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Orbital Velocity Question

  1. Jul 14, 2010 #1
    I am fairly new to astronomy, having just finished a course this past semester in high school, and I have a question about orbital velocity.

    Now, I know that the event horizon is the boundary at which the escape velocity is equal to the speed of light, but is there any significance to the boundary at which the orbital velocity is equal to the speed of light?

    I quickly messed around with the math to see where such a boundary would lie and determined this:
    Vorb = √((Gm)/r)

    substitute Vorb for C

    C = √((Gm)/r)
    r = (Gm)/(C2)

    This is obviously very similar to determining the Schwarzchild radius and this boudary would actually be inside the event horizon, light orbiting the singularity, but never moving out past the radius once in orbit.

    I looked online to see if such a boundary was important or existed and I think I found something similar the photon sphere, but the equation is different and this sphere is outside the event horizon.

    Does the boundary where r = (Gm)/(C2) mean anything?
     
  2. jcsd
  3. Jul 15, 2010 #2

    Ich

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    No. It's a coincidence that a Newtonian calculation gives the correct result for the Schwarzschild radius (he's not a child, btw). You can't expect that it works for other qunatities.
    If you do the calculations in GR, you get the photon sphere result. For illustration, think that the centrifugal force reverses direction near the horizon, where relativity becomes important. Thus, it's even harder to orbit than to escape.
     
  4. Jul 15, 2010 #3
    Ah ok that makes sense. Thanks for clarifying that up :)
     
  5. Jul 16, 2010 #4
    Just thinking, I do not know too much about newtonian mechanics or GR, but does this mean that NM breaks down under extreme conditions?
     
  6. Jul 17, 2010 #5

    Ich

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    Exactly. It works if you have small velocities and weak gravity. It's like a first order approximation to the exact GR laws.
    You can make it work more exactly or for more extreme circumstances if you include some corrections, like a second order approximation to GR. That's called the Post-Newtonian formalism.
    Under the most extreme conditions, you use full GR.
     
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