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Quantum Theory of Earth Satellite Analogy to Bohr Model

  1. Sep 9, 2015 #1
    Problem:
    In analogy to the Bohr Theory of the hydrogen atom, develop a quantum theory of Earth satellites, obtaining expressions for the orbit radius (r) and the energy (E) in terms of the quantum number (n) and the other relevant parameters. A satellite of mass 1000 kg is in a circular orbit of radius 7000 km, to what value of n does this correspond?

    Equations and Constants:
    Bohr Model: E = -R*h/n^2
    E = 1/2 * m * r^2 * ω*2
    ω = v/r
    v^2 = G*M/r
    R = 1.0973 x 10^7 m^-1
    h = 6.626x10^34 kg*m^2/s
    M = 5.972x10^24 kg
    G = 6.674x10^-11 m^3/(kg*s^2)

    Attempt:
    E = -R*h/n^2
    1/2 * m * r^2 * ω*2 = -Rh/n^2
    r^2 = -2*R*h/(n^2 * ω^2 * m)
    r^2 = -2*R*h*r^3/(n^2 * G* M * m)
    1/r = -2*R*h/(n^2 * G* M * m)
    r = (n^2 * G* M * m) / (-2*R*h)

    Just by parsing the units I know I've taken a wrong turn. I've tried multiple times and I appear to be missing a velocity term somewhere with the (-2*R*h). I think I'm missing something simple, I just don't see it.

    Thanks.
     
  2. jcsd
  3. Sep 9, 2015 #2

    mfb

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    If R is the Rydberg constant: the potential of Earth is not the same as the one of an electron in an atom.

    Edit: I moved the thread to the homework section.
     
    Last edited: Sep 9, 2015
  4. Sep 9, 2015 #3

    fzero

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    It's probably simpler in this case to use the Sommerfeld-Wilson quantization condition on the angular momentum of the satellite. You will find that since ##h## is so small compared to the other scales of the problem that the quantum number is huge and different quantized orbits are so close together as to be beyond distinguishing within the precision of the measurements that we could make.
     
  5. Sep 9, 2015 #4
    I'm not sure what to use for R then because my definition of R includes the mass of an electron and the charge. I can see replacing the mass of an electron with the mass of my satellite, but then what do I do with the charge? I tried to analyse the units to come up with a value but I got nothing that made any sense.

    Oh, and thanks for moving it! :)
     
  6. Sep 9, 2015 #5
    Oh, that does look more promising. Let me give it a stab.
     
  7. Sep 10, 2015 #6

    mfb

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    You cannot use the Rydberg constant - it is specific to the electromagnetism problem in an atom. You can see how it is derived based on the potential energy, however, and use this with the gravitational potential instead.
     
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