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According to the Wikipedia Lunar Theory article, the first term of the Moon's equation of the centre is

$$22,639^{\prime\prime}\sin l,$$where $l$ is the Moon's mean anomaly. According to the Wikipedia Equation of the Centre article, the equivalent term of the equation of the centre is$$2e\sin M,$$where $e$ is the eccentricity of the Moon's orbit around the Earth, and $M=l$. Substituting $e=0.0549$ (from the Lunar Theory article) and converting radians to arcseconds gives$$22,648^{\prime\prime}\sin M.$$So I'm wondering why there is a discrepancy between the two coefficients. My best guess is that the $2e$ coefficient is an approximation and is itself part of a series expansion. Does that sound a reasonable explanation? Thanks.

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    $\begingroup$ Yes, that sounds reasonable. FWIW, I have plots (and code) here of the Moon's osculating eccentricity. Of course, the mean eccentricity is more stable, and doesn't reflect the perturbations shown in those graphs. $\endgroup$
    – PM 2Ring
    Mar 17, 2023 at 11:55
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    $\begingroup$ @PM2Ring - To me, it's amazing how in the days before spreadsheets and code people were able to make sense of the Moon's orbit. Imagine what Newton could have done with Excel! $\endgroup$
    – Peter
    Mar 17, 2023 at 13:31
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    $\begingroup$ @Peter slightly related to your comment in History of Science and Mathematics SE: Tools of the trade: were early scientists and mathematicians really "writing with feathers using light from burning animal fat?" $\endgroup$
    – uhoh
    Mar 17, 2023 at 21:30

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It's Wikipedia. What do you expect? Perfection and consistency? Those are bad expectations. There are many Wikipedia articles that have internal inconsistencies, let alone multiple Wikipedia articles that are inconsistent with one another.

Another way to look at it is $22639'' \approx 22648''$. In other words, the two articles are more or less consistent with one another.

Yet another way to look at it: You are using a value of $e$ with three places of precision. That means your calculated value of $2e$ converted to arcseconds should have three places of precision. Anything beyond that is meaningless. Your calculated value of $2e$, converted to arcseconds, $22648''$, is completely consistent to within three places of precision with the $22639''$ value in the Wikipedia Lunar Theory article. In particular, $\frac{22648−22639}{22639}*1000≈0.3975$. If this had been greater than 1 or less than -1 there might be a consistency issue. But 0.3975 falls nicely in that $[-1,1]$ interval. There is no problem here.

One final way to look at it: You omitted higher order terms in $e$ in the $\sin M$ term in the equation of center. A more correct form is $(2e - \frac14e^3 + O(e^5))\sin M$. Adding in the $-\frac14e^3$ factor and converting to arcseconds results in $22639''$ arcseconds. There is no discrepancy at all.

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    $\begingroup$ Shortly after posting my question, I found the more accurate $2e-\frac{1}{4}e^{3}$ form of the $\sin M$ coefficient on page 127 of Brown's "An Introductory Treatise On The Lunar Theory". I did the calculation and out popped 22,639 arcseconds, which I was most pleased with. Brown's is an old book. Could you give me your source for $(2e - \frac14e^3 + O(e^5))\sin M$? Thanks. $\endgroup$
    – Peter
    Mar 17, 2023 at 12:52
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    $\begingroup$ @Peter The $2e - \frac14e^3 + O(e^5)$ is directly from the Wikipedia article you cited. The $O(e^5)$ simply means I am ignoring the $e^5$ and higher order terms. It's fairly standard notation. It would have been better if I had used $\mathcal O(e^5)$, but I was lazy and just used $O(e^5)$. Besides, \mathcal doesn't always look good. $\endgroup$ Mar 17, 2023 at 13:05
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    $\begingroup$ Hidden in plain sight. I should have read the Wikipedia article more thoroughly. $\endgroup$
    – Peter
    Mar 17, 2023 at 13:22
  • $\begingroup$ @DavidHammen: I don’t know if it’s only me, but a character doesn’t display well after “better if I had used”; I see the “glyph not found” square before ($e^5$). $\endgroup$ Mar 18, 2023 at 0:34
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    $\begingroup$ @PierrePaquette It displays fine for me; it's \$\mathcal O\$, or $\mathcal O$. The intent of the mathcal font library is to look as if the symbol was handwritten. $\endgroup$ Mar 18, 2023 at 2:51

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