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To calculate the amount of daily time dilation experienced by GPS satellites relative to Earth we need to separately determine the amounts due to [[special relativity]] (velocity) and [[general relativity]] (gravity) and add them together.
==== Special Relativity (SR) ====
The amount due to velocity will be determined using the [[Lorentz transformation]]. This will be:
: <math> \frac{1}{\gamma } = \sqrt{1-\frac{v^2}{c^2}} </math>
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: Note that this speed of {{val|3874|u=m/s}} is measured relative to Earth's center rather than its surface where the GPS receivers (and users) are. This is because Earth's equipotential makes net time dilation equal across its geodesic surface.<ref>{{Cite web |last=S. P. Drake |date=January 2006 |title=The equivalence principle as a stepping stone from special to general relativity |url=http://www.phys.unsw.edu.au/einsteinlight/jw/2006AJP.pdf |website=Am. J. Phys., Vol. 74, No. 1 |pages=22–25}}</ref> That is, the combination of Special and General effects make the net time dilation at the equator equal to that of the poles, which in turn are at rest relative to the center. Hence we use the center as a reference point to represent the entire surface.
==== General Relativity (GR) ====
The amount of dilation due to gravity will be determined using the [[gravitational time dilation]] equation:
: <math> \frac{1}{\gamma } =\sqrt{1-\frac{2G M}{r c^2}} </math>
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: <math> \Delta \left(\frac{1}{\gamma }\right) \approx 5.307\times 10^{-10} </math>
This represents the fraction by which the satellites' clocks move faster than Earth's. It is then multiplied by the number of nanoseconds in a day:
: <math> 5.307\times 10^{-10}\times 60\times 60\times 24\times 10^9\approx 45850 \text{ ns} </math> That is, the satellites' clocks gain 45,850 nanoseconds a day due to [[general relativity]] effects.
==== Combined SR and GR ====
These effects are added together to give (rounded to 10 ns):
: 45850 – 7210 = 38640 ns
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