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GPS and Special relativity (Graph)

  1. Dec 17, 2007 #1

    p4h

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    [SOLVED] GPS and Special relativity (Graph)

    1. The problem statement, all variables and given/known data
    The image attached is a graph I'm supposed to be able to figure out. I'm supposed to be able to figure out the distance between a Satelite and a GPS reciever.
    X axis: Time / seconds
    Y axis: Signal strength / no unit
    There is no legend to the graph, so I don't know which curve represents the satellite and which represents the reciever.

    2. Relevant equations

    Don't Know

    3. The attempt at a solution

    I simply don't know what to do? I've messed around with this graph for the last 2 days and it's the last thing I need done in my project on Special relativity and GPS.

    PS: Sorry for the double post, but need this done by today and haven't gotten an answer in the thread in the SR / GR forums yet for some time.
     

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    Last edited: Dec 17, 2007
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  3. Dec 17, 2007 #2

    Kurdt

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    The curves are exactly the same, except slightly shifted on the time axis. I think you can use a simple [itex] d =st[/itex] on this.
     
  4. Dec 17, 2007 #3

    p4h

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    Could you explain that equation a bit more? I'm foreign so we use some other letters for some of our equations.
     
  5. Dec 17, 2007 #4

    Kurdt

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    Sure. Its just distance = speed x time.
     
  6. Dec 17, 2007 #5

    p4h

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    Where do you get speed from? If it's the signal strength, then how do you convert that directly to speed?
     
  7. Dec 17, 2007 #6

    Kurdt

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    The signal will be microwave or radio wave or some sort of EM wave so one does not need to take the strength into account, as the speed of the signal is just the speed of light.
     
  8. Dec 17, 2007 #7

    p4h

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    But as the microwaves pass through the Ionosphere and the Tropossphere, they are delayed by the conditions in these, hence they can't move at the sped of light? Or am I overthinking it?
     
  9. Dec 17, 2007 #8

    Kurdt

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    Thats a good observation, but the refractive index of the atmosphere is so very nearly 1 that its a good approximation. I think at your level that might be a bit of over thinking unless your tutor has specifically asked you to take it into account.
     
  10. Dec 17, 2007 #9

    p4h

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    Well he didn't, but as I am writing a report on special relativity and GPS, so I guess the ideal thing for me would be to comment on my data saying, that since the refractive index is very near 1, I have ignored it completely.

    And another quick question. The shift, does it simply tell me the time between the reciever and the satellite?
     
  11. Dec 17, 2007 #10

    Kurdt

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    Yes a brief note that you understand the real value is buried in more complicated mathematics would be prudent.

    Yes the shift just tells you the time between the signal being sent by the satellite and received by the receiver (or the other way round).
     
  12. Dec 17, 2007 #11

    p4h

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    As my assignment is to just measure the distance between the two, which curve being which wouldn't really matter, no?
     
  13. Dec 17, 2007 #12

    Kurdt

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    No it wouldn't really matter. All that matters is that you have a measure of the time taken.
     
  14. Dec 17, 2007 #13

    p4h

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    And also, one last question.

    Since I have the equation d = s * t, then it doesn't really matter at all how the graph looks (where the peaks are etc.) does it? And thereby that signal strength does not have an impact on the actual distance?
     
  15. Dec 17, 2007 #14

    Kurdt

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    Thats right. The signal strength doesn't really have anything to do with the distance. The peaks are useful to us as we can compare the two graphs at the same point to get the time, but that is all.
     
  16. Dec 17, 2007 #15

    p4h

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    Okay thanks for all the help :) Will mark it as solved later, as I might have other questions regarding it
     
  17. Dec 17, 2007 #16

    p4h

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    Okay another question arose in my mind. I have loosely measured the shift of the graph, and it's roughly 0,7 seconds. Why isn't it the 0,7 seconds I have to use as my time in the d = st, as it's the time it takes for the signal to get from the satellite to the reciever?
     
  18. Dec 17, 2007 #17

    Kurdt

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    I think that should be 0.07s. I'm not sure what you're trying to ask could you clarify a bit?

    If you use 0.07s in the equation for distance you'll find that it comes out at ~21000km which is pretty much the orbital altitude of GPS satellites so it seems to be a good value.
     
  19. Dec 17, 2007 #18

    p4h

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    Yeah I meant 0,07.

    It was just, that I was under the impression from earlier, that I was to simply look at the time on the x-axis and look at what results that would give me, but that made no sense when I tried it, as time would flow by, I'd have an increase in distance that would be linear.

    Oh well.. I'm grateful that you wanted to help me so if you ever need help figuring out what 1 + 1 is, I'll gladly help!
     
  20. Dec 17, 2007 #19

    p4h

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    Another question, how do you mark this as solved? I tried editing the title but it doesn't change on the board?
     
  21. Dec 17, 2007 #20

    Kurdt

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    Under thread tools button at the beginning of the thread. I've done it for you.
     
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