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So, you can determine the location of an object in an orbit at any given time via Kepler's laws, but how do you determine the orbital inclination at any given point in time?

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    $\begingroup$ What do you mean by the orbital inclination at a given time? That's like asking what is the orbital period at a given time... $\endgroup$ – Rob Jeffries Oct 25 '15 at 11:24
  • $\begingroup$ Ok, sorry, what I mean : sometimes Pluto is above the ecliptic, sometimes it's below. Is there an equation for how inclined it (the object itself) is at a given point of time.... $\endgroup$ – Damon Blevins Oct 25 '15 at 13:08
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    $\begingroup$ DamonBlevins Do you mean how far it is above the ecliptic? An object's orbit has an inclination here, which as Rob Jeffries said, is constant. $\endgroup$ – HDE 226868 Oct 25 '15 at 13:37
  • $\begingroup$ Yes how far it is $\endgroup$ – Damon Blevins Oct 25 '15 at 13:38
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    $\begingroup$ @DamonBlevins, please edit your Q to state that. Distance perpendicular to the ecliptic is what you are looking for. That would actually make an interesting, if rather simple, question. Inclination is an angle, and does not change. $\endgroup$ – Aabaakawad Oct 27 '15 at 2:12
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I am assuming that the question you want answered is how to calculate the elevation of an orbit above a reference plane given the orbital inclination with this plane. If so, please update your question to reflect this, heeding the advice given in the comments.

Kepler's first law tells us that planets move in elliptical orbits, which we can define as follows,

$$r = \frac{p}{1+\epsilon \, \cos\theta}$$

where $p$ is the semilatus rectum, $r$ is the orbital distance from the central body, $\epsilon$ is the orbital eccentricity, and $\theta$ is the angle to the orbiting bodies current position from its closest approach.

Using $r$ along with the orbital inclination with respect to a given plane, we can calculate the height above this plane at any point in the orbit using simple trigonometry.

enter image description here

The image above shows the relevant variables for a given orbit when the body is both above and below the reference plane. $r$ is again the orbital distance, $i$ is the orbital inclination, and $d$ is the height above the reference plane asked for in the question. $d$ is then given by

$$d = r \, \sin \, i$$

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  • $\begingroup$ The vacuum of space allows geometry pure and simple to rule. $\endgroup$ – LocalFluff Jan 28 '16 at 6:28
  • $\begingroup$ This is a great answer that stars with the true anomaly θ and gets to the final result - height above the ecliptic d as asked (or intended at least). But the question is asking for the moon (so to speak), ending with "... at any given point in time?" Any chance this can be combined with M = E - e sinE? $\endgroup$ – uhoh Apr 30 '16 at 9:27
  • $\begingroup$ @uhoh the question doesn't mention the moon. What equation are you referencing? The inclination angle depends on the orbital distance, which depends on which point in the elliptical orbit it is. To introduce time you need to fix an initial reference time, which if you're talking about a specific object you can look up, but this isn't clear from the question. $\endgroup$ – christopherlovell Apr 30 '16 at 10:23
  • $\begingroup$ I like your answer, I think it really isolates the parameter the question is looking for - labeled 'd'. The equation is usually called "Kepler's equation" when it's singular rather than plural. "Asking for the moon" is an expression meaning asking for a lot, for very much. I shouldn't have used it without clarification, sorry! OK I'll go back and think more about it. $\endgroup$ – uhoh Apr 30 '16 at 10:54
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    $\begingroup$ @uhoh ah! My bad. yeah, perhaps using a turn of phrase involving a celestial object in a group about astronomy could get a bit confusing! $\endgroup$ – christopherlovell Apr 30 '16 at 11:08
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So, basically you're trying to merge orbital inclination to Kepler's laws. The simplest way is to take the measured orbital inclination of the planet, which is constant, and apply the Pythagorean theorem to any given location and that gives you 3 dimensions of distance from the 2 dimensions defined in Kepler's laws. That's probably what Kepler did. (see picture from the Wikipedia link above)

A much too simple explanation of orbital inclination here.

And a much more complex one, factoring in orbital movement here.

Kepler's laws only work in 2 dimensions because one object orbiting another is a 2 dimensional calculation and bringing a 3rd dimension into his law complicates it unnecessarily. 3 dimensions is essential for accurate planetary tracking, but it's unnecessarily in a 2 body gravitational system that takes into account distance, eccentricity and velocity. Kepler was likely aware of this.

Kepler, in fact, discovered orbital inclination by looking at Copernicus' observations, and that was something he was very proud of and it was a very important discovery. Source here. Kepler deduced that orbital inclination explained away previously not understood and overly complicated orbital oscillations. He didn't just undo Copernicus' circles, he worked things out in 3 dimensions and he might never have figured out his laws without understanding and measuring orbital inclination.

Nice question.

(footnote, my answer is more history of science and mathematics, but I think your question, the way you asked it is more astronomy, so, I think the question should stay here, but the powers that be can decide otherwise if they like).

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The orbital inclination is constant -- does not change with time. Except over extremely long timescales due to interaction with other bodies, and that interaction would be considered non-Keplerian.

So, measure orbital inclination by observation, then expect it not to change over time.

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  • $\begingroup$ Yeah I get that lol I'm changing the question $\endgroup$ – Damon Blevins Oct 27 '15 at 12:19
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The position of a planet is determined by 6 numbers

First there are the eccentricity and semi-major-axis. These give the shape and size of the ellipse. By Kepler Law, the sun is at one focus of this ellipse, Then there is the plane in which the orbit is placed. This can be conveniently determined by the angle between the planet's or it and the elliptical, and the position of the point where the orbit crosses the plane: the longitude of the ascending node. Finally there is the position of the planet on this path. The angular position the ginger to me hard at some reference time(the mean anomaly at epoch) and the amount of time passing since the epoch.

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  • $\begingroup$ But how does this give you the it's position relative to the elliptic at a given time $\endgroup$ – Damon Blevins Oct 27 '15 at 1:01
  • $\begingroup$ I think it's closer to seven if you include argument of periapsis - the other omega $\endgroup$ – uhoh Apr 30 '16 at 8:53

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