# Time flow difference for satellites [duplicate]

Clocks in satellites have to be adjusted due to the effects of relativity; but does time for satellites (GPS) flow slower, due to the relative motion, or faster, due to the weaker amount of gravitation received due to the increased distance to earth?

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• Yes, time does need to be adjusted. I suspect it flows slower. Mostly from a faint recall of reading some articles in the past. It is relevant for GPS which used very precise timing. – Hennes May 12 '15 at 20:09
• The time for the satellite goes faster relative to time on earth – Alexander May 12 '15 at 20:11

We'll consider the relevant terms in the Schwarzschild metric, and substitute $d\phi = \frac{v}{r} dt$ with $\theta = \pi/2$ to get (assuming circular orbits): $$d\tau^2 = \left(1-\frac{r_s}{r}\right)dt^2 - \frac{v^2}{c^2}dt^2$$ where $\tau$ is the proper time (time as measured by the object), and $v$ is the orbital speed.

The relevant quantity to measure time dilation is: $$\frac{d\tau}{dt} = \sqrt{1 - \frac{r_s}{r} - \frac{v^2}{c^2}}$$

We need to plug in values to compare. The Schwarzschild radius of the earth is $r_s = 8.9\text{mm}$. An observer on the Earth has an angular velocity of $\frac{v}{r} = \frac{2\pi}{3600\times24}\text{rad/s} \approx 7\times10^{-5}\text{rad/s}$. Combined with a radius of $r = 6400\text{km}$, we get $v = 465\text{m/s}$: $$\frac{d\tau_{E}}{dt} = \sqrt{1 - \frac{8.9}{6400,000,000} - \frac{465^2}{299792458^2}} \approx \sqrt{1 - 1.4\times10^{-8}}$$ where the contribution due to the speed is negligible.

For a GPS satellite, $r \approx 26800\text{km}$, and $\frac{v}{r} \approx 14\times10^{-5}\text{rad/s}$. This means that $v = 3897\text{m/s}$. Plugging this in: $$\frac{d\tau_{G}}{dt} = \sqrt{1 - \frac{8.9}{26800,000,000} - \frac{3897^2}{299792458^2}} \approx \sqrt{1-5\times10^{-10}}$$

We see that it is the effect of the weak gravitational field that dominates, and the passage of time is therefore faster for a satellite than someone on the surface (where the notion of simultaneity is given by the $t$ coordinate).

• AV23: "the passage of time is therefore faster for [...]" -- What an ugly, possibly coordinate dependent choice of phrase! (Half the blame goes to the OP question.) Better to make a (plain) comparison of (definite) durations; e.g. "It takes a GPS satellite $\approx~(1 + 1.35 \times 10^{-8})$ as long to complete a full revolution around Earth than a constituent of Earth's surface." p.s. Any thoughts on the difference (albeit it as good as negligible for the problem at hand) between the exact expressions of your solution and of this accepted answer? – user12262 May 13 '15 at 5:36
• p.p.s. Oops! ... Well, in my preceding comment I had mistakenly been thinking of geostationary satellites. So, instead, express "the answer" rather as: $\qquad~\qquad~\qquad~\qquad~\qquad~\qquad~\qquad~\qquad~\qquad~\qquad$ "It takes a GPS satellite $\approx~(1 + 1.35 \times 10^{-8})$ as long from one passage right above a particular constituent of Earth's surface, until the next, as it takes this constituent from one passage right underneath the GPS satellite, until the next." – user12262 May 13 '15 at 5:45
• @user12262 I've updated in a way that I believe clarifies the sentence in a simpler way. And I'm not sure of the reasons behind the difference, unless that answer considers the 'radial' velocity, which seems to not be the case. It may have to do with the definition of $v$, which, in my case, is circumference (true distance as measured at the same $t$) covered per unit change in $t$. I'll appeal to at least one source (though it's Wikipedia: en.wikipedia.org/wiki/…) that agrees with this version. – AV23 May 13 '15 at 9:23
• AV23: "[...] not sure of the reasons behind the difference [...]" -- And I didn't mean to be critical of your derivation as much as of the (seeming) lack of care and responsiveness of certain other contributors. "I've updated in a way that I believe clarifies [...]" -- I still recognize the offending choice of phrase, and you didn't use my suggested substitution. But you added: "where the notion of simultaneity is given by the $t$ coordinate" -- Hmm: "simultaneity" in a region which is explicitly not flat?? (I'd like to pursue this further but I'll get a chance only in about a week). – user12262 May 13 '15 at 14:40