>Why is Nodal precession affected by the rotational period of the planet?

I'm going to venture an answer. It doesn't, at least not directly.

The first equation [there][1] for the rate of precession $\omega_p$:

$$\omega_p = -\frac{3}{2} \frac{R_E^2}{(a(1-\epsilon^2))} J_2 \omega \cos(i)$$

depends on the parameters of the orbit ($a, \epsilon, \omega$, i) and the Earth's equatorial radius $R_E$ and its "second dynamic form factor ($-\sqrt{5}C_{20}$)" or $J_2$ term. There is no explicit dependence on the rotational rate of the earth nor should one be expected, as $J_2$ is an expression related to an axisymmetric term.

But the Wikipedia article goes on to use a different equation that *tries to predict* the value $J_2$ term based on some equilibrium model of a rotating body. The article says:

>This last quantity is related to the oblateness as follows:

>$$J_2 = \frac{2 \epsilon_E}{3} - \frac{R_E^3 \omega_E^2}{3 GM_E}$$

without really giving a source for this or an explanation. 

In my opinion the Wikipedia article could be improved by explaining that this equation isn't necessarily fundamental, *nor should it be used* to generate a value for $J_2$ that would then be used to propagate a spacecraft's orbit.

The oblateness or $\epsilon_E$ in the equation comes from the surface of the Earth, whereas an accurate measurement of $J_2$ will only come from careful, experimental measurements of Earth's gravitational field, which results from the real 3D distribution of mass within the entire volume of the Earth.

A small rocky body's $J_2$ would be unrelated to it's rotation rate since centrifugal forces would not define it's shape. The larger the body and the more that it could flow in order to reach an equilibrium mass distribution however, the closer the link would be between its rotation rate and the mass distribution and therefore $J_2$.


[1]: https://en.wikipedia.org/wiki/Nodal_precession