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The simplest solution would be a tetrahedron, but as the satellites all need to orbit in circles, it seems this configuration would only guarantee full coverage at one time.

For simplicity, there's no minimum resolution requirement, the Earth is a smooth oblate ellipsoid, and the satellites must be in orbit around Earth.

What's the true minimum number, and what would be their configuration?

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    $\begingroup$ What's your required detection range/resolution? $\endgroup$
    – Deer Hunter
    Mar 11, 2015 at 13:52
  • $\begingroup$ I guess with infinite resolution you could place two at infinity distance, but let's say they have to be in orbit around Earth. $\endgroup$
    – Guillochon
    Mar 11, 2015 at 14:19
  • $\begingroup$ Define "full coverage". Do we need to concern ourselves with for example occlusion by cloud cover, aircraft, buildings, ...? Or are you seeking to more-or-less duplicate The Blue Marble except provide an image showing the reverse side of the Earth as well? $\endgroup$
    – user
    Mar 11, 2015 at 14:49
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    $\begingroup$ Well the blue marble is impressively high resolution, I'm more just interested in the geometrical configuration required for all points on Earth's surface to have a line of sight to at least one satellite at all times. Let's ignore buildings and surface topology, and presume the the earth is an oblate ellipsoid. $\endgroup$
    – Guillochon
    Mar 11, 2015 at 14:59
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    $\begingroup$ @Guillochon For that purpose, two spacecraft are enough, placed sufficiently far away, 180° apart in the same orbital plane. $\endgroup$
    – gerrit
    Mar 11, 2015 at 15:21

3 Answers 3

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4 is actually the number, as is documented in Patent US4854527. It is a tetrahedral constellation using elliptical orbits. In one hemisphere two have their further point, while the other two have the opposite coverage. The orbital period is 27 hours.

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    $\begingroup$ The diagrams in that patent are great! Unless there's some clever way to do it with three I doubt there's a better answer. $\endgroup$
    – Guillochon
    Mar 11, 2015 at 21:30
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    $\begingroup$ It's easy to see that it can't be done with three. Three points define a plane. If that plane doesn't go through the center of the body, then you're definitely missing the side of the body opposite the plane. If it is going through the center, then at a finite distance you are missing a little bit of both poles relative to the plane. So the answer has to be >= 4. Since the patent shows a way to do four (I haven't analyzed it, but I presume it does), then that is the answer. $\endgroup$
    – Mark Adler
    Mar 12, 2015 at 0:19
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    $\begingroup$ I find it appalling that this is patentable. $\endgroup$
    – gosnold
    Mar 12, 2015 at 12:04
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    $\begingroup$ @gosnold: Personally I don't, but I think patentable orbits would make an excellent question, if you want to take it;-) $\endgroup$
    – PearsonArtPhoto ♦
    Mar 12, 2015 at 12:21
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    $\begingroup$ Arthur C. Clarke specifically declined patenting the Clakre orbit (aka geosynchronous) (link). And an interesting recent post on patenting orbits: How Satellite Companies Patent Their Orbits as some food for thought. $\endgroup$
    – user5892
    Mar 12, 2015 at 16:47
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Five. 3 in polar orbits at 120º intervals, and two above the equator at opposite longitudes, at points perpendicular to the plane in which the polar satellites orbit. You'd have to synchronize the precession of the polar satellites with the equatorial satellites.
Two will get you close, but not quite there: you'll miss a tiny section where (viewed from Earth) the satellite is too low above the horizon to see the surface.
Here's a calculator which shows what fraction of the Earth's surface is visible at each altitude. Even at 1.5 million km, coverage is below 50%.
See also this related question.

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  • $\begingroup$ Since the two equatorial satellites will be circling the Earth, they will occasionally line up with the orbital plane of the polar satellites, so wouldn't you need 3 or even 4 in the equatorial plane to ensure 100% coverage at all times? $\endgroup$
    – Nickolai
    Mar 11, 2015 at 19:25
  • $\begingroup$ The patent in @PearsonArtPhoto's answer covers the 5 and 6 satellite scenarios as well. Take a look at the paragraph that starts with "In the late 1960's, it was thought that six satellites were required to provide continuous global coverage." For a patent it's actually quite readable if you are familiar with basic orbital mechanics. There's not a lot of legalese. $\endgroup$
    – Nickolai
    Mar 11, 2015 at 22:16
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You just need two of them, placed in the same orbit, but at different (opposite) positions. Wether they move relative to the surface or not (see Geosynchronous orbit is irrelevant because they will move at the same speed, thus being always at the exact opposite to each other.

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    $\begingroup$ You can't see an entire half of the Earth from any finite distance, so this doesn't work. $\endgroup$
    – Mark Adler
    Mar 11, 2015 at 18:15
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    $\begingroup$ To add to what @MarkAdler wrote, suppose we had one spacecraft in a pseudo orbit about the Sun-Earth L1 point, another in a pseudo orbit about the Sun-Earth L2 point. Both those points are about 1.5 million kilometers from Earth and are diametrically opposed to one another. The question asks for 100% coverage. At best those two satellites will provide 99.6% coverage, and only when they happen to be diametrically opposed. (Most of the time, they won't be.) What if they orbit further from the Earth than that? They can't. That's the furthest an object can "orbit" the Earth. $\endgroup$
    – David Hammen
    Mar 11, 2015 at 21:55
  • $\begingroup$ David, is it easy to get good image of surface from 1.5 (+-0.15) million kilometers away? What is diffraction limit for this distance (for 1 meter mirror telescope, what will be the best pixel size in surface meters; from 800 km there are satellites with 0.5 m per pixel) $\endgroup$
    – osgx
    Mar 12, 2015 at 2:50
  • $\begingroup$ @osgx That does not sound like a clarification or request for improvement of this answer, and thus should not be a comment. It does however sound like a decent question for perhaps Astronomy. $\endgroup$
    – user
    Mar 12, 2015 at 15:33

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