You have a planet that's a perfect sphere.

[ Editor: the postings give arrangements with six, five and (perhaps)
four satellites; no proof of the minimum was found in them. ]

Andrew C. Plotkin <ap1i+@andrew.cmu.edu> wrote:
>I can arrange six orbiting satellites to do it. Put three in an
>equatorial orbit (equally spaced), far enough away that they constantly
>watch all the ground between 45 north and 45 south latitude. Then put
>another three in (the same) polar orbit (also equally spaced) at the
>same distance. That does it. (You might want to move the second trio a

[ Editor: the rest of Plotkin's posting is not in the collected
material. ]

I haven't figured out any way with 4 satellites, nor a proof of its
impossibility, but here's a way to do it with 5 satellites:

Position 4 satellites equally spaced 90 degrees away from each other in
a far away equatorial orbit.  Call them A, B, C, and D, so that A is
opposite C and B is opposite D.  Call the period of their orbits a
"day", divided into 24 hours in the obvious way.  Make the orbit radius
far away enough such that they have a 178.9999 degree field of view.

Modify the orbits (rotate their axes by 1 degree) such that the
satellites oscillate north and south of the equator.  Note that this
oscillation will naturally have a period of a day.  Note that no matter
how these modifications occur, the planet's surface between the 88
degrees S and 88 degrees N latitude will always be covered.  Call the
surface south of 88 degrees S and north of 88 degrees N the polar caps.

Make it such that A and C are northmost at 12 AM, and such that B and D
are northmost at 6 AM.

A and C will cover the north polar cap for more than 7 PM until 5 AM.
B and D will cover the north polar cap for more than 1 AM until 11 AM.
Thus, the north polar cap is covered for more than 7 PM until 11 AM.

A and C will cover the south polar cap for more than 7 AM until 5 PM.
B and D will cover the south polar cap for more than 1 PM until 11 PM.
Thus, the south polar cap is covered for more than 7 AM until 11 PM.

The only gaps left are the north polar cap from 11 AM until 7 PM and
the south polar cap from 11 PM until 7 AM.  This is covered by a fifth
satellite, E.  This satellite is at the same orbital radius as the
others, but is in a polar orbit.  It is northmost at 3 PM.  It covers
the north polar cap for more than 10 AM until 8 PM.  It covers the
south polar cap for more than 10 PM until 8 AM.

These 5 satellites will cover the whole surface all the time.

-- Isaac Kuo <isaackuo@math.berkeley.edu>

In article <1994Aug17.025333.11215@cs.cornell.edu>, karr@cs.cornell.edu
(David Karr) writes:
|> I'd really like to see a proof that total continuous coverage *can't*
|> be accomplished with fewer than 5 satellites, though.  None of us
|> seems to have come up with such a proof yet.

I think it might be possible to accomplish it with 4:

Let all the satellites be distant enough to see a near half-sphere.
All satellites have same period in nearly circular, nearly equatorial
orbits, with:
A farthest away at 0 EW, at a N latitude
B closest at 0 EW at a S latitude
C farthest away at 90 E at a S latitude
D closest at 90 E at a N latitude

A and B is approximately opposite C and D, as in this polar view
(+ means northern latitude, - means southern, no sign is approx. 0 lat):

view:   1                   2                   3                   4

        A+
      . . .               . . .               .C D.               . . .
   .    B-   .         .         .         .         .         .         .
 .             .     .             .     .             .     .             .
.               .   .               A   .               .   A               .
.       o       .  C-D+     o       .   .       o       .   .       o      C+D-
.               .   .               B   .               .   B               .
 .             .     .             .     .             .     .             .
   .         .         .         .         .    A-   .         .         .
      .C D.               . . .               . . .               . . .
                                                B+

(Due to the elliptical orbit B will be ahead of A when they cross the
equator in view 2, etc.)

Because the orbits are far away we treat the earth as a point o, and
the entire surface is visible if this point is within the irregular
tetrahedron defined by ABCD, this is obviously the case in all the four
snapshots above, in between these we have the situation like this
(symmetrical for other in-between situations):

      . . .
   .         .A+
 .             B-
.               .
.       o       .
.               .
 C-             .
  D+         .
      . . .

Line AD is obviously 'above' o, and BC obviously below o, now if ABCD
can be shown to always enclose o in this projection then we're done.

Let A' be the A mirrored around o into C's orbit in the projection.
Initially at view 1 C will be ahead of A', and as A is at its slowest
orbital speed, C will gain on A' to start with.  Later, A's speed
increases and C's speed increases, but due to symmetry (both A and C
are in the slow far half of their orbits) they will travel the same
distance between view1 and view 2.  Thus AC will stay to the
left/above of o throughout this quarterperiod.

Similarly let B' be the mirror image of B.  Initially B' is ahead of D,
and as B's speed is at its greatest it will gain on D to start with.
Similar symmetry applies, and line BD will stay to the right/below o
throughout.

Thus ABCD encloses o at all times, and thus these orbits should allow
the entire planet to be seen at all times.

I'll let someone else figure out exactly how far out these orbits must
be, and if they are physically realisable.

(... or, perhaps more likely, I'll let someone else figure out just
where I goofed :-)

-- Stein Kulseth <stein@hal.nta.no>