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I'm calculating the azimuth and elevation in a ground point to see some satellites, but if for example the error in orbit for a satellite grows 1 km how is that distance (error) represented in radians ( for azimuth/elevation)?

UPDATE 1: For example I'm calculating the position of GPS satellites, that constellation are to 20200 kms of altitude [ECEF]. If the perpendicular error is around 1km in the orbit, how many is the error represented in radians, when I'm seeing azimuth/elevation from an observer?

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  • $\begingroup$ There may not be enough information in your question. Can you add some more details about how you are doing your calculation? Either show the equations you are using, or at least a link to them? Are you looking for a rough guideline? For example, using a small angle approximation, if the satellite is 800 km away from you, and it's position changes by 1 km perpendicular to the direction you are viewing it, the error will be about 1/800 radians, or about 0.07 degrees. $\endgroup$ – uhoh Mar 27 '17 at 2:33
  • $\begingroup$ Thank you uhoh, I just to updated the question; If I understand you: GPS satellites distance is to 20200 Kms, so, If the error in orbit is 1Km then, the error in Radians is 1/20200, right? $\endgroup$ – mikesneider Mar 27 '17 at 13:32
  • $\begingroup$ That's right - in radians. Multiply by $180/\pi$ to get degrees. Altitude = distance only if you are right under the satellite. Usually they are a bit farther, but as long as they are well above the horizon, it's only about a 20 or 30% difference. Next time, put "@" in front of the name of the person you'd like to see your comment. I don't have to put yours because these comments are under your question, so you see a flag automatically. $\endgroup$ – uhoh Mar 27 '17 at 13:44
  • $\begingroup$ I can give your question a +1 since you've made some effort to improve your question. Once you edit, whoever gave you a -1 is now able to reverse their down vote if they happen to notice. Don't worry too much about a few votes, hopefully you'll ask more questions :) $\endgroup$ – uhoh Mar 27 '17 at 13:45
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Firstly, this depends on the angle between the orbit and the reference plane (horizon). For instance, a perpendicular error in an orbit perpendicular to the horizon will result in a almost purely azimuthal error, unless close to zenith. In contrast, an orbit with a low maximal elevation will have almost only elevation errors.

Secondly, it is more convenient to use an angle instead of a distance to represent the error as that makes the calculation independent of altitude.

Looking at the problem now, we can see that using the initial orbit as the reference plane, the error is just adding elevation. The problem is now reduced to a simple transform between two spherical coordinate systems. To keep this transform as simple as possible you should use the ascending node of the orbital plane as your reference direction for azimuth in both systems. Getting a final azimuth value can then be done as a final rotational transfer along the polar axis, meaning you are back to your initial system by subtracting the longitude of the ascending node.

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