GPS is a Global Positioning System. it uses satellites in orbit to identify precise location on a globe. Several posts here talk about it being used in Low Earth Orbit (LEO) with success. About 30 satellites are used in the GPS system

If we had satellites in orbit around the Sun used to identify precise location in the solar system, how many satellites would be required?

If distance impacts the answer, I am interested in locations between the orbits of Mecury and Saturn, locations closer to or farther from the Sun can be ignored.

Locations in the shadow of a planet, can be ignored as well. The current GPS system does not provide coverage in some shadow areas, like deep valleys and large city streets.

  • $\begingroup$ i guess 4 in most cases, but let it be 5 to have some redundance :-) $\endgroup$ – szulat Apr 4 '17 at 13:03
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    $\begingroup$ GPS needs not only satellites in earth orbit, ground stations at fixed known locations on earth are needed to measure the very precise orbit data of all satellites. Without that orbit data, exact position of a GPS receievr could not be estimated. A solar positioning system would not work if the orbit of the satellites is not measured from "ground" stations. If a threedimensional position should be estimated, satellites in three different orbit planes are necessary. At least two satellites in each plane. 6 satellites for all planes. $\endgroup$ – Uwe Apr 4 '17 at 18:08
  • $\begingroup$ To answer this question we must first determine the range of the theoretical satellite signals. If they can received at 160 light seconds range, then only 4 would be needed as mentioned by @szulat. But at some point the signal will be too faint to be useful, and they will need to be omni directional broadcasts, increasing the total power requirement. $\endgroup$ – OrangePeel52 Apr 4 '17 at 18:46
  • $\begingroup$ 4 sats at 90º, 4 sats at 45º, 4 sats at 0º, 4 sats at -45º, each one with a decent solar panel array to power the omni radio signals. $\endgroup$ – CptEric Apr 5 '17 at 6:57
  • $\begingroup$ I believe power and signal strength would be the two largest hurdles. Assuming you have sufficient power to send a signal from anyplace in the orbit (3D sphere not just the plane) of Saturn to Mercury and vice versa, you would need 2 groups of satellites, one group in Mercury orbit the other in Saturn orbit. If you are 5 light minutes out from Mercury sphere and 70 light minutes in from Saturn sphere, you are about in Earth Orbit. As power is easier to get nearer the sun, instead use Venus for your second orbit sphere, with an accurate clock on your ship in theory 8 to 12 satellites. $\endgroup$ – James Jenkins Apr 5 '17 at 13:40

@szulat is right. mathematically speaking, 4 spheres cenetered about satellites with known position, and varying radius will intersect at one point; we can calculate that point and be done with it. As far as engineering an entire system, you have to provide more constraints

the radius you want to cover (sun to saturn) is about 80 light minutes.

if you want to cover the entire volume enclosed by 80 light minutes, that's about 2.14E6 cubic light minutes. The ratio of the distance to saturn vs the radius of the earth is 1.9E9 : 6.4E3 = .2968 E 6.

cubing that, it's .0261 E 18 = 2.6 E 16 that is to say, the volume you have to cover is 10^16 times as big as earth! to ensure that each satelite communicates within one minute of the others, you'd need about 2 million satelites. That's 1000 times more satelites than currently orbit the earth

they'd be orbitting in rings around the sun, much like our planets are. But in all three dimensions

so yah, I'd say 2 million is the upper limit. If you're ok with light-minute accuracy

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    $\begingroup$ I am not sure, about this. Why do we care about if they are within a light minute of each other? Also you need at a minimum, four satellites must be in view of the receiver for it to compute four unknown quantities (three position coordinates and clock deviation from satellite time) $\endgroup$ – James Jenkins Apr 4 '17 at 17:46
  • $\begingroup$ one light minute was arbitrary (also easy to calculate). And I'm not disagreeing with the minimum, that's easy to establish. I'm establishing an upper limit. $\endgroup$ – Mohammad Athar Apr 4 '17 at 17:51
  • $\begingroup$ Yes but why the light minute separation? Maybe I don't understand something about GPS? If a satellite is 10.35678 light seconds away adjust your calculations for it. If you have signals from 4 satellites and they are all 10 minutes behind your GPS clock, do the math. Why do you need one that is only one light minute away? $\endgroup$ – James Jenkins Apr 4 '17 at 17:56
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    $\begingroup$ There are other more important practical constraints: you need to ensure enough satellites are visible (not behind the Sun or a planet), you need spares for unexpected outages, so the real number would be several times over the mathematical minimum. But I don't understand where this constraint is coming from. $\endgroup$ – Lesser Hedgehog Apr 5 '17 at 0:26
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    $\begingroup$ Each GPS transmission is precision timestamped and includes each satellite's ephemeris data, so as long as the satellites' time is all synchronized and they are continuously transmitting, they do not have any distance constraints. On Earth, the intersection of 3 spheres produces two points; since only one is on the surface of the Earth, you can eliminate the other. In space, you don't have that constraint, so you would need 4 satellites for a reliable fix...5 if you wanted to prevent against shadowing from a body. Also note that at least one would need to be out of the solar plane. $\endgroup$ – ereisch Apr 5 '17 at 14:29

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