Wikipedia's Geostationary orbit; Communications says:

Geostationary communication satellites are useful because they are visible from a large area of the earth's surface, extending 81° away in both latitude and longitude.

But how to calculate that above 81° they are not visible? And also how do I know exactly at what height will I see a satellite in some city (ex. Moscow - 55.7558° N, 37.6173° E)?

  • $\begingroup$ At the end of the sentence you've quoted there is footnote 22 Determination of Look Angles to Geostationary Communications Satellites which contains a lot of math. It is probably more than you need to get started. Instead, you can search this site (using the search box) for "look angle" and try the search with and without "geostationary". $\endgroup$
    – uhoh
    Commented Sep 12, 2021 at 23:09
  • $\begingroup$ @AJN yes you are of course right. For some reason I was trying to apply to subsatellite points at arbitrary latitudes rather than on the equator. I've now performed some personal station keeping and coffee making corrective actions. $\endgroup$
    – uhoh
    Commented Sep 12, 2021 at 23:13

1 Answer 1


A diagram showing the cross section (longitudinal section) of the earth at the same longitude as the geostationary satellite is shown below.

cross section of the earth showing a geostationary satellite

A location from the the geo stationary satellite aligns with the horizon is marked. Its latitude is also marked in the picture. At higher latitudes, the satellite is below the horizon.

$$ cos(\mathrm{lat}) = \frac{6400}{36000+6400}\\ \mathrm{lat} \approx 81\deg $$

Instead if we imagine that the view in the picture is by an observer in space above north pole, then the diagram shows longitude and so the 81 deg is applicable to longitude also; but only at the equator since the cross section was made at the equator.

In fact, a tangent line drawn from the satellite to the earth's surface traces out a cone of half angle 90 - 81 = 9 deg.

  • 1
    $\begingroup$ Fig.1 in the link you provided seems to show the calculation for the look angles. $\endgroup$
    – AJN
    Commented Sep 12, 2021 at 17:29

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