I considered posting this to astronomy.stackexchange.com, but I've bugged them enough for today...

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Let $p(t)$ be a parametric function that traverses an ellipse such that it sweeps out equal areas in equal time, as per the diagram above. In other words, $p(t)$ is the ideal elliptical orbit (no 3rd party peturbations). Additionally:

  • The ellipse's semimajor axis has length $m$.

  • The ellipse's eccentricity is $e$.

  • $p(t)$'s period is $T$. In other words, $p(T) = p(0)$

Question: what circle best fits this ellipse? More specifically, if we parametrize a circle as:

$$c(t) = \big(x_0+r\cos(bt-c), y_0+r\sin(bt-c)\big) $$

what values of $x_0$, $y_0$, $b$, and $c$ would minimize:

$$\int_0^T d(c(t),p(t)) \, dt$$

where $d$ is the linear distance between $c(t)$ and $p(t)$?

By introducing $x_0$, $y_0$, $b$, and $c$, I'm allowing for the possibility that:

  • The best fit circle's center is different from the ellipse's center.

  • The best fit circle is "out of phase" with the ellipse.

  • The best fit circle's period is not $T$, the ellipse's period.

These all seem unlikely (especially the last 2), but I want to allow for the most general parametric circle.

This ultimately answers the question: if we ARE going to pretend a planet's orbit is circular, what's the best circle to use?

(as a humorous note, stackexchange suggested my question was subjective and would most likely be closed, possibly because I used the word "best". Of course, in mathematics "best fit" is a perfectly valid, non-subjective concept)

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    $\begingroup$ If you were to minimize $\int_0^T d(c(t),p(t))^2\,\mathrm dt$ instead, it would make the problem more separable in $x$ and $y$. Then the Fourier transform might be useful. $\endgroup$ – Rahul Dec 10 '13 at 21:33
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    $\begingroup$ Are you willing to entertain nonuniform motion on the circle? $\endgroup$ – hardmath Dec 10 '13 at 21:36
  • $\begingroup$ @RahulNarain OK, I'm willing to do that (ie, accept an answer that minimizes d^2, not d). It seems traditional to minimize the distance squared, even though it's different from minimizing the distance itself. $\endgroup$ – barrycarter Dec 10 '13 at 21:38
  • $\begingroup$ @hardmath I'd be interested in seeing a solution like that, but the idea of using a circle is to make the math easy. If the nonuniform motion were ugly enough, it would defeat the purpose. However, I'll upvote (but not approve) an answer like that, just to see what it looks like. $\endgroup$ – barrycarter Dec 10 '13 at 21:40
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    $\begingroup$ According to your requirements it is indeed possible that the best “circle” does not share period with the ellipse. Two possible scenarios: the circle is open in the aphelion or the circle is open in the perihelion. $\endgroup$ – Carlos Eugenio Thompson Pinzón Dec 10 '13 at 21:46

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