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GPS clocks

  1. Oct 23, 2007 #1
    Suppose i have two clocks on earth showing 12:00 hours.
    After, one of the clocks is put in a gps satellite for a year.
    After one year that clock returns to earth.
    Due to relativistic effect what time that clock shows? Less than 12:00?
    If so why?
     
  2. jcsd
  3. Oct 23, 2007 #2
    A GPS satellite IS a clock, isn't it? A fair few communications systems rely on them for sync. So is the question

    If a terrestrial system was taking its clocking from a GPS satellite and you could bring the clock source to Earth would there be a relativistic deviation in sync between master and slave?

    The problem with just two disparate clocks is that they're independent of one another, one is not a sync source for the other so I'm not sure anything gainful could be deduced from the experiment.
     
  4. Oct 23, 2007 #3

    russ_watters

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    I'm not really sure what you mean, Colin1 - GPS clocks are synchronized to ground stations and are used to provide accurate time signals for various purposes including but not limited to navigation. The gps signal itself is a coded time signal.

    GPS clocks are preset before launch to run at a rate that will enable them to stay roughly synchronized with ground stations. Anyway, that correction is....

    http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

    So, if GPS satellietes didn't have their rates altered, they would gain 38 microseconds a day. An uncorrected clock in a satellite brought back to earth a year later would show (at noon) 12:00:13.87 seconds. It would be almost 14 seconds fast!
     
  5. Oct 23, 2007 #4
    So in reality when my physics teacher uses the example of astronaughts going to space and comming back aging slower than us on earth really isn't true because of general relativity and they would actually age faster... or do they reach speeds fast enough to compensate for this?
     
  6. Oct 23, 2007 #5
    No - the physics teacher is right - it is just in the case of GPS that the altitude effect is greater than the velocity effect - if the clock is not in orbit but traveling at half the speed of light for a long period, the travling clock would show less time
     
  7. Oct 23, 2007 #6

    russ_watters

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    GPS Satellites orbit unusually high - 20,000 km - and slow - 14,000 km/h - compared with 240 km and 27,000 km/h for most normal satellites and real astronauts. I suspect, but don't really know, that the SR correction would be the bigger factor for astronauts. They'd age slower.

    Yogi, I'm not sure if Sorry!'s teacher was talking about real astronauts or the hypothetical ones from the twins paradox...
     
  8. Oct 24, 2007 #7
    Ok! Now if the clock that stay on earth, after a year, is put on the gps satellite what time did a eventual gps being reads when it arrives there?
     
  9. Oct 24, 2007 #8

    russ_watters

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    That's a jumbled mess of an attempted sentence. Are you asking how a ground-based clock would compare to an uncorrected GPS clock when sent into orbit after a year on the ground? It would read 14 seconds slower than the GPS clock - same answer as the previous question (for our purposes it's really the same question).
     
  10. Oct 24, 2007 #9

    Jorrie

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    GPS clock correction

    You're right. I calculated that a 240 km altitude satellite clock loses about 10 ms/year relative to a ground clock.

    BTW, I think you made a typo in a previous post - an uncorrected GPS satellite clock would gain about 14 ms/year relative to a ground clock, not 14 seconds.
     
  11. Oct 24, 2007 #10
    When a clock is in orbit, the slow-down due to velocity is always going to be less than the greater rate required to adjust for the height assuming the reference is taken as the non-moving earth centered frame. One could propose an object circling the earth at o.5 c which results in a greater loss of time due to velocity than the increase required for the gravitional potential adjusment - but such an object is not in orbit - the frame of the object is no longer a free fall inertial frame - we are back to a traveling twin type of analysis or some such thing
     
  12. Oct 24, 2007 #11

    Jorrie

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    Agreed, but do you agree that for a LEO satellite, the slowdown due to velocity dominates when referenced to a terrestrial clock? I find that to be the case below about 3200 km altitude, where the velocity and gravitational effects cancel out.
     
  13. Oct 24, 2007 #12
    yeah it's real astronaughts :p thanks for the answers :)
     
  14. Oct 24, 2007 #13

    russ_watters

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    I wouldn't say "always" - there is an orbital velocity that corresponds to sea level, afterall. So the question is, is the altitude below which the SR correction is bigger than the GR correction high enough to be above the atmosphere? Jorrie seems to have calculated that that altitude is somewhere above LEO.
     
  15. Oct 24, 2007 #14

    russ_watters

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    You're right - I confused my prefixes (micro with mili). I thought 14 seconds seemed awfully large.
     
  16. Oct 24, 2007 #15
    Correct Russ - if you reference all clocks to an equipotential defined by sea level, then there is no correction for height for an object traveling in an evacuated tunnel at orbit velocity - I had in mind a different reference point.
     
  17. Oct 25, 2007 #16

    Jorrie

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    Ground-orbit time equality

    Here's my calculation. I think the simplest way to compare clocks is to express both relative to the time (t) of a distant static observer in asymptotically flat spacetime, considering Earth as an isolated body at rest at the origin (i.e. Schwarzschild coordinates).

    On Earth's surface, radius R and taking the rotation speed as insignificant, the clock rate relative to t is:

    [tex]d\tau_0/dt \approx \sqrt{1-2GM/(R c^2)} [/tex]

    For a satellite orbiting at constant radius r, the clock rate relative to t is:

    [tex]d\tau_1/dt = \sqrt{1-2GM/(r c^2)-v_o^2/c^2} = \sqrt{1-3GM/(r c^2)}, [/tex]

    since circular orbital velocity [itex]v_o^2 = GM/r[/itex].

    It is now easy to spot that [itex]d\tau_0/dt = d\tau_1/dt [/itex] when r = 1.5R, giving an altitude of ~3190 km above MSL. Below this altitude uncorrected satellite clocks lose time and above it they gain time relative to a terrestrial clock.
     
    Last edited: Oct 25, 2007
  18. Oct 27, 2007 #17

    Jorrie

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    If the tangent velocity for an equatorial clock ([itex]v \approx 463[/itex] m/s due to Earth's rotation) is taken into account, the altitude for ground-orbit clock rate equality is about 3170 km, from:

    [tex]1 - \frac{2GM}{Rc^2} - \frac{v^2}{c^2} =1 - \frac{3GM}{rc^2}[/tex]

    giving

    [tex]r=\frac{3GMR}{2GM+Rv^2}[/tex]
     
    Last edited: Oct 27, 2007
  19. Jan 15, 2012 #18
    Has anyone ever brought a clock back from orbit and compared it with a clock on the ground?
     
  20. Jan 15, 2012 #19

    russ_watters

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    I don't think so -- but why would that matter?
     
  21. Jan 15, 2012 #20
    if when they are both on the earth surface they read p.e. 12 oclock, and then one is put in orbit and after a certain period it came to earth surface did they show the same number (time)?
     
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