<p>I found a convenient formula to determine Eclipse (satellite in shadow). From Ismail, et al., Equation 10 <a href="https://www.sciencedirect.com/science/article/pii/S2090997715000206" rel="nofollow noreferrer">here</a>.</p>

<p>Image from same article:</p>

<p><a href="https://i.sstatic.net/x97EB.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/x97EB.jpg" alt="Theta-E is the angle of eclipse"></a></p>

<p>So time in shadow = ((Theta-E)/360)*Orbital Period.</p>

<p>But how do you calculate Theta-E? Well, for my part, I used the law of similar triangles. You will have two similar isosceles triangles - one from the Earth to the point of the umbra cone and one from the Sun to the point of the umbra cone. Cut the isosceles triangles in half and use the right triangles. The sides of similar triangles are proportional and using the distance of 1 AU and the radii of the Sun and Earth, you can find the distance to the point of the umbra cone with a little algebra. Then use the arc-tangent to find the half-angle to the distance at GEO, then doubled it to find Theta-E.</p>

<p>You should be aware, however, that you are dealing with (1) circular orbits with (2) 0 inclination. The time of year also can change things.</p>

<p>My result was very close to that of James K. with only a few seconds difference.</p>

CC BY-SA 4.0
2019-04-14T11:24:53.713Z
2019-04-14T11:24:53.713Z
Friejek
<p>At the equinox this is quite easy to calculate, at least approximately.</p>

<p>A geostationary orbit is 265000km long, but the Earth's shadow is only 12700km wide (since the Earth-Sun distance is much larger than the orbit). </p>

<p>At an equinox, the sun is in the plane of the equator, and since GEO satellites orbit on the equator, the satellite passes through the middle of the shadow. So it spends 12700/265000 of its orbit in shadow. Each orbit takes 1 day, so that fraction of one day is about 1 hour 10 minutes.</p>

<p>At a solstice, the shadow deviates by 23.5 degrees = 0.41 radians from the equator. This means that at solstice the centre of the shadow will pass about 0.41\times 36000km =15000km from the satellite, and the satellite won't pass through the shadow at all.</p>

<p>At dates in between the calculation is harder. Approximations can be calculated with trigonometry if required.</p>

CC BY-SA 4.0
2018-08-15T19:43:11.847Z
2018-08-15T19:43:11.847Z
James K
<p>If a satellite is revolving in a GEO orbit, then what will be the eclipse duration for it? How to calculate when the satellite will enter the dark side of the Earth? </p>

<p>Any formula or online tool would be appreciated.</p>

<earth><solar-eclipse><satellite>
Calculate eclipse days and duration of satellite
iLLkeeN Nemo
# Calculate eclipse days and duration of satellite

(Asked by: username_0 on 2018-07-26)

If a satellite is revolving in a GEO orbit, then what will be the eclipse duration for it? How to calculate when the satellite will enter the dark side of the Earth?

Any formula or online tool would be appreciated.

**Tags:** earth, solar-eclipse, satellite


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## 2 answers

### Answer by username_1 on 2019-04-14. Score: 3

I found a convenient formula to determine Eclipse (satellite in shadow). From Ismail, et al., Equation 10 [here](https://www.sciencedirect.com/science/article/pii/S2090997715000206).

Image from same article:

[![Theta-E is the angle of eclipse](https://i.sstatic.net/x97EB.jpg)](https://i.sstatic.net/x97EB.jpg)

So time in shadow = ((Theta-E)/360)\*Orbital Period.

But how do you calculate Theta-E? Well, for my part, I used the law of similar triangles. You will have two similar isosceles triangles - one from the Earth to the point of the umbra cone and one from the Sun to the point of the umbra cone. Cut the isosceles triangles in half and use the right triangles. The sides of similar triangles are proportional and using the distance of 1 AU and the radii of the Sun and Earth, you can find the distance to the point of the umbra cone with a little algebra. Then use the arc-tangent to find the half-angle to the distance at GEO, then doubled it to find Theta-E.

You should be aware, however, that you are dealing with (1) circular orbits with (2) 0 inclination. The time of year also can change things.

My result was very close to that of username_2. with only a few seconds difference.


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### Answer by username_2 on 2018-08-15. Score: 2

At the equinox this is quite easy to calculate, at least approximately.

A geostationary orbit is 265000km long, but the Earth's shadow is only 12700km wide (since the Earth-Sun distance is much larger than the orbit).

At an equinox, the sun is in the plane of the equator, and since GEO satellites orbit on the equator, the satellite passes through the middle of the shadow. So it spends 12700/265000 of its orbit in shadow. Each orbit takes 1 day, so that fraction of one day is about 1 hour 10 minutes.

At a solstice, the shadow deviates by 23.5 degrees = 0.41 radians from the equator. This means that at solstice the centre of the shadow will pass about 0.41\times 36000km =15000km from the satellite, and the satellite won't pass through the shadow at all.

At dates in between the calculation is harder. Approximations can be calculated with trigonometry if required.


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