actual rocket, but it could be imagined arranging for a staged multi-fission process. One large particle violently fissions into two at it's apogee, and this then repeats for the remainder part(s) that fly outwards. Hugely unlikely of any practical realization. If you had in mind single or multi-stage rockets, I'd like to know what conceivable fuel would do the trick!
     
  17. Jul 2, 2012 #16

    PAllen

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    In Newtonian physics light is not a limiting velocity and can't be: Galilean relativity only allows infinite speed as a limiting velocity, else the velocity addition rule must change. Thus, there is a finite, large, acceleration which will allow you to overtake and pass light on the way out of a dark star's pull.
     
  18. Jul 2, 2012 #17
    Admittedly I hadn't allowed for an infinite speed in Newtonian picture. So ok large and finite acceleration could do it there, but there is still the big question mark as to how that could be sustained in a self-contained system for long enough. Someone's bound to suggest 'matter/anti-matter annihilation' as propellant, but I'd like to see a worked example!
     
  19. Jul 2, 2012 #18

    PAllen

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    Simple: The surface gravity of a dark star is inversely proportional to M. Make M large enough, and 1 g thrust, continued long enough, will get you away (and take you beyond the speed of light).
     
  20. Jul 2, 2012 #19
    Maybe not so simple. Agreed G can be made arbitrarily small, but the problem again is the total energy needed to achieve escape. We are asssuming a fully self-contained system - rocket = payload + propellant. At a stab I'd say minimum theoretical configuration would be that hypothetical 'matter/anti-matter annihilator engine'. Just not sure off-hand if a single-stage design could do the trick. Any rocket scientists on-board? :biggrin:
     
  21. Jul 2, 2012 #20

    PAllen

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    Everything is still governed by Newonian gravity. The potential energy difference from any r>0, for any M, to infinity, is finite. Now if you want to propose the Newtonian physics is bound by 'energy has mass' and the E=mc^2 equivalence (which all, obviously, comes from relativity), then it still happens to be no problem (if I haven't screwed up a factor): the energy needed to take a mass m from the dark star surface (at critical radius) to infinity is mc^2/2 - so you can do it. You just need to convert half the mass to energy.
     
    Last edited: Jul 2, 2012
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