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My task is to calculate current declination and right ascension for some artificial Earth satellite. For example ISS.

I am using latest TLE data for extracting keplerian orbital elements and then calculate current cartesian orbital state vectors in ECI frame. From the position vector I directly calculate declination and right ascension, using cartesian to spherical coordinates conversion.

Right ascension gets calculated correctly but I am having problems getting the same result for declination as the one displayed by satellite trackers.

Speaking of satellite trackers, I don't understand how is their declination value even correct.

Declination is an angle measured north or south of the celestial equator. declination visually

Celestial equator of course is on the same plane as the equator of Earth.

Then how is it possible that, when a satellite (ISS in this example) is almost directly above Earth's equator, satellite trackers calculate declination to be around -29° ?

tracker 1 tracker 2 tracker 3

Shouldn't the declination angle be zero? My calculations for this "ISS above equator" example did give declination to be close to 0°.

Who is doing the calculations wrong and why?

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You are forgetting parallax.

Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator.

From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear to be in the southern sky, and your declination coordinate will likely be south of 0°.

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  • $\begingroup$ Thank you, you helped me finally pinpoint the essence of the problem. However, I knew there would have to be some corrections for locating the satellite in observer's frame (altitude and azimuth coordinates) because of the fact that satellite isn't a very distant object, but I thought that right ascension and declination were in earth-centered-inertial frame, spherical equivalent of cartesian position vector, independent of the observer's location. "Declination in astronomy is comparable to geographic latitude, projected onto the celestial sphere" en.wikipedia.org/wiki/Declination $\endgroup$
    – Vekszor
    Jan 24, 2018 at 17:04
  • $\begingroup$ The software routines by D. Vallado may come in useful to cross-check the parallax correction described by @Mick. celestrak.com/software/vallado-sw.asp $\endgroup$
    – Ela
    Jan 24, 2018 at 17:34
  • $\begingroup$ @Vekszor For distant objects that is certainly true but on Solar System scales corrections do need to be made for the observer location, and for Earth-orbit position calculations even more so. $\endgroup$
    – Mick
    Jan 25, 2018 at 0:48

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