Calculating GPS Locations: The Impact of Special Relativity

In summary, the conversation revolves around the topic of special relativity and its effects on GPS calculations. The speaker is writing a report on the opposite perspective of the commonly accepted belief and is uncertain about which sources to believe. They ask for clarification on the equations and effects involved, and also mention a questionable website they came across. The conversation ends with a helpful member providing a simplified explanation and calculations for the effects of special relativity on GPS.
  • #36
Kinda afraid to post after the verbal spanking Randall gave me, but got another question:

contraction = sqrt( 1 - v^2/c^2 ) = sqrt ( 1 - 1.6e-10 ) = 0.99999999991699995
times 24hours = 7 microsecs

I'm sorry to seem stupid, but I've tried two hours now to get 0.99999999991699995
times 24hours = 7 microsecs.. I just don't see how?
 
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  • #37
p4h said:
I'm sorry to seem stupid, but I've tried two hours now to get 0.99999999991699995
times 24hours = 7 microsecs.. I just don't see how?
Remember it's the fraction of a day so,
1 - 0.99999999991699995 = 8.3 E-11 of a day

And 24 * 60 * 60 * 8.3E-11 = 7.2 E-6 seconds
 
  • #38
Ah! Thanks again..

If I do:

[tex]\sqrt{1-\frac{3900}{299792458}} = .9999934955[/tex]

So we don't get the same. >.<

Where do you get the 1 - .9999934955 from?
 
  • #39
The result of the sqrt is the contraction - this is the fraction of a day that the satelite expereinces, so a satelite's view of a day is 0.9999934955 of an Earth's view of a day.
The difference between these is simply 1 - 0.9999934955
or if you prefer, satelite day = 0.9999934955 * 24 * 3600 seconds, Earth day = 1 * 24 * 3600 seconds.
The difference is 7.2 us.

edit sorry just realized you are asking about the actual numerical value
You forgot to square the v and c, it should be sqrt( 1 - (3900^2)/(3E8^2))
 
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  • #40
mgb_phys said:
The result of the sqrt is the contraction - this is the fraction of a day that the satelite expereinces, so a satelite's view of a day is 0.9999934955 of an Earth's view of a day.
The difference between these is simply 1 - 0.9999934955
or if you prefer, satelite day = 0.9999934955 * 24 * 3600 seconds, Earth day = 1 * 24 * 3600 seconds.
The difference is 7.2 us.

edit sorry just realized you are asking about the actual numerical value
You forgot to square the v and c, it should be sqrt( 1 - (3900^2)/(3E8^2))

Yeah thanks it was the top part I wanted..

And ye forgot to square them, but that only got my answer closer to 1

But thanks yet again!
 
  • #41
p4h said:
Kinda afraid to post after the verbal spanking Randall gave me, but got another question:
No reason to fear posting – often newbes accidentally work on homework in the discussion forums instead of the homework forum. If someone doesn’t tell you, you won’t learn how to best use PF.

By the way if a Mentor has not already (I am not one) welcome to the PF forums.

PS: Also, no need to abuse your teacher behind their back (post #28), at least be sure what they want you to learn from doing a report. Remember your writing a report not a textbook.
 
  • #42
Well I just got alittle frustrated because I couldn't see my way out of the project, since I also have this graph I can't figure out.
 
  • #43
Okay I got two more things left before I stop reviving this thread, as I figure I might as well use this one, instead of starting a new one.

1. When thinking of what impact the theory of relativity (both SR and GR) have on things around us, I couldn't think of anything except Redshift / Doppler effect. I'm sure there are a lot more areas but I can't think of any?

2. 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.

I'd appreciate it if only hints is dropped, so I can fiddle with it myself, but I may ask again if I'm lost :)

PS: I realize this should be in the homework section, but as I already have a thread with this subject, I might as well use it.
 

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  • #44
Cute problem.
I assume you can translate the fairly intuitive answer of about “0.065 tid/s” into the appropriate units of distance.
The correct value answer of course is not the important part.
I’m sure they expect you to report on what detailed information has to be known at the receiver to use in observing this one satellite.
And why using SR & GR in so important to maintaining that information.

A good problem in that it is easy to answer, once you understand the issues of running a GPS.
Hint: think simple

Mentors can move the thread if they feel it is needed, interesting approch at least.
 
  • #45
I assume I have to make a connection between length and signal strength, but to me, when I think of signal strength, I just think that it has to vary a lot due to different factors, being it the ionozation in the Ionosphere, or being it the humidity in the Troposphere? I may be overthinking it.
 
  • #46
Signal strength has nothing to do with it.
You are making it harder than it is.
Radio photons are no different than light photons.
Think simple & how do you measure distance with them.
 

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