I have come across a problem that has had me stumped for a while now.

A spacecraft is in a Sun-synchronous Earth orbit with a =1.40 R⊕ and e = 0.2. The argument of periapsis is ω = 0◦ , and the Right Ascension of the Ascending Node, Ω, is equal to the Sun’s Right Ascension plus 12 hours. Calculate the length of time per orbit for which the spacecraft is in Earth’s shadow, assuming that the Earth is at an equinox.

I'm not fully sure how to interpret this, as everywhere I read, Sun-synchronous orbits are always sunlit?

Any help or a point in the right direction would be much appreciated.

  • $\begingroup$ How sun-synchronous orbits really work is not a trivial thing to understand, or even explain, especially because of the substantial inclination of the Earth's axis relative to its orbit around the Sun. But I'm sure that nowhere reasonable have you read that "...Sun-synchronous orbits are always sunlit" though they are possible. If you can find a link, or cite a paragraph with block-quotes describing what it is that you've read, maybe we can figure out the disconnect there. A high orbit that follows the terminator (dusk/dawn) can in principle remain in Sunlight essentially all of the time. $\endgroup$
    – uhoh
    Commented Jan 7, 2018 at 2:47
  • $\begingroup$ There may be a very rare eclipse of the Sun by the Moon, but those can be predicted and the orbit planned to avoid them. Take a look here for more info... en.wikipedia.org/wiki/Sun-synchronous_orbit#Technical_details Also check out this video youtu.be/4K5FyNbV0nA It is possible that this simulator (where you can enter your data) can "shed some light" on the problem, but I am not sure how to "turn on" the Sun and visualize daylight with it yet. orbitalmechanics.info $\endgroup$
    – uhoh
    Commented Jan 7, 2018 at 2:51
  • $\begingroup$ Also, Quora; quora.com/What-is-a-Sun-synchronous-orbit and also quora.com/… where a beta-angle of about 90 degrees would put the orbit near the terminator and therefore more likely to remain in sunlight. Also this; wmo.int/pages/prog/sat/globalplanning_en.php $\endgroup$
    – uhoh
    Commented Jan 7, 2018 at 2:58
  • $\begingroup$ "the Right Ascension of the Ascending Node, Ω, is equal to the Sun’s Right Ascension plus 12 hours." This means that when the satellite crosses the equator going North, it is midnight local time. $\endgroup$
    – Ghedipunk
    Commented Dec 27, 2019 at 16:34

3 Answers 3


Sun-synchronous orbits are inclined by a few degrees away from going directly over the poles.

Because of these, they can go “behind” the Earth at either the north or South Pole, going into shadow. (At equinox has the easiest geometry for this, because the terminator is also aligned with the poles)

The first step for this problem would be to work out the exact inclination needed by the given orbit. The Wikipedia article explains how to do this.

Then take that inclined orbit as an ellipse, project it onto the plane perpendicular to the Sum, and see where it goes behind the Earth.


as everywhere I read, Sun-synchronous orbits are always sunlit?

No. What you probably read is that the only orbit that can be always sunlit is a sun-synchronous orbit. A sun-synchronous orbit has an inclination from Earth's rotational axis, generally about eight degrees but the precise amount depends on the altitude of the orbit. Because of Earth's equatorial bulge, the orbit precesses at about one degree per day, so it is always over the same local time in the same point of its orbit. If the orbit follows the terminator, that local time happens to be dawn / dusk, and so the satellite is never in the Earth's shadow. But if that local time happens to be noon/midnight, or most any other time, then it will of course be eclipsed once per orbit.

It's possible that the moon may eclipse the sun while the satellite is in orbit. This can be planned for and avoided by properly choosing the orbit, or, I imagine, later adjusting it. Or it can just be ignored because it wouldn't last that long for a fast moving satellite and the batteries can carry it through.

Here are some very informative links provided in @uhoh's comments:

  • 1
    $\begingroup$ Argh, how can I not embed the youtube video but just provide a link?? $\endgroup$ Commented Dec 27, 2019 at 16:26
  • 2
    $\begingroup$ @RossPresser [link text here](http://example.com)... So in your case, "The video [Space: Prograde, Retrograde, and Sun-Synchronous Orbits](https://youtu.be/4K5FyNbV0nA) explains and visualizes many kinds..." $\endgroup$
    – Ghedipunk
    Commented Dec 27, 2019 at 16:36
  • 1
    $\begingroup$ @Ghedipunk Thanks! $\endgroup$ Commented Dec 27, 2019 at 17:41

A sun-synchronous orbit is not an orbit that is always lit.

Always being in sunlight is one potential benefit of a sun-synchronous orbit. These types of sun-synchronous orbits are often used with satellites that use power constantly, such as radar and lidar surveys.

The main benefit of a sun-synchronous orbit, though, is that at the same period of your orbit, for every orbit, the sun will be in the same direction.

If you're passively imaging the Earth, like the Landsat satellites do, you want shadows to be as shallow as possible, so you will want to have the sun directly above you each and every pass. You'll want as much of your sun-synchronous orbit to go over areas during local Noon as possible... which means that roughly half of your orbit will also be over an area that is in local midnight, as well.

This homework question packs in a lot of jargon into a short paragraph. As this question is over a year old, so won't help with anyone's homework tonight, but would help people understand orbits in general, let's break it apart:

a = 1.40 R⊕ and e = 0.2. The argument of periapsis is ω = 0◦, and the Right Ascension of the Ascending Node, Ω, is equal to the Sun’s Right Ascension plus 12 hours.

Altitude = 1.4 * Earth's radius. (radius = 8,904 km; altitude above sea level = 2,544 km)
Eccentricity = 0.2 (significant)
Argument of periapsis (how close the periapsis is to the Ascending Node) = 0°
Right Ascension of the Ascending Node = Sun's Right Ascension + 12 hours.

From the average altitude, we can tell the orbital period: 2h19m47s.

With an eccentricity of 0.2, periapsis will have a radius at 7,123 km (altitude: 763 km), and apoapsis will have a radius at 10,684 km (altitude: 4,324 km).

With an Argument of Periapsis at 0°, our periapsis will occur right at the ascending node. That is, periapsis is right at the point in our orbit where we cross the equator while going north.

The Right Ascension is a location relative to "the celestial sphere." Specifically, how far East we are of the westernmost point of the constellation Aries. Right Ascension is measured in time, with the celestial sphere separated into 24 hours, which are divided into 60 minutes, which are divided into 60 seconds. 6 hours = 90°... 12 hours = 180°, etc.

Fortunately for us, our orbit is a sun-synchronous orbit. The Right Ascension of any point in our orbit will always have the same offset relative to the sun (plus or minus a little bit of wiggle room, since the Earth's gravity field is a lumpy oblate spheroid (continents are "heavier" than the ocean floor)).

The Right Ascension of our Ascending Node is what we care about the most: Since our Right Ascension is measured relative to the Sun, we can know the local time below our craft, just by looking at our offset. In our case, our offset is 12 hours, so our Ascending Node will always be at local midnight (since, by definition, the sun is always located at local noon). We will be passing through the widest part of Earth's shadow.

Our speed at periapsis is going to be 8,194 m/s, and at apoapsis will be 5,463 m/s.

The circumference of our orbit will be 55,917 km.

Since we're not actually launching a satellite, we aren't too worried about how big our batteries will need to be, or how much light our solar panels will need to gather. Our satellite is getting the same treatment as those pesky spherical cows that graze on frictionless, infinite planes. We'll handwave away the penumbra and the scattering of light from the atmosphere, and say that we have a binary condition for in light or in shadow, and that condition happens right at the center of the penumbra. (That is, right when the limb of the Earth covers the centerpoint of the sun.)

With the Earth's radius at the poles being 6,356 km, we have a shadow 12,712 km wide. Since we're traveling along a curved path, we need to know how much distance we're traveling through that 12,712 km wide shadow... Our adjacent leg of our right triangle is the radius at periapsis: 7123 km. Our opposite leg is the radius of the Earth at its poles: 6356 km... tan(7123/6356) = 45.3°.

So, about 90° of our orbit, about 1/4th of our 55,917 km orbit ... about 14,000 km will be spent in shadow.

Since that time is also during our periapsis, we're traveling at our fastest, about 1.5 times faster than at apoapsis.

Finding the exact speed along the entire curve will take some calculus. As I'm a software developer who got tricked into going to a diploma mill "for profit higher education institution", and has found a gap in my education that I have not yet filled on my own, I will leave that as an exercise for the reader...

However, a back-of-the-envelope estimate would put us at spending 1/6th of our orbital period in shadow. With our total orbital period of 2h19m47s, that means we spend about 23 minutes in shadow each orbit.

Our hypothetical satellite is probably a passive earth-observing mission. Perhaps with some DOD funding behind it.


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