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In this answer @DavidHammen showed me how to get coordinates from JPL's Horizons using the Earth's equatorial plane as a reference plane by selecting

reference plane: Earth mean equator and equinox of reference epoch
reference system: ICRF/J2000.0

What I would like to do is take positions in that frame from Horizons that I have in a python script and quickly make an approximate projection of the location of one item on the celestial sphere as seen from another. I know this is an option from within Horizons, but I want to do this using existing data, and I don't need the absolute highest accuracy.

Question: If I have a vector x, y, z that is drawn from the observer to the observed, can I get RA and Dec to say a few arcminutes of accuracy with the simple equations:

$$RA = \arctan2(y, x) $$

$$Dec = \arctan2(z, (x^2 + y^2)^{1/2}) $$

Implicit in this is the idea that the $\hat{\mathbf{x}}$ axis points in the direction of $RA, Dec = 0, 0$, and I'm ignoring light time issues, aberration, and motion of the Earth's axis since J2000.0

UPDATE:

I've given this a try and there is a small difference between the calculated RA/Dec based on subtraction of state vectors, and the astrometric position seen from the Geocenter. But there is a huge difference between those and the apparent positions, tenths of a degree!

This is much larger than I expected, and I don't know why!

All data is for January 1, 2018, 00:00 UTC.

I've compared the results of applying those equations to the state vectors from Horizons, then also displayed astrometric (ICRF J2000.0) and apparent positions as viewed from the Geocenter taken from Horizons in Observer Mode.

I've chosen bodies as close as the Moon and as far as the Voyagers, and used Body centers rather than barycenters.

       astrometric  Equation  apparent     astrometric  Equation   apparent
         --------- --------- ---------      ---------  ----------  ---------
Voygr 1  258.25159 258.25184 258.45314        11.97035  11.97037  11.95155 
Voygr 2  300.65126 300.6518  300.99952       -57.29555 -57.29610 -57.24509 
Nw Horiz 286.63319 286.63353 286.89020       -20.64794 -20.6479  -20.61809 
Pluto    290.00861 290.00968 290.26684       -21.70644 -21.70662 -21.67124 
Neptune  343.43280 343.43376 343.66156        -8.03381  -8.03345  -7.93949 
Uranus    22.73512  22.73629  22.97047         8.8979    8.89837   8.98873 
Saturn   271.24249 271.24428 271.50497       -22.53535 -22.5354  -22.53099 
Jupiter  224.53584 224.53805 224.78021       -15.8163  -15.81697 -15.88433 
Mars     221.74196 221.74522 221.98477       -15.15716 -15.15828 -15.22858 
Venus    279.04685 279.05300 279.31079       -23.64835 -23.64834 -23.63092 
Sun      281.16631 281.16543 281.42877       -23.04004 -23.04010 -23.01910
Moon      84.2338  84.22705   84.50195        19.31174  19.31069  19.31952 

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  • $\begingroup$ @DavidHammen any thoughts on this? If you can at least send me looking in the right direction, that would be great! $\endgroup$ – uhoh May 29 '18 at 8:16
  • $\begingroup$ The celestial sphere has a radius of 1 with no defined units. Real objects will be at varying distances from each other. That might be the issue here. $\endgroup$ – barrycarter May 29 '18 at 18:40
  • $\begingroup$ @barrycarter I don't see how. I'm comparing three methods that all express the direction of vectors from the Geocenter to single objects. $\endgroup$ – uhoh May 29 '18 at 22:39
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    $\begingroup$ OK, I got it. It is precession. Roughly speaking, points on the equator precess through 360 degrees over 24000 years. That works out to 0.015 degrees per year or 0.27 degrees since 2000. For other declinations, the precession is something like 0.27*cos(dec) degrees. $\endgroup$ – barrycarter Jul 10 '18 at 17:53
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    $\begingroup$ Feel free to post my answer as your own (you can "verify" it by looking at Stellarium's difference between "RA/DE(J2000)" and "RA/DE(of date)" for a star like Regulus). The apparent and astrometric difference is explained by the * next to the "Apparent RA & Dec". On ssd.jpl.nasa.gov/horizons.cgi?s_tset=1#top it notes * affected by optional atmospheric refraction setting (below) $\endgroup$ – barrycarter Jul 11 '18 at 0:32

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