On Tw*tter, Mike Lawler linked to a nice Physics Today article that points out that the planet whose average distance to Earth over any long period of time is the shortest is Mercury, not Venus.
I agree with their conclusions … but I wonder if there is an error in the specific formula they give for the average distance as an elliptic integral.
If we specialise to the case where one orbit is a unit circle and the other has radius r, Mathematica gives a somewhat different formula than theirs (actually two different formulas, for r>1 and r<1, blue and gold in the plot), which give close results to the Physics Today formula (green in the plot), but not exactly the same.
Numerical integration seems to confirm the formulas Mathematica gives.
Have I made some dumb mistake in the way I’ve set up the problem, or is the formula in the article wrong?
@buster and @duetosymmetry; the conventions for Mathematica’s EllipticE function and the elliptic integral function in the article are different; Mathematica’s function expects an argument that is the square of the one used in the article.]
On Tw*tter, Mike Lawler linked to a nice Physics Today article that points out that the planet whose average distance to Earth over any long period of time is Mercury, not Venus.
You mean the average distance is shorter; you missed a predicate. :)
The Mathematica 'EllipticE' takes the "parameter" 𝑚 as its argument, whereas in the Physics Today article, the convention seems to be that 𝐸 takes the "eccentricity" 𝑘; you should square the eccentricity 𝑘=2√𝑟/(1+𝑟) to get 𝑚=4𝑟/(1+2𝑟+𝑟²) before passing it to EllipticE.
@davidr@darabos We are approximating both orbits as circular, with constant velocity. But to find the average distance between the planets, we still need to perform an integral over one complete orbit of one of the planets.
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