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Satellite Paradox

  1. Sep 10, 2009 #1
    1.
    {This paradox denotes the fact that a satellite in a near circular orbit suffers an increase in velocity when subject to a drag force.}

    The Specific energy of the satellite is K + U = E where K =v^2/2 is specific kinetic energy. U= -u/r is specific energy and E is specific total energy. If satellite in circular orbit then u=(v^2)*r.

    If orbital radius changes by a small amount dr, what is the resulting change dE of the total energy?

    3. The attempt at a solution

    What I figured to do is take derivative with respect to r but if I do that it will cancel out the distance completely when I plug in the u into the E equation. I i'm not sure if that's the approach I should take. Or should I take derivative with respect to time, because as time changes distance also changes. Not to sure...
     
  2. jcsd
  3. Sep 10, 2009 #2

    kuruman

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    Can you show what you mean by this?
     
  4. Sep 10, 2009 #3
    So if I take the derivative of the specific total energy E with respect to the distance value r something like this:

    dE/dr = -[tex]\mu[/tex]/2r --> i plug in the value of [tex]\mu[/tex] = (v^2)*r

    it turns into -(v^2)*r/2r the r will cancel in this case...(or maybe i should leave one alone and continue to eval the derivative....

    The other case would be to take dE/dt to see the change in dr.
     
  5. Sep 10, 2009 #4

    D H

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    This is paradoxically true -- assuming the drag force is small, of course.

    What is the total energy for a circular orbit as a function of orbital radius?
     
  6. Sep 10, 2009 #5
    That is what I show up there its E=-[tex]\mu[/tex]/2r where [tex]\mu[/tex] = (v^2)r
     
  7. Sep 10, 2009 #6

    kuruman

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    If -μ/r has dimensions of energy, then how can dE/dr = -μ/2r? The dimensions don't match.
     
  8. Sep 10, 2009 #7
    Ok then that would mean dE/dr wouldn't work so, it has to be dE/dt in order to get my answer; so in order to take the derivative with respect to time I would have to plug in v=d/t into the equation right...
     
  9. Sep 10, 2009 #8

    D H

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    dE/dr is the correct approach.
     
  10. Sep 10, 2009 #9
    Okay thanks I got it
     
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