Perimeter of Ellipse

[Gerson W. Barbosa]

(12-04-2019 06:28 PM)Valentin Albillo Wrote:  An error "less than 5 meters" in such huge value seems unrealistic from a purely physical point of view. From a purely mathematical point of view, that's another matter and your error estimation may be correct, though again the orbit's eccentricity is not known to high accuracy either (~0.2488, quite large when compared with the much more circular orbit of Earth, say).

Hello, Valentin,

This has been done only to assess the quality of the approximation, if any. Anyway, despite the uncertainties involved, the order of magnitude of the difference should be correct.

Some doublechecking:

semi-major axis: a = 5906376272 km
eccentricity: e = 0.24880766 ->
semi-minor axis: b = 5720637952.8 km,

for which the Free42 program returns the exact result

36529672878.01583840603514193230844 km

Now, on the wp34s,

5906376272 ENTER

5720637952.8

AGM ->

5813136193.07

3 * 5906376272 ENTER

5720637952.8 * √ 2 * -

2 * π *

-> 36529672878.01109446182 km

Difference:

0.00474394421514193230844 km

or

4.74 m

———-

This first Ramanujan approximation brings the error down to under one meter:

π[3(a + b) - √((3a + b)(a + 3b))]

His second approximation is even better, with an error less than one micrometer:

https://www.johndcook.com/blog/2013/05/0...-ellipse/

Best regards,

Gerson.