I am reading A Brief History of Time by Stephen Hawking, and in it he mentions that without compensating for relativity, GPS devices would be out by miles. Why is this? (I am not sure which relativity he means as I am several chapters ahead now and the question just came to me.)

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    $\begingroup$ astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html $\endgroup$
    – kennytm
    Commented Nov 18, 2010 at 13:52
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    $\begingroup$ I'm trying to locate my sources on this, but I have read that even if you don't account for general relativity (by slowing down the clocks prior to launch) your GPS would work just fine because the error is the same for all satelites. The only issue would be that the clocks would not be synchronized with the ground, but that is not necessary for calculating your current position. Can anyone confirm this? $\endgroup$ Commented Nov 13, 2012 at 11:27
  • $\begingroup$ Found something: physicsmyths.org.uk/gps.htm can anyone comment on this? $\endgroup$ Commented Nov 13, 2012 at 11:29
  • $\begingroup$ found something else in this same site: physics.stackexchange.com/q/17814/3177 (some answers mention this) $\endgroup$ Commented Nov 13, 2012 at 11:38
  • $\begingroup$ I looked at that uk site hurriedly and there seem to be some crank "disproofs" of special relativity, so I doubt that that site is trustworthy. There are cranks on stack exchange, too, of course....and on Wikipedia, and in academia, and .....yours truly, $\endgroup$ Commented Dec 5, 2015 at 23:11

5 Answers 5


Error margin for position predicted by GPS is $15\text{m}$. So GPS system must keep time with accuracy of at least $15\text{m}/c$ which is roughly $50\text{ns}$.

So $50\text{ns}$ error in timekeeping corresponds to $15\text{m}$ error in distance prediction.
Hence, for $38\text{μs}$ error in timekeeping corresponds to $11\text{km}$ error in distance prediction.

If we do not apply corrections using GR to GPS then $38\text{μs}$ error in timekeeping is introduced per day.

You can check it yourself by using following formulas

$T_1 = \frac{T_0}{\sqrt{1-\frac{v^2}{c^2}}}$ ...clock runs relatively slower if it is moving at high velocity.

$T_2 = \frac{T_0}{\sqrt{1-\frac{2GM}{c^2 R}}}$ ...clock runs relatively faster because of weak gravity.

$T_1$ = 7 microseconds/day

$T_2$ = 45 microseconds/day

$T_2 - T_1$ = 38 microseconds/day

use values given in this very good article.

And for equations refer to HyperPhysics.

So Stephen Hawking is right! :-)

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    $\begingroup$ Is $R$ the radius of the earth, or the orbit radius? $\endgroup$ Commented May 11, 2014 at 16:32
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    $\begingroup$ But what's relevant for GPS is the difference between timestamps from different satellites, right? And since they are on the same altitude they should be time shifted by the same amount, so the differences should be basically the same as without relativity. I mean it doesn't matter how much the error in the clocks is after a day, since the localization error is not cumulative, because the satellites' clocks don't drift away from each other. $\endgroup$
    – isarandi
    Commented Jun 14, 2015 at 23:31
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    $\begingroup$ @Dims 15/300000000 != 100*10^(-6), it equals 5*10^(-8). I got my answer by just typing it into google, but it should be easy to see that 15 divided by 3 is going to be a leading 5, not a leading 1. $\endgroup$ Commented Apr 27, 2018 at 14:21
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    $\begingroup$ Lots of misinformation here. As per the US Naval Observatory (the creators of GPS to replace LORAN): GPS does NOT use relativity calculations at all (repeat, it does NOT use relativity calculations). $\endgroup$
    – MC9000
    Commented Jan 16, 2019 at 0:26
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    $\begingroup$ @MC9000 - This is actually a common misconception about GPS. See for example here. The point is that instead of calculations directly based on general relativity, much simpler corrections are used to approximate those, as long as the receiver is only slowly moving on the surface of the planet. $\endgroup$ Commented May 6, 2019 at 10:03

There's the article from Ohio State University http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html which explains quite well why the clocks on a GPS satellite are faster by about 38 microseconds every day. The article then claims that not compensation for these 38 microseconds per day would cause a GPS to be off by about 11 km per day, plainly unusable, and claims that this (the fact that we need to compensate for the 38 microseconds to get GPS working) is proof for General Relativity.

The problem is that while the clocks are indeed off by 38 microseconds per day and General Relativity is all fine, we wouldn't actually have to compensate for it. The GPS in your car or your phone doesn't have an atomic clock. It doesn't have any clock precise enough to help with GPS. It doesn't measure how long the signal took to get from satellite A to GPS. It measures the difference between the signal from satellite A and the signal from satellite B (and two more satellites). This works if the clocks are fast: As long as they are all fast by the exact same amounts, we still get the right results.

That is, almost. Satellites don't stand still. So if we rely on a clock that is 38 microseconds fast per day, we do the calculations based on the position of a satellite that is off by 38 microseconds per day. So the error is not (speed of light times 38 microseconds times days), it is (speed of satellite times 38 microseconds times day). This is about 15 cm per day. Well, satellite positions get corrected once a week. I hope nobody thinks we could predict the position of a satellite for long time without any error.

Back to the original assumption, that without compensation the error would be 11km per day: The satellite clocks are multiplied by a factor just shy of 1 so that they go at the correct speed. But that wouldn't work. The effect that produces 38 microseconds per day isn't constant. When the satellite flies over an ocean, gravity is lower. The satellite speed changes all the time because the satellite doesn't fly on a perfect circle around a perfectly round earth made of perfectly homogenous material. If GR created an error of 11km per day uncompensated, then it is quite unconceivable that a simple multiplication of the clock speed would be good enough to reduce this to make GPS usable.

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    $\begingroup$ Nice. But I have to say that from a the philosophical position of an experimenter, a machine that makes it operators tear their hair out (which GPS would in the absence of of GR) isn't working until those behaviors are understood (which would happen when someone invented GR to explain the anomaly). But that's a philosophical point. $\endgroup$ Commented Dec 6, 2015 at 1:07
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    $\begingroup$ This is the one correct answer on this page. GPS was significant evidence for GR because we can compare the speed of clocks in orbit to those on earth. However, the accuracy of the GPS system doesn't depend on the satellites keeping exact time. As long as they keep the same time, the system works. $\endgroup$
    – user235504
    Commented Dec 6, 2015 at 3:39
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    $\begingroup$ Actually, GPS is a poor "proof" of GR for the reason you state. gnasher has the correct answer - Einstein field equations are not used in GPS at all (imagine the number crunching involved and the computer power necessary wasting all that energy - not to mention added weight to satellites - especially a few decades ago) $\endgroup$
    – MC9000
    Commented Jan 16, 2019 at 0:29
  • $\begingroup$ It's true that the only thing needed to determine the GPS receiver position relative to the satellites is that the satellite clocks be synced and the speed of transmission be the same. But that's relative to the satellites. The user wants the GPS receiver to calculate where it is on the Earth, which requires accounting for where the satellites are in orbit and how the Earth has rotated. That's why the satellite clocks have to be kept synced to clocks on the ground and why they are adjusted to keep them synced. $\endgroup$ Commented Oct 30, 2019 at 0:08
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    $\begingroup$ @MC9000: No one ever claimed that the Einstein field equations are solved on the fly by the GPS satellites' computers. The geometry of spacetime near Earth is approximated well enough by Schwarzschild spacetime, so solving the field equations all over again is not necessary. In particular, time dilation in Schwarzschild is described by rather simple formulae, so no extensive number crunching would be necessary in the first place. $\endgroup$
    – balu
    Commented Jun 18, 2020 at 13:57

You can find out about this in great detail in the excellent summary over here: What the Global Positioning System Tells Us about Relativity?

In a nutshell:

  1. General Relativity predicts that clocks go slower in a higher gravitational field. That is the clock aboard the GPS satellites "clicks" faster than the clock down on Earth.
  2. Also, Special Relativity predicts that a moving clock is slower than the stationary one. So this effect will slow the clock compared to the one down on Earth.

As you see, in this case the two effects are acting in opposite direction but their magnitude is not equal, thus don't cancel each other out.

Now, you find out your position by comparing the time signal from a number of satellites. They are at different distance from you and it then takes different time for the signal to reach you. Thus the signal of "Satellite A says right now it is 22:31:12" will be different from what you'll hear Satellite B at the same moment). From the time difference of the signal and knowing the satellites positions (your GPS knows that) you can triangulate your position on the ground.

If one does not compensate for the different clock speeds, the distance measurement would be wrong and the position estimation could be hundreds or thousands of meters or more off, making the GPS system essentially useless.


The effect of gravitational time dilation can even be measured if you go from the surface of the earth to an orbit around the earth. Therefore, as GPS satellites measure the time it's messages take to reach you and come back, it is important to account for the real time that the signal takes to reach the target.

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    $\begingroup$ GPS signals do not return to the satellite, they only go to the receiver AFAIK... $\endgroup$
    – Thomas O
    Commented Nov 18, 2010 at 13:53
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    $\begingroup$ But the main point still holds, and it is that more time passes on Satellite's clock than your clock back on earth, with respect to either one of you. $\endgroup$
    – Cem
    Commented Nov 18, 2010 at 13:59
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    $\begingroup$ Interestingly general relativity is not use per se in calculations for GPS systems. Rather, a nice little trick involving special relativity (applying a series of Lorentz transformations in infinitesimal steps) is what it does. This turns out to be sufficiently accurate and a lot easier computationally. $\endgroup$
    – Noldorin
    Commented Nov 18, 2010 at 14:22
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    $\begingroup$ You can detect time dilation just by spending a few days in the mountains. leapsecond.com/great2005/index.htm $\endgroup$
    – endolith
    Commented Nov 18, 2010 at 15:16
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    $\begingroup$ @endolith : ... if you bring an atomic clock with you ! $\endgroup$ Commented Nov 18, 2010 at 18:14

I don't think that GPS "depends on relativity" in the sense that a technological civilization that never discovered special/general relativity would be unable to make a working GPS system. You can always compare the clock in a satellite to clocks on the ground and adjust the rate until they don't drift out of sync, whether or not you understand why they were drifting out of sync. In fact, they do synchronize them empirically, not by blindly trusting a theoretical calculation.

Asking what would happen if the clocks drifted by 38 μs/day (for any reason) is a strange counterfactual because it suggests that no one is maintaining the system, in which case it would presumably quickly succumb to various other problems of non-relativistic origin. If someone is keeping some parts of the system in sync, you'd probably have to specify which parts. For example if the satellites accurately know their positions with respect to an inertial frame moving with the center of the earth, but the orientation of the earth is calculated from the time of day, then you'd have an accumulating position error of 38 μs worth of earth rotation, or a couple of centimeters at the equator, per day. But if the satellites accurately know their position with respect to a corotating reference frame, then the error would be much smaller.


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