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Wikipedia has a formula for the nodal period of a near-Earth satellite, taking into account the oblateness of the Earth (J2), and neglecting other effects (https://en.wikipedia.org/wiki/Nodal_period). From this formula, if the eccentricity is very small (say of the order of 1e-3 or less), I have concluded that the nodal period (Tn in Wikipedia notation) would necessarily be smaller than the period of the ideal Kepler model (no aspherity for Earth), irrespective of the value of the inclination angle i in the formula. But, this seems to contradict the teachings on Celestial Mechanics (eg https://farside.ph.utexas.edu/teaching/celestial/Celestial/node93.html, cf discussion after Eq. (10.129)). Did I miss something, or Wikipedia's formula is wrong?

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    $\begingroup$ There is a link to the paper by King-Hele in the Wikipedia note. The Capderou reference is not linkable because it is a textbook (Capderou Orbit and missions - Springer 2005). But this link provides more or less the same teaching:(farside.ph.utexas.edu/teaching/celestial/Celestial/node93.html). Read the discussions after Eq. (10.129). This discussion concludes that the J2 makes the satellite fly faster OR SLOWER -depending on i-, than when the satellite were orbiting a perfectly spherical (and homogeneous) planet - the Kepler case. (Thanks for pointing to the related question). $\endgroup$
    – Ng Ph
    May 23, 2021 at 18:25
  • $\begingroup$ Different but related: Equation for orbital period around oblate bodies, based on J2? I seem to remember that I'd actually tried a few cases to verify, but I'm not sure now. It's the equation in the answer there that you're asking about, right? $\endgroup$
    – uhoh
    Nov 1, 2021 at 12:31
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    $\begingroup$ @uhoh, yes. According to this formula (and my calculations) $T_n(i)<T_0$, independently of the inclination i. And this independence to i is what I am trying to establish. $\endgroup$
    – Ng Ph
    Nov 1, 2021 at 13:50
  • $\begingroup$ I see, it's the "always-less-than-ness", ya that's a good question! $\endgroup$
    – uhoh
    Nov 1, 2021 at 14:13

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This is not a complete answer, but, as far as I can tell:

First, the angular speed in the linked paper is based on the anomalistic period (i.e., the time between two periapsis passages), not the nodal period. To get the angular speed based on the nodal period, one would have to add the expression for the time derivative of the argument of periapsis (formula 10.127). Still, the result it gives is quite different from the Wikipedia's.

Second, the linked paper gives the formula for the instantaneous angular speed in terms of the current osculating elements, while the paper referred by the Wikipedia gives the formula for the nodal period (the inverse to which is the average angular speed over a period) in terms of the osculating elements at the ascending node at the period's beginning. This may explain the difference.

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    $\begingroup$ I am not familiar with these various concepts so it will take me some time to digest them. It is quite clear though that I have overlooked many things.Thanks for the hints. $\endgroup$
    – Ng Ph
    Nov 3, 2021 at 12:12

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