I used to know how to do this, but for the life of me I can't remember the source of the equations. This is clearly a question of geometry rather that any specific space science however I still feel this is the correct place to ask the question. For more detail:

Orbit Altitude: 500km
Inclination: 0
Eccentricity: 0
Orbital Period: 5677s
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    $\begingroup$ 5677 seconds. No wonder Hannelore got sick of having to listen to Zarathustra for each and every sunrise :-) $\endgroup$ Sep 1, 2015 at 16:36
  • $\begingroup$ @JyrkiLahtonen I had to listen to your link to Richard Strauss - Also sprach Zarathustra, Op. 30 before I got it. ;-) $\endgroup$
    – uhoh
    Mar 25, 2018 at 13:04

3 Answers 3


enter image description here

In this pic the angle $\phi$ is asin(6378/6878) or about 68º. It stays in the shade over 2 * 68º or about 136º. That's about 38% of the period. That's 2145 seconds or about 36 minutes.

This is what the orbit would be during the fall or spring equinox. Other times of the year it could be shorter.

In this pic I assume the sun's ray's are parallel. A more accurate drawing would have the shadow boundary about half a degree from horizontal, but it gives an approximation.

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    $\begingroup$ Can it actually be longer? It seem to me like the case shown here, where the orbit passes directly through the center of the shadow, should maximize the time spent in shadow (at least assuming a circular orbit and a cylindrical shadow, like here). $\endgroup$ Sep 1, 2015 at 15:23
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    $\begingroup$ @IlmariKaronen I think this is assuming an orbit parallel to the plane of the solar system. In a polar orbit, for instance, I think it may be possible that it would vary over the time of the year. $\endgroup$
    – Dan
    Sep 1, 2015 at 17:02
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    $\begingroup$ @Dan: Of course it may vary (all the way down to zero, for some orbits), but I'm pretty sure that the time in shadow calculated by David above is actually the maximum for the given orbital radius. $\endgroup$ Sep 1, 2015 at 21:10
  • $\begingroup$ @IlmariKaronen Assuming an orbit in the plane of the solar system, I think you're right, this is maximum nighttime. Which is probably the most useful figure (compared to average or minimum) for design considerations. $\endgroup$
    – Dan
    Sep 1, 2015 at 21:17
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    $\begingroup$ @Dan: Even for inclined (circular) orbits, it's still the maximum. In fact, for circular orbits in the same plane as the Earth's orbit around the sun, it's also the minimum. $\endgroup$ Sep 1, 2015 at 21:22

It will vary dependent on the time of the year. According to STK, the exact time is about 2123 s. The time won't vary that much over the course of the year, I don't expect it would be much more than a 100 s variation. The low inclination and LEO altitude results in the low variation.


We can roughly approximate the answer by considering how wide a swath of shade the Earth casts, relative to the total length of the orbit.

At 500km altitude + 6371 km Earth radius (= 6871km), the satellite's circular orbit describes a circle $2 π r$ in circumference, or 43,172 km.

Assume rays of sunlight at Earth orbit to be essentially parallel, so the width of the shadow cast on the orbit is the width of Earth, 12,742 km.

So the sat will spend about 12742/43172 of its time in shadow, or 1676 seconds (27.93 minutes) on each orbit.

That's a severe underestimate, though. For more accurate computation you'd need to take into account that the shadow width is a chord across the orbit rather than a circular arc -- and since it's a large fraction of the orbit, that difference is pretty significant. (If the orbital altitude were much higher, the chord would be much closer to the arc, and this method would be more accurate.)

For that, you can do a little bit of trig; if c is the shadow width (Earth diameter) and r is the orbital radius (Earth radius plus orbital altitude), then the angular width of the shadowed arc is $$2 \arcsin \frac c {2r} $$ or 136 degrees; 0.378 of the circle, or 2145 seconds (35.74 minutes). See HopDavid's answer for the excellent diagram corresponding to this.

For still more accurate computation you'd need to take into account the sun's distance and radius, and make decision about umbra vs. penumbra.

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    $\begingroup$ 6371 km is the mean radius. As the inclination of the satellite's orbit is 0, the equatorial radius, 6378 km would be more accurate. $\endgroup$ Sep 1, 2015 at 10:47

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