# How to derive ecliptic in simplified solar system, with vector math?

Assume the Earth orbits the sun in a fixed circle along the ecliptic (I'd be interested in answers that consider an elliptic orbit as well, but this is the simplest version).

What's the simplest way to derive the vector normal to the ecliptic $\vec e$, given the following information?

• $\vec s$, the direction toward the sun in a local Cartesian coordinate frame "somewhere" on Earth, at a single point in time
• $\vec r$, the Earth's rotation axis pointing through the north pole, in the same coordinate frame
• Earth's tilt $\mathcal E = 23.44° = 0.4091 \ \mathrm{rad}$
• $\vec s \cdot \vec r$ is most positive at the June solstice, most negative at the December solstice, and 0 at the equinoxes

We do not have the following information:

• The longitude of our location on Earth
• The year, time of year, etc.
• The local direction of the sun at any time other than now (except rotated around the Earth's axis, which doesn't help)

I can get this far:

• Local latitude = $elevation(\vec e) = asin(\vec e_y)$, where y is the local vertical component
• $\vec e = (\vec s \times \vec r) \times \vec s$ at the solstices
• We can in theory tell where we are in orbit by applying some function to $\vec s \cdot \vec r$, but I haven't derived this function, and there are two solutions at every point except the solstices.
• We can rotate $\vec r$ counter-clockwise around $\vec s$ by $\mathcal E$ at the March equinox to get $\vec e$, or clockwise at the September equinox, though I wouldn't know how to differentiate them without more information.

Do we have enough information to solve the problem? It seems like the answer is right in front of me, but the ambiguity of "which equinox we're closer to" gives me pause.

Followup question: Furthermore, if we can derive $\vec e$, do we have enough information to define unique basis vectors for an ecliptic coordinate frame, such that the non-$\vec e$ vectors point from solstice-to-solstice and equinox-to-equinox?

• I don’t think you should be worrying about the location of the observer on the Earth’s surface. Only one thing is clear to me, that $\vec e\perp\vec s$, and I guess your equation for $\vec e$ at solstice is correct. I think your analysis thus far is as much as can be said. Do you have reason to think that this problem is solvable? – Lubin Feb 24 '18 at 21:51
• Also: We can assume $/vec e$ is fixed in ecliptic coordinates, i.e. ignore precession. – Mike S Feb 24 '18 at 21:52
• $\vec s\cdot\vec r$ varies sinusoidally with time. The equinox/solstice ambiguity only affects the sign of $\vec e$. – amd Feb 24 '18 at 22:05
• @Lubin, no, I don't have any special reason to believe that it's solvable, aside from knowing $cos(\theta) = \vec s \cdot \vec r$ is "almost" unique. For each $cos(\theta)$ there are only two possible locations we could be in the Earth's orbit around the sun, and if we could somehow distinguish between them we could [probably] derive a unique angle with which to rotate $\vec r$ around $\vec s$, such that the cross product of the result with $\vec s$ would be in the plane of the ecliptic. – Mike S Feb 24 '18 at 22:10
• @amd, thank you. That makes sense, and it implies we can get the Earth's angle from "an" equinox around $\vec e$ via $asin(\vec s \cdot \vec r)$. How would you then proceed to find $\vec e$? – Mike S Feb 24 '18 at 22:16

Let $\vec w$ be the projection of $\vec r$ onto the ecliptic plane: we know that the angle between $\vec w$ and $\vec r$ is $\alpha=90°-23.44°=66.56°$, hence $\vec w$ lies on a cone of axis $\vec r$ and half-width $\alpha$.
On the other hand, from the given data we can compute the angle $\theta$ between $\vec r$ and $\vec s$: $$\theta=\arccos\left({\vec r\cdot\vec s\over|\vec r||\vec s|}\right).$$ The angle $\beta$ between $\vec w$ and $\vec s$ is then given by $$\beta=\arccos\left({\cos\theta\over\cos\alpha}\right),$$ hence $\vec w$ lies on a cone of axis $\vec s$ and half-width $\beta$.
But those two cones have two rays in common: one of them contains $\vec w$ but the other doesn't. Without some additional info I'm afraid you cannot tell which ray is the good one, that is you have in general two possible choices for the ecliptic plane.