Time dilation

On this image, if I understand correctly, the time dilation diagram is shown depending on the height for circular orbits. First in low orbit, time slows down relative to the surface due to high speed, but for higher orbits, speed and gravitational dilation is lowering, so at roughly 3,200 km elevation, time passes almost at same rate as on the surface!

However, GPS satellites are mostly located at about 20,000 km. In that case, why don't we place GPS satellites at an altitude where time passes at the same rate as on the surface? (Of course, in this case, too, corrections would have to be made for different speeds of the Earth's surface at different latitudes, but perhaps at an altitude of 3,200 km, this would be much easier to do?)

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    $\begingroup$ Perhaps time dilation is not relevant for most of the satellites purposes. $\endgroup$ May 7, 2023 at 17:45
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    $\begingroup$ Time always passes at the same rate. Merely the synchronization of clocks is affected by effects like Doppler and gravitational time dilation. Since Doppler depends on the velocity vector, one can't avoid it either way. $\endgroup$ May 7, 2023 at 17:47
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    $\begingroup$ @clau It's very much related, but the effect is corrected. $\endgroup$
    – Mithoron
    May 7, 2023 at 21:21
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    $\begingroup$ What is your definition of "almost no time dilation", that apparently is not satisfied by the less than one part per billion of the maximum point on your figure? $\endgroup$ May 8, 2023 at 17:30
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    $\begingroup$ Time dilation is most definitely important for GPS. But it's not that difficult to account for, if you have a tame physicist on your crew. So GPS satellites are located where they can do the most use for the least cost. $\endgroup$
    – TonyK
    May 10, 2023 at 19:09

3 Answers 3


The higher the satellite's orbit, the more of the Earth it can see (and hence the fewer satellites you need to ensure complete coverage of the Earth). The particular orbits chosen for navigation satellites also have the advantage that they take an integral fraction of a day to complete, so the ground track of the satellites is easier to predict.

Furthermore, in a low or medium earth orbit, there is still a non-trivial amount of atmospheric drag. So these satellites would require significant amounts of propellant or frequent refueling to ensure that their orbits don't decay. Given the expected lifespan of GPS satellites, this would be extremely expensive. This further complicates matters by making the location of the satellite less predictable.

The cost of adjusting for time dilation is really trivial (it's a simple matter of scaling the clock), whereas the costs of using the lower orbit where time dilation is not a factor would be very high -- the ground track would be harder to calculate, and more satellites would be required. The cost of adding even a single satellite would be vastly, vastly higher than the cost of putting a simple scaling circuit (or software) into all of the satellites.

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    $\begingroup$ Geostationary (and other geosynchronous) orbits are also crucial for many applications and require a higher altitude. The fact is time dilation is far down on the list of priorities. $\endgroup$
    – Puk
    May 7, 2023 at 20:48
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    $\begingroup$ Good answer, though not all navigation satellites have a period equal to the Earth's axial rotation period (sidereal day) divided by an integer. While GPS satellites have an orbital period of half the Earth's rotational period, GLONASS and Galileo satellites have different periods (11.25 and 14.08 hours respectively). $\endgroup$
    – FTT
    May 7, 2023 at 23:30
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    $\begingroup$ @FTT : whoops, you're right, thank you for correcting me. The GLONASS, Galileo, and BeiDou satellites have orbital periods that are not integral fractions of a day, but they are in orbits such that the ground tracks repeat, e.g. the Galileo satellites repeat their tracks after 10 days / 17 revolutions. $\endgroup$
    – Eric Smith
    May 8, 2023 at 1:30
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    $\begingroup$ Real-world example: you need 3 geostationary satellites to cover the entire surface of Earth, whereas a low-LEO constellation like Starlink currently has 4000 and is expected to have between 10000 and up to 40000. The total number of satellites launched in the whole world in the entire history of humanity until 2022 was only ~7800, so a single low-LEO constellation like Starlink needs more satellites than have ever been launched in to space by anyone. $\endgroup$ May 8, 2023 at 12:38
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    $\begingroup$ @safesphere if you instead space them equally around the equator, they cover everything but the north and south poles, which is probably enough for many applications. Keep in mind that if a "geostationary" satellite has an inclined north/south orbit, then it isn't really geostationary. It oscillates up and down while staying at a particular latitude. So even four might not be enough to cover all of the earth at all times. $\endgroup$
    – AXensen
    May 10, 2023 at 8:40

You have to differentiate between corrections that are sources of error and corrections that are well understood and can be accounted for.

Time dilation is a well understood effect, and if you wanted to do something like sync two clocks on earth by having a satellite simultaneously send signals to both clocks (a pretty common timing-metrology technique), you definitely need to account for the time dilation between the two earth clocks and the satellite. And you can do that by tracking the satellite's position and using the relevant equations to find the expected time dilation effect. We aren't in the business of spending several million to put a satellite into orbit, but putting it at 3200km because "ehh... I don't want to write the code to account for time dilation."

However this can become a source of error if there is any uncertainty on the position of the satellite. In which case, actually, you'd want your satellite to sit in a place where the slope of this curve is smallest - not where the total effect is smallest. Since the horizontal axis is logarithmic, it's always better to be higher.

To prove this, say we had a measurement of the satellite's height $h_0$, and some error on that height $\delta_h$. And let's say the time dilation at a height $h$ is $T(h)$. Then the error on the time dilation effect is $T'(h_0)\delta_h$.

All that being said - I do agree it's interesting, and surprising to me, that there's a height at which circularly orbiting satellites have zero time dilation relative to Earth's surface.


It's probably worth adding that the altitude you're talking about is in the inner Van Allen radiation belt, which extends from about 1,000 to 12,000 km above the surface. Any satellites at that altitude would need additional radiation shielding for their electronics.


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