The polar regions won't be missed as they will pass under the sun-synchronous orbit twice a year. – SF. Nov 19 '15 at 9:56

@SF. - Your comment is incorrect. Sun synchronous satellites never pass directly over the poles. The A-train satellites, for example, have an orbital inclination of 98.2 degrees, and hence never go north of 81.8 degrees north latitude. The poles are in view from the northernmost point of the orbit, but are never directly below. – David Hammen 4 hours ago

Could someone explain it like I was 5?

Earth axial tilt of 23.4° sends the poles 23.4° away from the terminator, into the night side, or the day side, at solstices. So, at summer solstice, the north pole is 23.4 degrees away from the terminator.

The "terminator-riding" satellites move at 80 to 82 degrees inclination, that means at the time of equinox, they are 10 to 12 degrees from the pole, which is exactly at the terminator line. At summer solstice, if the satellite is to pass over the terminator line over the equator, if it retains the 80 degrees inclination (relative to Earth equator), that would mean it's 20 degrees from the pole + 23.4 degrees between the pole and the terminator = 43 degrees away from the terminator... that's hardly "riding the terminator", almost halfway to perpendicular to it... (and near winter solstice, they actually ride exactly over the terminator line).

or am I missing something? Could someone, say, show a picture of such a satellite's orbit at summer solstice, in relation to north pole and the terminator line? Something like this but for summer solstice date?

  • $\begingroup$ It's a good question! Like I said I am sooooo confused. I think a tl;dr of an answer might be that as long as the equatorial bulge has certain symmetries, including that about the Earth's axis, the direction of the bulge's perturbation will be in the plane of the bulge (e.g. the equator). There are no such things as "terminator-riding orbits" The orbit responds to the shape of the bulge, and has no idea where the sun or the terminator are and doesn't care. A good thought experiment would be if the axis were tilted at 90 degrees to the ecliptic. $\endgroup$ – uhoh May 4 '17 at 7:06
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    $\begingroup$ No. It is not possible to explain this to a 5 year old. $\endgroup$ – gerrit May 4 '17 at 13:05
  • $\begingroup$ Part of the problem here is an ELI5 misconception of a "terminator-riding" satellite. Such satellites don't exist. It's a lie-to-children. $\endgroup$ – David Hammen May 4 '17 at 17:03

The satellite does not run parallel to the terminator, it crosses the terminator twice on each orbit. So on each orbit, half the path underneath the orbit is in shadow, the other half is in daylight.

The satellite's inclination is measured relative to the equator (98.2°) and poles (8.2°), and does not change with the seasons. The northernmost point of the orbit revolves around the pole.

  • $\begingroup$ Maybe I should rename my comment ts;dr ("too-short, didn't read") :) $\endgroup$ – uhoh May 4 '17 at 9:39

Could someone explain it like I was 5?

TL;DR: A true "terminator-riding" satellite doesn't exist. The fuel maintenance costs for such an orbit would be ridiculously high.

Sun synchronous satellite orbits are not quite Keplerian. The right ascension of ascending node is constant in a Keplerian orbit. The right ascension of ascending node instead precesses in sun synchronous orbits to match the Earth's orbit about the Sun.

This precession is a plane change. Plane changes in Keplerian orbits are rather expensive operations, delta-V wise. These continuous plane changes would make a sun synchronous orbit about a spherical planet be ridiculously expensive. Those continuous plane changes of a sun synchronous orbit about the non-spherical Earth come free of charge because the Earth's oblateness makes any inclined orbit precess. The rate at which an orbit precesses is a function of the orbit's inclination, semi-major axis length, and eccentricity.

Satellites in a sun synchronous have orbits with inclination, semi-major axis length, and eccentricity specifically designed to make this natural precession rate match the Earth's orbit about the Sun. Such a satellite with it's orbital plane more or less normal to the direction to the Sun (non-ELI5 term: large solar beta angle) and it will more or less keep that orientation constant. The satellite will more or less ride the terminator.

What about a true terminator-riding orbit? Such an orbit would require constant plane changes, both in inclination and in right ascension. The Earth's oblateness does not affect inclination. Those plane changes would be the responsibility of the satellite. Moreover, that constantly changing inclination would mean that the natural precession due to oblateness would almost never matches the precession needed to keep the satellite sun-synchronous. A true terminator riding satellite would also have to expend immense amounts of fuel for those plane changes. For these reasons, a satellite in a true terminator-riding orbit does not exist.

  • $\begingroup$ So... in that "more or less ride the terminator", in some cases "definitely less", as with my 43 degrees instance? The inclination relative to terminator changes over the year, with about zero near winter solstice and over 40 degrees in summer? $\endgroup$ – SF. May 4 '17 at 16:20

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