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Special Relativity: Time Dilation

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    An Earth satellite used in the Global Positioning System moves in a circular orbit with period 11
    hours and 58 min. The satellitle contains an oscillator producing the principal nonmilitary GPS signal. Its frequency is 1575.42 MHz in the reference frame of the satellite. When it is received on the Earth's surface, what is the fractional change in this frequency due to time dilation, as described by special relativity?

    2. Relevant equations

    Fractional change = 1/2 (v/c)^2

    3. The attempt at a solution

    Answer: 1/2 ( 3880m/s / c )^2 = 8.36*10^-11


    Not sure if I'm even on the right track at all... some guidance would be much appreciated.
     
  2. jcsd
  3. Jan 31, 2012 #2

    Simon Bridge

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    Well, you can check your logic: the fractional frequency change is [tex]\frac{\Delta\nu}{\nu}[/tex] ... so do you know how that relates to the formula you used? To time dilation?
     
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