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> b = a·sqrt(1−e²), so corr(a,b) = 1.00000 to 5 decimals
Isn't e different for each planet?
> Mercury is the only planet where they differ by more than 2%
I remember something about Mars been the planet with the most eccentric elipse
> *So fit quality degrades gracefully, symbolic recovery doesn't. Making that part noise-robust is pretty much the open frontier of the whole field, not just of my tool.
Nice. It's a hard problem. Which heuristic are you using to pick the "best" formula?
Thanks for the formatting tip, noted.
Isn't e different for each planet?
Yes, each planet got its own e (Mercury 0.206, Venus 0.007, Earth 0.017...). The correlation still comes out at 1.00000 because all eccentricities are small while a spans 0.39 to 30 AU — a ≤2% per-planet wobble is invisible to Pearson across two decades of range. That's exactly what makes it a nasty decoy: almost collinear, but not quite.
Mars been the planet with the most eccentric ellipse
Close — Mercury is actually the most eccentric (0.206), Mars is second (0.093). Funny enough, Mars is the famous one precisely because Kepler derived his laws fighting with Tycho's Mars data: its ellipse was just eccentric enough to kill every circular fit. If Tycho had handed him Venus data instead, we might have waited a while longer.
Which heuristic are you using to pick the "best" formula?
Hold-out validation during evolution, then the old CART-style one-standard-error rule: pick the smallest formula whose validation error is within ~1 SE of the best one. Constants get refit with Levenberg-Marquardt before that comparison, so small forms compete at their best. It also returns the full accuracy-vs-size Pareto front so you can override the choice. And to connect it to your noise question: under noise that 1-SE band is exactly where wrong-but-simpler formulas sneak in and tie the true one — the two problems are really the same problem.