# Calculate eclipse days and duration of satellite

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.

• You could try using HORIZONS or CSPICE for this (see astronomy.stackexchange.com/questions/13488 for links), and I think it would depend on the satellite. Feel free to contact me directly (contact info in profile) if you'd like.
– user21
Jul 27 '18 at 17:38

I found a convenient formula to determine Eclipse (satellite in shadow). From Ismail, et al., Equation 10 here.

Image from same article: 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 James K. with only a few seconds difference.

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.