I know that besides the effects of Newton's theory of Gravitation on the satellite's motion, one has to take account of the retardation of the satellite's clocks when compared to earth-fixed clocks. Is this just an effect of Special Relativity? Or are there other effects that must be taken into account, which would not be predicted by Special Relativity alone, but which only occur in General Relativity?
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$\begingroup$ Do you mean Einstein, not Newton, in your first line? It is an important distinction. $\endgroup$– user81619Jul 28, 2015 at 18:23
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/1061/2451 and links therein. $\endgroup$– Qmechanic ♦Jul 28, 2015 at 18:39
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$\begingroup$ Special relativity: clock goes slower (satellite is fast): 7$\mu$s/day. General relativity: clock goes faster (lower gravity): 45 $\mu$s/day. Both matter. Details in the duplicate noted by Qmechanic. $\endgroup$– FlorisJul 28, 2015 at 18:41
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$\begingroup$ @Floris: I am 99.9% sure that the numbers in the accepted answer in that duplicate question are wrong; if you read Clifford Will's article (who knows what he's talking about), that's not actually what he says. The general conclusion that GR matters is correct, though. See my answer below. $\endgroup$– Michael SeifertJul 28, 2015 at 18:47
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$\begingroup$ Upon further reflection, the answers given by @Floris are correct (though I think that the formulas given in the linked answer are misleading at best.) $\endgroup$– Michael SeifertJul 28, 2015 at 19:26
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The short answer is yes. An object moving at non-relativistic velocity $\vec{v}$ in a weak gravitational field will have a proper time $\Delta \tau$ elapse that is related to the time $\Delta t$ on distant clocks (far from the Earth) by the equation $$ \frac{\Delta \tau}{\Delta t} \approx 1 - \left( \frac{1}{2} \frac{v^2}{c^2} + \frac{G M_E}{r c^2} \right) $$ where $r$ is the distance from the center of the Earth and $M_E$ is the Earth's mass. The first term in parentheses is just the change in the Lorentz factor $\gamma = (1 - v^2/c^2)^{-1/2}$ to lowest non-vanishing order in $v$. The second term, however, can be seen to have something to do with gravity (in fact, the general expression would be $\Phi/c^2$, where $\Phi$ is the Newtonian gravitational potential.) For a satellite in circular orbit, it's not too hard to show that $v^2 = G M_E/r$; thus, you can see that the correction due to special relativity is one-half as large as the correction due to gravity.
Of course, we don't measure the time difference relative to clocks at infinity; we measure the time difference relative to clocks on the ground, which are themselves moving (due to rotation of the Earth) and in a gravitational field. To find the true correction effects, you could combine the correction factor for orbital clocks relative to infinity with the correction for Earth-based clocks relative to infinity. This would lead to a fractional correction relative to Earth of something like $$ \frac{\Delta \tau_s}{\Delta \tau_g} \approx 1 - \left( \frac{1}{2} \frac{v_s^2}{c^2} + \frac{G M_E}{r_s c^2} \right) + \left( \frac{1}{2} \frac{v_g^2}{c^2} + \frac{G M_E}{r_g c^2} \right) $$ where the subscripts $g$ and $s$ stand for "ground" and "satellite" respectively. If we assume a station at the Earth's equator, then we have $v_s \approx 464$ m/s and $r_s \approx 6380$ km; for GPS satellites, we have $r_s = 26 580$ km (an altitude of approximately 20,200 km) and, since they are in approximately circular orbits, $v_s = 3 872$ m/s (using the formula above.) Thus, the overall correction factor due to special-relativistic effects would be $$ \frac{1}{2} \left( \frac{v_g^2}{c^2} - \frac{v_s^2}{c^2} \right) \approx -8.2 \times 10^{-11} \approx - 7 \, \mu\text{s / day} $$ while the correction factor due to gravitational effects would be $$ \frac{G M_E}{r_g c^2} - \frac{G M_E}{r_s c^2} \approx 5.2 \times 10^{-10} \approx 46\, \mu \text{s / day}. $$ Thus, the correction effects work in opposite directions, but you can see that the gravitational effects cannot be neglected compared to the special-relativistic effects.
answered Jul 28, 2015 at 18:52
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$\begingroup$ May I suggest that you take a second to show where your expression comes from - and plug in some values for the orbital height and velocity to show the magnitude of the relative effects? You are right that the time difference should be calculated relative to a clock on earth (which is itself moving at a certain speed, and experiences a certain gravity). It would be good to do the whole derivation if you want some credibility in contradicting an accepted answer with 37 upvotes. I for one am keeping my finger off the vote button for now... $\endgroup$– FlorisJul 28, 2015 at 19:01
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2$\begingroup$ @Floris: Well, I'll be; the numerical answers in that answer were in fact correct. In my paltry defense, the formulas in said answer were incorrect; to get the correct factors, you have to take the differences in the Lorentz factors, not just the Lorentz factors themselves (which is what the answerer appeared to be doing.) Still, my critique is cheerfully withdrawn. $\endgroup$ Jul 28, 2015 at 19:25
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$\begingroup$ And NOW I can cheerfully click that up-arrow thingy... - and I have added a link to this answer under the other one since you bring up some very important points. $\endgroup$– FlorisJul 28, 2015 at 19:38
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$\begingroup$ Thanks for all this information. I assume then, that the correction you describe, due to gravitational effects precribed by general relativity, is actually incorporated into the computations made to operate the GPS system. $\endgroup$ Jul 29, 2015 at 19:14