Special Relativity: Time Dilation

  • Thread starter JennV
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  • #1
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Homework Statement



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?

Homework Equations



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

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.
 

Answers and Replies

  • #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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