# Ecliptic reference frame for satellite orbits in JPL Horizons

I'm trying to get orbital elements for natural satellites of all the plates with respect to their system's barycenters and I don't see how it is possible that using the ecliptic as a reference plane always produces a viable result.

Here is an example of an input:

Notice below how Pluto's orbit goes way off the ecliptic (considering heliocentric ecliptic here but the argument also works for geocentric ecliptic). How is it possible then to obtain Charon's orbital elements wrt to the ecliptic if the entire Charon orbit ellipse never crosses the ecliptic plane? What does the longitude of the ascending node that Horizons returns even mean, if the orbit plane and the reference plane never cross?

Keep in mind that using the other option in the drop down for body mean equator doesn't really work since the reference body is just a barycenter point

Here is the output for the input above:

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Revised: Jan 26, 2018          Charon / (Pluto)                            901

Fit to all available observations including New Horizons encounter tracking.

SATELLITE GENERAL PHYSICAL PROPERTIES (PLU042, IAU2009, and derived values):
GM (km^3/s^2)           = 102.271      Density (g cm^-3)      = 1.65
Radius (km)             = 605          Geometric Albedo       = 0.372 +- .012
Mass (10^21 kg )        =   1.53       Hill Sphere radius, km = 664
Surface gravity (cm/s^2)=  27.9        Escape velocity, km/s  = 0.581

SATELLITE GENERAL ORBITAL DATA:
Semi-major axis, a (km) =  19596       Orbital period, days   = 6.38723
Eccentricity, e         =  0.0002
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Ephemeris / WWW_USER Thu Feb  8 12:42:39 2018 Pasadena, USA      / Horizons
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Target body name: Charon (901)                    {source: plu055l_merged}
Center body name: Pluto Barycenter (9)            {source: plu055l_merged}
Center-site name: BODY CENTER
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Start time      : A.D. 2000-Jan-01 12:00:00.0000 TDB
Stop  time      : A.D. 2000-Jan-01 12:01:00.0000 TDB
Step-size       : 1440 minutes
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Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}
Center radii    : (undefined)
Keplerian GM    : 6.9165565958516879E+02 km^3/s^2
Output units    : KM-S, deg, Julian Day Number (Tp)
Output type     : GEOMETRIC osculating elements
Output format   : 10
Reference frame : ICRF/J2000.0
Coordinate systm: Ecliptic and Mean Equinox of Reference Epoch
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JDTDB
EC    QR   IN
OM    W    Tp
N     MA   TA
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$$SOE 2451545.000000000 = A.D. 2000-Jan-01 12:00:00.0000 TDB EC= 2.072743604774027E-03 QR= 1.739208931057318E+04 IN= 1.128984926230046E+02 OM= 2.274012844469266E+02 W = 1.445907325672654E+02 Tp= 2451541.877754761837 N = 6.549200579117892E-04 MA= 1.766725371869484E+02 TA= 1.766862878182623E+02 A = 1.742821352870745E+04 AD= 1.746433774684171E+04 PR= 5.496854091594919E+05$$EOE
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Coordinate system description:

Ecliptic and Mean Equinox of Reference Epoch

Reference epoch: J2000.0
XY-plane: plane of the Earth's orbit at the reference epoch
Note: obliquity of 84381.448 arcseconds wrt ICRF equator (IAU76)
X-axis  : out along ascending node of instantaneous plane of the Earth's
orbit and the Earth's mean equator at the reference epoch
Z-axis  : perpendicular to the xy-plane in the directional (+ or -) sense
of Earth's north pole at the reference epoch.

Symbol meaning:

JDTDB    Julian Day Number, Barycentric Dynamical Time
EC     Eccentricity, e
QR     Periapsis distance, q (km)
IN     Inclination w.r.t XY-plane, i (degrees)
OM     Longitude of Ascending Node, OMEGA, (degrees)
W      Argument of Perifocus, w (degrees)
Tp     Time of periapsis (Julian Day Number)
N      Mean motion, n (degrees/sec)
MA     Mean anomaly, M (degrees)
TA     True anomaly, nu (degrees)
A      Semi-major axis, a (km)
AD     Apoapsis distance (km)
PR     Sidereal orbit period (sec)

Geometric states/elements have no aberrations applied.

Computations by ...
Solar System Dynamics Group, Horizons On-Line Ephemeris System
4800 Oak Grove Drive, Jet Propulsion Laboratory
Pasadena, CA  91109   USA
Information: http://ssd.jpl.nasa.gov/
Connect    : telnet://ssd.jpl.nasa.gov:6775  (via browser)
http://ssd.jpl.nasa.gov/?horizons
telnet ssd.jpl.nasa.gov 6775    (via command-line)
Author     : Jon.D.Giorgini@jpl.nasa.gov
*******************************************************************************

• I can't spend much time here right now, but "wrt ecliptic" is only related to the direction of the xyz axes used, not to which body. You'll get correct results using either Ecliptic or Earth Equator reference planes. In other words, if you extract positions for Charon using Pluto as the center, then calculate the distance from Charon to Pluto, you'll get the same distance no matter which plane they are referenced to. It's just that the axes are rotated by about 23 degrees from one to the other. Try posting some numerical results and explaining why you believe they are wrong or don't work... – uhoh Feb 8 '18 at 14:11
• ...and you'll start to realize that they are not wrong, and it does work. If not, then post those numerical results that you believe to be wrong in your question explicitly, rather than just say "I don't think it works" or "I can't get what I need". Horizons works extremely well. You can also review other questions here with the Horizons tag. Sorry I can't post more right now. – uhoh Feb 8 '18 at 14:13
• If there is any way you can avoid using the osculating orbital elements, and work with the more fundamental xyz data as state vectors, it will make a lot more sense, at least it would to me. – uhoh Feb 8 '18 at 14:20
• So basically what you're saying is that "Ecliptic and mean equinox" just dictate the plane/axes directions of the coordinate system, but the center point of that coordinate system is always set to the reference center (the Pluto system barycenter in my case)? I see, I'll give it a shot. I just decided to ask since I don't think I would be able to tell if the orbits were all off because I considered a wrong center. Also, I mean to draw the orbit ellipses, so the orbital elements get me a bit closer than using state vectors – Daniel Feb 8 '18 at 14:39
• Yep, that's what I'm saying. I think it will be OK. If it works out, you can consider posting an answer yourself, and even accepting it. That way if future readers find your question, they'll immediately see that there is an answer posted and it has been accepted. You also will get past 50 reputation points quicker, which allows you to leave comments elsewhere. – uhoh Feb 8 '18 at 14:45