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According to the orbital plane XYZ, I understand that the axial tilt is the angle between the planet axis rotation and any normal of the plane (actually Z in right handed system). BUT, this is not sufficient to define a vector direction, I need another angle into the XY plane. Where do I find this information? Actually I believe it's relative to the direction of the periapsis. Am I right?

Following all the nice conversation on the answers, I maybe need to ask

How to get the axial tilt vector(x,y,z) relative to ecliptic or orbital plane, given ONLY ONE angle, i.e. 23.439281 deg for the Earth, 25.19 for Mars, etc.

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  • $\begingroup$ Does this post answer your question? $\endgroup$
    – called2voyage ♦
    Commented Sep 1, 2016 at 16:48
  • $\begingroup$ nope, still discussion from specialist and no one give simple definition.. one is speaking about the velocity vector, the other to another referential.. nothing clear for a newbiz like me. BTW many thank to taking the time to answer. $\endgroup$
    – Guillaume Pelletier
    Commented Sep 1, 2016 at 17:19
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    $\begingroup$ This is the answer: astronomy.stackexchange.com/a/8800/6 Unfortunately I'm not sure how to simply it from there. $\endgroup$
    – called2voyage ♦
    Commented Sep 1, 2016 at 17:20
  • $\begingroup$ When you say "pole" do you mean "How do they define north and south?" $\endgroup$
    – called2voyage ♦
    Commented Sep 1, 2016 at 17:45
  • $\begingroup$ no, i just want to obtain the vector of the axial tile relative to the ecliptic plane (z to the eccliptic pole) or relativ to the planet orbital plane, which must have the form of (alpha, theta) or (x,y,z). Actually, for the earth, it's seem a rotation X is doing the job. but for others planet with orbital plane different than the eccliptic i do not find any clear explanation anywhere.. $\endgroup$
    – Guillaume Pelletier
    Commented Sep 1, 2016 at 17:58

1 Answer 1

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Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit. Earth's north pole vector in equatorial coordinates is $$\vec N_{\oplus,eq} = (0, 0, 1)$$ To transform this to ecliptic coordinates, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get $$\vec N_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$ In a spherical coordinate system, two angles define a unique direction. In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$. This IAU report, table 1, gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years. To convert these to rectangular form $$(x_{eq}, y_{eq}, z_{eq}) = (\cos \alpha \cos \delta, \sin \alpha \cos \delta, \sin \delta)$$ Then if you want J2000 ecliptic coordinates $$(x_{ecl}, y_{ecl}, z_{ecl}) = (x_{eq}, y_{eq} \cos \varepsilon + z_{eq} \sin \varepsilon, z_{eq} \cos \varepsilon - y_{eq} \sin \varepsilon)$$ But if you want another planet's orbital plane to be the $xy$-plane, then you also need the orbit's inclination and longitude of ascending node; I leave that transformation as an exercise for the motivated reader.

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  • $\begingroup$ Mike, thank's for the answer and to correct my english (which is not my mother langage). If i understand well, the norm is : rotation of obliquity counter clockwise around the vector defining the ascending node (x in the case of the eath). I already done the work to translate planet position thank to their orbit using kepler formula, what i missed was the norm used to define the rotation axis for the title... maybe a misunderstanding of the concept of ascending node. I start studying astronomy to no more than a week.. in all case many thanks. $\endgroup$
    – Guillaume Pelletier
    Commented Sep 2, 2016 at 10:06
  • $\begingroup$ Take care not to confuse orbital elements with rotational elements. Another planet's ascending node relates its orbital plane to Earth's orbital plane, independent of either planet's rotation. $\endgroup$
    – Mike G
    Commented Sep 2, 2016 at 13:22
  • $\begingroup$ Yes i do not confuse, but for a planet the axis tilt is defined thanks to its orbital plane by a rotation of obliquity using X vector ( X defined into the orbit plane, toward the point of ascending node ). Is that right ?? $\endgroup$
    – Guillaume Pelletier
    Commented Sep 3, 2016 at 8:44
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    $\begingroup$ No, that is only valid for Earth. In general there is no such shortcut; a single angle is insufficient to define a unique direction. Rotating the ecliptic pole about another planet's ascending node (not the x axis) by that planet's orbital inclination gives you the pole of the other planet's orbit. The obliquity of the planet's rotational axis gives you an angle away from that orbital axis but does not determine direction. You really need both $\alpha$ and $\delta$ as above. $\endgroup$
    – Mike G
    Commented Sep 3, 2016 at 14:13
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    $\begingroup$ @uhoh Table 1 of the IAU WGCCRE report gives most planets' polar $\alpha_0$ and $\delta_0$ [ICRF] as functions of $T$ [Julian centuries; T(J2000.0) = 0]. $\endgroup$
    – Mike G
    Commented Dec 9, 2019 at 3:30

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