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Inclination of planetary orbits

  1. Aug 30, 2009 #1
    Does anyone know the inclination of each planetary orbit (in degrees) with respect to the plane of the Sun's equator-plane?All planetary inclinations are known with respest to the ecliptic, and the ecliptic's inclination is known with rescpect to the Sun's equator (7.25 degrees). Such data, however, seem to me not sufficient for a precise calculation of the answer I need.
     
  2. jcsd
  3. Aug 30, 2009 #2
    Hi Leo - The authority on these things is the JPL Horizons system.

    I'm not sure if you can use the sun's equator, but you'll certainly be able to find the inclination of each planetary orbit with respect to the ecliptic, then do the arithmetic by adding/subtracting the 7.25 deg. (At least it seems that simple at first look . . . )

    Note that inclination, like other orbital elements, actually osculates, meaning it changes ever so slightly over time. So look in the osculating elements of your results after you submit your object data. The inclination is given by the quantity "IN", which is in degrees by default.

    hth
     
  4. Aug 31, 2009 #3
    What you suggest doesn't work, as I've already pointed out, when stating that those data are not sufficient; because there is no criterion to chose whether each particular orbit inclination has to be added or subtracted with/from 7.25 degrees.
     
  5. Aug 31, 2009 #4

    D H

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    Use the inclination and the longitude of ascending node.
     
  6. Aug 31, 2009 #5
    Yup, that's the answer. You have to specify a complete reference system, and if you throw out the ecliptic as your reference frame, you also throw out Earth's ascending node as the reference zero-degree direction. You can either retain it, or look at each orbit independently in terms of its ascending node, or something more clever.

    Without knowing more about what you're trying to do, I'm thinking that's the best we're going to be able to do for you.

    I referred you to JPL's horizons because you said you were seeking precision.

    You're welcome.
     
  7. Sep 2, 2009 #6
    Thanks to Spacester and DH. I got the answer.
     
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