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I was speaking with some friends of mine, one of whom was an aerospace engineer. He posited the infeasibility of a hypothetical "Margaritaville Satellite" that orbited earth in such a way that wherever it was above, it was six-o'clock (the idea being, the Margaritaville Satellite is in Happy Hour 24/7), matching Earth's rotation exactly.

However, there was disagreement on whether or not the satellite could actually be put in such an orbit, as if it were rotating at the exact speed of earth, it would effectively just be "falling" straight towards Earth.

Is it possible to sync a satellite's orbiting with the time of day of the earth without some sort of propulsion to keep it aloft?

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@Pearsonartphoto: What does "rotating at the exact speed of earth" mean here? Satellites do this all the time; that's what a geostationary orbit is. A "fixed time" orbit would have to orbit at the angular speed of the Sun, as seen from Earth (or equivalently, the angular speed of Earth in its orbit around the Sun), which could be achieved at an altitude of about 2.1x10^6 km. –  raxacoricofallapatorius Nov 18 '12 at 19:37
It's roughly equivalent to a jet moving east at the speed of the earth's rotation. Essentially, that jet would remain at the same solar local time until it landed. The eastward vector in this case remains the same, with the north/south component allowing it to orbit correctly. –  PearsonArtPhoto Nov 18 '12 at 20:18
@Pearsonartphoto: That's the jet's speed relative to the ground: westward at the speed of the Earth and slightly eastward at the speed of the sun across the sky. For an orbiting satellite, only the latter is involved. –  raxacoricofallapatorius Nov 18 '12 at 21:39
Way too tired today to be answering anything sensible... Good catch on my mixing east/west... But I still hold that the vector of the satellite would be the such that the cross product of the westward moving jet and the north vector equals orbital velocity. Essentially, as the satellite moves by the ground, the ground time is changing relative to the speed of the satellite. –  PearsonArtPhoto Nov 18 '12 at 21:42
@Pearsonartphoto: Yes, seen from the Earth, the satellite and the plane would achieve the same desired effect of appearing stationary with respect to the Sun. But the satellite's orbital velocity is not measured relative to a point on the revolving surface of the Earth. –  raxacoricofallapatorius Nov 18 '12 at 21:48
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3 Answers

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There is definitely satellites that achieve this. The type of orbit is called "Sun synchronous". It's impossible to always be 6:00 PM, but it can always be either 6:00 PM or 6:00 AM. The trick is that it's a slight retrograde polar orbit, that allows for this feat to be achieved.

It's very common for earth observing (And for that matter, Mars, Moon, Venus, and Mercury observing) satellites to follow this type of orbit, due to the illumination remaining consistent for all photos.

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But not quite simultaneous, the quickest (lowest) LEO is about 80mins, so it would take a half-hour from the north-south. Of course after a couple of Margarita's you wouldn't really care –  Martin Beckett Apr 13 '11 at 17:38
@Martin: Actually, it is, at least, given solar mean time. The trick is that the orbit shifts over half an hour for your half-hour movement on the ground. It's kind of wierd, but it does work. –  PearsonArtPhoto Apr 13 '11 at 18:45
yes, I had thought that since everywhere on the meridian had the same instantaneousness local sun time it didn't work. But of course the 'place' under the satelite also changes east-west. It was a liquid relativistic effect (ie. Too late+not enough coffee!) –  Martin Beckett Apr 13 '11 at 18:49
Note that whether this works depends on what "time" is being sought. If it's mean solar time, then the solution described here works (and inclining the orbit a bit to compensate for the sun's movement during an orbital pass is very clever). I'm not sure about apparent solar time. For that to work the plane of the satellite's orbit would need to precess (around an axis perpendicular to the plane of the ecliptic) at a varying rate through the year. Perhaps there's a way to achieve the same effect with an eccentric orbit that precesses around an axis perpendicular to the plane of that orbit. –  raxacoricofallapatorius Nov 19 '12 at 2:28
If it's clock time, it's impossible, of course since it is only 6:00 for an instant in each time zone. –  raxacoricofallapatorius Nov 19 '12 at 2:29
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Agreed with Ted Bunn's answer, that it is possible to keep a satellite over a fixed time point on earth by using the Lagrange points. One of those L-points is on the opposite side of the sun, which doesn't help much. There are two problems with the other four. One is that there are exactly four solutions, corresponding to exactly 4 local times on Earth: high noon; midnight; ROUGHLY 7am; and ROUGHLY 5pm. The other problem is that all such L-points are far, FAR outside of LEO or geosynchronous orbit.

Another possibility, which would take very little station-keeping fuel, would be to send the satellite into exactly the same orbit around the Sun as Earth. Thus, it would appear to an Earth-bound observer as if the satellite was always above the dividing point of day and night (6:00am or 6:00pm, depending on whether the satellite was leading or lagging earth in the shared orbit around the Sun).

The satellite would have to be much, much farther away than geosynchronous orbit in order to be able to fight the constant pull of Earth. It might be possible to place it far enough out, and close enough inside lunar orbit, to gain enough periodic lunar pull to aid in stationkeeping.

So, six solutions have surfaced, for stationkeeping over a fixed local Earth time with little or no stationkeeping fuel burn: ~7:00 am; 6:00 am; noon; ~5:00 pm; 6:00 pm; and midnight. Happily, one of those solutions coincides with "Margaritaville time."

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There's another possibility, aside from the sun-synchronous orbit mentioned by @Pearsonartphoto. If you want your satellite to maintain a constant position relative to both Earth and Sun, you can put it at one of the Earth-Sun Lagrange points. It's always noon at L1 and always midnight at L2. L4 is always directly above a point on Earth corresponding to some time in the morning, and L5 is always directly above a point corresponding to some time in the afternoon, although in the latter cases the points are so far away that it's a bit funny to identify those terrestrial times of day with times on board the satellite.

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