Re: SRT GPS formula
- From: John C. Polasek <jpolasek@xxxxxxxxxx>
- Date: Sat, 04 Aug 2007 15:33:15 -0400
On Sat, 4 Aug 2007 07:35:14 +0200, "qbit" <qbit@xxxxxxxxxxxxxxxxx>
wrote:
"Special relativity predicts that atomic clocks moving at GPSUse 1 minus the Lorentz(x) or call it f(x) = 1 - sqrt(1 - x^2/c^2) ~ .
orbital speeds will tick more slowly than stationary ground
clocks by about 7,200 ns per day."
( http://en.wikipedia.org/wiki/Global_Positioning_System )
Which formula must be used to calculate this value?
The orbital speed is 3.868e^3 m/s so f(3868) = 8.33e-11
Multiply this by K = 86.4*10^9 nsec/day = 7.20 usec/day. That is the
value ignoring equatorial velocty.
The equatorial speed is 463 m/s, so f(463)xK = .103 usec/day
So the orbiting clock runs slower from both causes by
7.20 - .103 = 7.097 usec/day
just due to relative velocity.
In relativity they use
[1 - Lorentz(3868)/Lorentz(463)] x K = 7.09usec/day.
John Polasek
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