How long will a 500 km altitude satellite spend in Earth - shade per orbit

Question

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

5677 seconds. No wonder Hannelore got sick of having to listen to Zarathustra for each and every sunrise :-) — Jyrki Lahtonen, Sep 01 '15 at 16:36
@JyrkiLahtonen I had to listen to your link to Richard Strauss - Also sprach Zarathustra, Op. 30 before I got it. ;-) — uhoh, Mar 25 '18 at 13:04

HopDavid · Accepted Answer · 2015-09-01T22:59:42.750

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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.

edited Sep 01 '15 at 22:59

answered Aug 31 '15 at 23:11

HopDavid

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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). – Ilmari Karonen Sep 01 '15 at 15:23
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@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. – Dan Sep 01 '15 at 17:02
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@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. – Ilmari Karonen Sep 01 '15 at 21:10
@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. – Dan Sep 01 '15 at 21:17
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@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. – Ilmari Karonen Sep 01 '15 at 21:22
@IlmariKaronen I see your point, and agree. And as you indicated, for orbits in the solar system's plane, it would be constant. – Dan Sep 01 '15 at 21:26
@IlmariKaronen Thinking about it, I believe you're correct. The time I gave is maximum time in shadow. I've amended my answer. – HopDavid Sep 01 '15 at 23:01

PearsonArtPhoto · Answer 2 · 2018-03-25T10:11:10.127

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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.

edited Mar 25 '18 at 10:11

answered Aug 31 '15 at 22:50

PearsonArtPhoto

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There is no such thing as an "exact time" precise to 1 millisecond, especially if there is seasonal variation. I am sure STK will give you a variation much larger than a millisecond for various epochs within one sidereal day, just due to $J_2^2$ alone. Eclipse takes almost 10 seconds to actually happen as well. See How fast is the onset of periodic eclipse for a spacecraft in LEO? – uhoh Mar 25 '18 at 09:16
And so I've just asked How large would be the seasonal variation of eclipse duration for a spacecraft in LEO? – uhoh Mar 25 '18 at 09:54

Russell Borogove · Answer 3 · 2015-08-31T23:08:33.273

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.

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. — Peter Mortensen, Sep 01 '15 at 10:47

How long will a 500 km altitude satellite spend in Earth - shade per orbit

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