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GPS and Relativity

  1. Aug 31, 2014 #1
    Due to the blogs being removed, I thought it might be worthwhile posting a few in the forums-

    https://www.youtube.com/watch?v=zQdIjwoi-u4


    Time dilation due to gravity (GR)-

    [tex]d\tau=dt\sqrt{1-\frac{2M}{r}}[/tex]

    where [itex]M=Gm/c^2[/itex]

    Time dilation due to velocity (SR)-

    [tex]d\tau=dt\sqrt{1-\frac{v^2}{c^2}}[/tex]

    Quantities-

    Earth's mass- 5.9736e+24 kg

    Earth's (mean) radius- 6.371e+6 m

    Satellite's altitude- 2e+7 m

    Satellite's velocity- 3.889e+3 m/s

    86400 seconds in a day


    Difference due to gravity-

    dT sat (r=6.371e+6 + 2e+7)- 0.999999999833
    dT Earth (r=6.371e+6)- 0.999999999304

    difference- 0.000000000529

    Time dilation on Earth relative to satellite is 4.562265e-05 seconds per day or 45.62265 microseconds


    Difference due to velocity-

    dT sat (v=3.889e+3)- 0.999999999916
    dT Earth- 1 (0.999999999999 at the equator, i.e. ~1)

    difference- 0.000000000084

    Time dilation on satellite relative to Earth is 7.26931e-06 seconds per day or 7.26931 microseconds


    Total time dilation- 38.35334 microseconds per day on Earth, atomic clocks on satellites need to be slowed down to match atomic clocks on Earth.



    The ISS-

    altitude- 2.78e+5 to 4.6e+5 m

    velocity- 7701.11 m/s

    The time dilation on the ISS relative to Earth is between 24.35197 and 25.8877 microseconds a day (depending on the altitude) meaning the atomic clocks on the ISS have to be speeded up to match those on Earth in contrast to the GPS satellites (which means that since its inception in 1998, the ISS has travelled forward in time by approx. 0.1 of a second).
     
  2. jcsd
  3. Aug 31, 2014 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    A few comments:

    (1) You didn't mention it, so I don't know if you are taking into account that an object on the Earth's equator has a nonzero velocity (at least in the frame you appear to be using for your calculations, in which the Earth is assumed to be spherical and non-rotating). Since ##v \approx 450## in SI units, the correction to dT(Earth) is only in the twelfth decimal place, but I think it's still worth mentioning.

    (2) For objects that are in free-fall orbits, you can use the formula for orbital velocity in terms of Earth's mass and the orbital radius to get a formula that takes both velocity and altitude into account:

    $$
    d\tau = \sqrt{1 - \frac{3 G m}{c^2 r}}
    $$

    (3) It's worth noting that the correction due to the non-sphericity of the Earth is just a smidgen too small to appear in your calculation; it appears in the thirteenth decimal place (this is the correction due to the Earth's quadrupole moment). This illustrates why approximating the Earth as spherical works so well; even in a calculation this precise, non-sphericity can be ignored for many purposes (since the correction is well under 1 percent of the overall value).
     
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