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Escaping a Black Hole's Event Horizon

  1. Feb 21, 2012 #1
    When classically deriving the Schwarzschild radius of a black hole, the kinetic energy of an outgoing particle (moving at the impossible-to-achieve maximum of the speed of light) is equated with the gravitational potential of the black hole at that point.

    [itex]\frac{1}{2}mc^2 = \frac{GMm}{r}[/itex]

    This seems reasonable for justification that light cannot escape, but what about objects that can do work or exert forces?

    The gravitational force of the black hole at the event horizon is:

    [itex] F = \frac{GMm}{r^2} [/itex]

    One could imagine a spaceship throwing off fuel in the opposite direction so that

    [itex] \frac{Δp}{Δt} = \frac{GMm}{r^2} [/itex].

    I'm well aware that this is not the case, so I'm curious as to what the reasoning is. Thanks for the help!
  2. jcsd
  3. Feb 22, 2012 #2
    the formula above does give the correct value of r but

    I would just like to point out that ½mc² is the nonrelativistic formula for kinetic energy (of a particle moving at c)

    In reality, for velocities close to c, one would have to use the relativistic forumula to find the actual kinetic energy.

  4. Feb 22, 2012 #3
    My issue is that I don't see what prevents a spaceship throwing off fuel at a constant rate A kg/s, so that:

    [itex]\frac{dp}{dt} = \frac{GMm}{r^2}[/itex]

    [itex]A v = \frac{GM(m - At)}{r^2}[/itex].

    What prevents this system (classically, so long as v is not too large) from escaping the event horizon?

    My initial thought would be that v would have to be greater than the speed of light, but I can't quite justify that mathematically.

    Any thoughts would be incredibly helpful.
  5. Feb 22, 2012 #4
    time stops at the event horizon

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
    Last edited by a moderator: May 5, 2017
  6. Feb 22, 2012 #5


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    In a Newtonian sense, you're right. But black holes are not Newtonian objects, and although you get the right answer for the Schwarzschild radius, it's completely by incorrect reasoning. It's not actually an issue with special relativity -- but with the much larger theory of General Relativity.

    If you imagine yourself going towards the event horizon of a real Einsteinian black hole, you would need to accelerate infinitely fast as you fell down towards the event horizon just to stay at the same radius! Curiously, once you've fallen past the event horizon, any amount of blasting your rockets you do will only hasten your eventual demise! If you want to live as long as possible, the best bet is to sit there and do nothing.
  7. Feb 23, 2012 #6
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