How would a constellation of GPS satellites differ for Mars? For Earth the system is supposed to work with at least 24 satellites orbiting at ~20.000 km. Would Martian equivalent need less of them? At what orbit height?

  • $\begingroup$ Good question! Other systems like Glonass, and Galileo, and BeiDou have or will have a similar number of satellites in roughly speaking similar orbits to GPS (groups in few planes of MEO). Earth atmosphere has a lot of water which varies the speed of the GPS signal. WIthout it maybe Mars would not need so many ground stations? $\endgroup$
    – uhoh
    Commented Feb 6, 2016 at 1:46

2 Answers 2


GPS requires four satellites being visible from any point on the surface at any time (one less if the receivers carry a high-precision atomic clock).

On a perfect spherical world this can (theoretically) be provided by eight satellites in a very high orbit - from far away every observer can see exactly half of the planet. This constellation would use four different orbits with two satellites positioned opposite to each other in each orbit.

High orbits might not be favorable due to the large distance to the receivers and accordingly high transmitting power needed and also larger measurement errors. Additionally, satellites that are barely visible above the horizon can be obstructed by mountains (do exist on Mars) or high buildings (don't exist on Mars yet). So, one might want to place 3 satellites in each orbit and use more than four different orbits.

A GPS that is available at any time at any point of the surface and that guarantees a minimum elevation (10 degrees according to my very rough estimate) above the horizon can likely be built with about 15 satellites (e.g. 5 orbits with 3 satellites each). The 24 satellites used for the Earth-GPS are due to lower orbits, high minimum elevation (atmosphere is thick at low elevation angles causing large errors) and redundancy. Note that four satellites visible is the absolute minimum to do any position estimation. A fifth one is already needed to make any estimation on precision.

For Mars I'd assume one is fine with a lower precision than on Earth and one does not want to spend too many satellites and hence goes with a lower count of satellites in higher orbits. 15 seem likely to me.

Also note that in no calculation the size of the planet plays a role - it's only the ration between planetary radius and orbital radius that determines the area a satellite can be seen from.

  • $\begingroup$ About the planet size - I am imagining that to get the same error-rate you can you se the same absolute height (or even higher because of smaller effects of the thinner atmosphere) so the ratio can be different? $\endgroup$
    – jkavalik
    Commented Feb 7, 2016 at 18:43
  • $\begingroup$ There is not too much difference here: The earth atmosphere causes quite some errors, but these can only be corrected by measuring the properties of the atmosphere (e.g. by stations at defined positions), height of satellites doesn't help, only elevation above the horizon (which means less atmosphere in between) plays a role. From this point of view, satellites close to the horizon can be used more reliably on Mars than on Earth. $\endgroup$
    – asdfex
    Commented Feb 7, 2016 at 18:52
  • $\begingroup$ Doesn't the planet size affect what would be considered a "high orbit"? I imagine there's a threshold where it's cheaper to use more satellites than to use a stronger signal, and that would depend on absolute distances rather than the ratio. GPS on Earth is about 26Mm, much less than the >70Mm radius of Jupiter; would the same ratio be just as good there? $\endgroup$ Commented Nov 30, 2016 at 20:42

From the 3rd Kepler's law you can find that the orbit radius for the satellite in a circle orbit can be found as $$ R = \frac{G^{1/3}}{(2\pi)^{2/3}} T^{2/3} M_{planet}^{1/3} $$ So if you take $G=6.67 \times 10^{-11}$ SI units, $M = 6.4171\times 10^{23}$ kg, and, like GPS satellites, $T$ is equal to half of the Martian sidereal day (which is 24h 37m 22s or 86644 seconds), you finally get $$ R = 12672~\text{km} $$ from the center of the planet.

It also means that the orbit should be about $9282~\text{km}$ above the average surface of Mars. In case of Earth, such an orbit is more than twice as far out since Earth is more massive.

The number of satellites may depend on actual configuration and requirements for the minimum number of satellites visible from a certain point of surface (if you need 24/7 performance), but it can't be much less than the same in case of Earth. I believe something about 18 (6 in three orbital planes) is a necessary minimum.

  • 2
    $\begingroup$ Where do you see the necessity to have an orbital period of half a day? $\endgroup$
    – asdfex
    Commented Feb 7, 2016 at 18:14

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