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.

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    $\begingroup$ i guess 4 in most cases, but let it be 5 to have some redundance :-) $\endgroup$
    – szulat
    Commented Apr 4, 2017 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
    Commented Apr 4, 2017 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. $\endgroup$ Commented Apr 4, 2017 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
    Commented Apr 5, 2017 at 6:57
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    $\begingroup$ 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. $\endgroup$ Commented Apr 5, 2017 at 13:40

2 Answers 2


How many satellites would be required for Solar System GPS?

Let's try six, along the lines of @szulat's comment but a few more.

The question says we can ignore planetary shadows for the purposes of this question, so ignoring the need for redundancy for now, let's put six in circular polar heliocentric orbits at say 6 AU and never mind the cost, in two planes at right angles, phased in some reasonable way.

Instead of transmitting at circa 1 GHz, let's use 375,000 GHz by using 800 nm lasers and photodetectors coupled to 30 cm diameter telescopes.

Doing a link budget calculation along the lines of this answer we see that at 800 nm, the gain $G$ of a 30 cm "dish" or telescope is 121.4 dB. With a 1 watt average power laser (it would be pulsed of course) and a distance of 10 AU, the received optical power between two points can be calculated as follows:

$$ P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX} $$

  • $P_{RX}$: Received Power
  • $P_{TX}$: Transmitted Power
  • $G_{TX}$: Gain of Transmitting antenna (compared to isotropic)
  • $L_{FS}$: "Free space Loss", what we usually call $1/r^2$ (but also has $R^2 / \lambda^2$) because receive gain is relative to isotropic)
  • $G_{RX}$: Gain of Earth's Receiving antenna (compared to isotropic)

$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right)$$

$$G_{Dish} \sim 20 \times \log_{10}\left( \frac{\pi d}{\lambda} \right)$$


$$ P_{RX} = 0 + 121.4 - 387.4 + 121.4 = -144.6 \ \text{dBw}$$

dBw is like dBm but using watts instead of milliwatts. That's a received power of 3.5E-15 Watts, which is over 10,000 photons per second. Assuming they are pulsed and you can average for a while, you should be able to determine distances quite accurately.

Yes, everything is moving at km per second, but once ephemerides (pronounciation) are established you can calculate trajectories precisely even though you are building up data with tens of meters precision over tens or hundreds of kilometers.

X-ray pulsars and atomic clocks

This kind of calculation on jittery, noisy, low-statistics data has already been done using X-ray pulses from pulsars instead of light pulses from spacecraft as outlined in the question Is NICER/SEXTANT the first civilian “spacecraft” to determine it's own position in space without GPS or uplinked data?.

As oulined in Where would one deploy deep space atomic clocks? and in How does an onboard atomic clock help interplanetary navigation? atomic clocks in deep-space spacedraft is on its way. We know that optical communications using modest telescopes is as well, so building these "interplanetary GPS satellites" should be straightforward.

Optical pulses from a few interplanetary spacecraft in well characterized heliocentric orbits may complement this technique by adding triangulation capability, and it could potentially compete with absolute navigation using NICER/SEXTANT absolute navigation using only pulsars and no Earth stations or "interplanetary gps satellites.

Telescope operation and scheduling

I suppose each navigation satellite could have dozens of telescopes, but because these deep space orbits can be well characterized over time and ephemerides built up, with good math they don't really need to be in constant communication with each other, and unless your spacecraft is in a critical near-planet enounter, you don't need to receive signals from all of them at the same time. So you would probably have to run some kind of scheduling service so that your interplanetary spacecraft can have signals from the six navigation satellites sent to it in rapid succession. Your spacecraft would require a good optical commuications telescope and a correspondingly active and agile attitude control system to pick up your signals when they are scheduled to arive.

Here's an illustration of a possible minimal 6 satellite 6 AU configuration, with Mercury through Saturn shown as well, and the Python 3 script that made it.

minimal solar system GPS

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]

semis     = np.array((0.387, 0.723, 1.0, 1.523, 5.204, 9.583))
args      = twopi * (np.linspace(0, 1, 301) + np.random.random(len(semis))[:, None])
funcs     = (np.cos, np.sin, np.zeros_like)
planets   = np.stack([f(args) for f in funcs], axis=1)*semis[:, None, None]

semis     = 6*np.ones(3)
args1     = twopi * (np.linspace(0, 1, 301) + 0.05 + np.arange(3)[:, None]/3.)
args2     = twopi * (np.linspace(0, 1, 301) + 0.05 + np.arange(3)[:, None]/3. +  1/6.)
funcs1    = (np.cos,        np.zeros_like, np.sin)
funcs2    = (np.zeros_like, np.cos,        np.sin)
navsats1  = np.stack([f(args1) for f in funcs1], axis=1)*semis[:, None, None]
navsats2  = np.stack([f(args2) for f in funcs2], axis=1)*semis[:, None, None]
navsats   = np.vstack((navsats1, navsats2))

if True:
    fig = plt.figure()
    fig.patch.set_facecolor('xkcd:mint green') # https://stackoverflow.com/q/14088687/3904031
    plt.subplots_adjust(top=0.95, bottom=0.05, left=0.05, right=0.95, hspace=0.2, wspace=0.2)
    plt.rcParams['axes.facecolor'] = 'black' # https://stackoverflow.com/a/40371037/3904031

    ax  = fig.add_subplot(1, 1, 1, projection='3d')

    for (x, y, z) in planets:
        ax.plot(x, y, z)
        ax.plot(x[:1], y[:1], z[:1], 'o')
    for (x, y, z) in navsats:
        ax.plot(x, y, z, '-c')
        ax.plot(x[:1], y[:1], z[:1], 'or')

@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) $\endgroup$ Commented Apr 4, 2017 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. $\endgroup$ Commented Apr 4, 2017 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? $\endgroup$ Commented Apr 4, 2017 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. $\endgroup$ Commented Apr 5, 2017 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
    Commented Apr 5, 2017 at 14:29

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