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?
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.