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GR and time dilation

  1. Dec 4, 2007 #1
    I am sorry for posting this problem again. I posted it in introductory physics and someone mensioned it might not be an introductory physics problem. Any way I still don't have an answer to it so thought of asking you all, thanks.

    1. The problem statement, all variables and given/known data
    A satellite is in circular orbit of radius r about the Earth (Radius R, mass M). Astandard clock C on the satellite is compared with an identical clock C0 at the south pole on Earth. Show that the ratio of the rate of the orbiting clock to that of the clock on earth is approximately:

    1+(GM/Rc^2)-(3GM/2rc^2).

    Note that the orbiting clock is faster only if r > 3/2 R, ir if r-R>3184 km.


    2. Relevant equations

    --------------

    The formula to find the rate is : 1+delta (potential)/c^2
    so I have to find the diffrence between the potential


    3. The attempt at a solution

    potential on earth = -GM/R
    potential of object in orbit = -GM/r

    diffrence = GM/R-GM/r

    answer should be : GM/R-3GM/2r --- so my answer is wrong
     
  2. jcsd
  3. Dec 4, 2007 #2

    dynamicsolo

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    I think I smell the rat. Don't forget that the satellite is also in motion relative to the Earth's surface. What is the orbital velocity at radius r? What is the time dilation effect for that velocity (as a linear approximation)? [I thought it seemed odd that the orbiting clock was faster only above a certain altitude... Note that the standard clock is at the South Pole: what is the linear speed of the clock there?]
     
    Last edited: Dec 4, 2007
  4. Dec 4, 2007 #3
    orbital v = (GM/r)^1/2

    so do we add 1/2 m v^2 + GmM/r = 3/2 G mM/r ~ 3/2 GM/r ? and then
    delta (phi) = 3/2 G M/r - GM/R? If that's correct then I missed up with the signs.
     
  5. Dec 4, 2007 #4

    dynamicsolo

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    You don't add the terms as such. Your calculation for the ratio of "time rates" at radius r is correct (to first order). This would give you the rate for a clock stationary with respect to the Earth's center (or the clock at the South Pole).

    You would now multiply this rate by the Lorentz factor, gamma, for a clock moving at the orbital velocity with respect to Earth's center. What will that factor look like? If you use the binomial approximation, what is this factor approximately? You would then multiply this binomial with the binomial approximation you have for the "gravitational" effect (and only keep the "low order" terms).
     
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