Orbital period and nodal precession & apsidal precession

Question

In classical orbital mechanics as per Newton/Kepler, the time taken to complete one orbit around a body of mass $M_e$ is:

$$ T = 2 \pi \cdot \sqrt { \frac {a^3} {G \cdot M_e} } $$

But also, for orbits around an oblate spheroid, we also have precession of the orbital plane itself, and the precession of the orbit's major axis which, for every turn of the orbit, are respectively:

$$ \Delta\Omega = -3\pi\cdot \cfrac {J_2} {G\cdot M_e} \cdot \left[\frac {1} {a \cdot \left(1-e^2\right)}\right]^2 \cdot \cos i $$

$$ \Delta\omega = -6\pi\cdot \cfrac {J_2} {G\cdot M_e} \cdot \left[\frac {1} {a \cdot \left(1-e^2\right)}\right]^2 \cdot \left(\frac 5 4 \cdot \sin^2 i - 1 \right) $$

What I'm unclear is, in presence of these factors, how is the orbital period $T$ defined?

Can it be defined as the time elapsed between two consecutive passes through the periapses (or apoapsis)? How does that work when the periapsis (or apoapsis) is itself shifting? Do I count the time to where the periapsis was when I started, or where the periapsis is now?

Specifically, what happens in an equatorial circular orbit? (Or, if we like, an orbit where $i$ and $e$ are just very, very small?) Substituting $i=0$ and $e=0$ gets us:

$$ \Delta\Omega = -3\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}\qquad \Delta\omega = +6\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2} $$ So if we assume the oblate spheroid is also rotating with time period $T$, and the orbiting body passed directly over a point $P$ on the spheroid's equator at time $t_0$, where will it be at time $t_0 + T$?

• Will it be directly over $P$?
• Or will it be $3\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}$ ahead (East) of $P$ in longitude (considering $\Delta\Omega + \Delta\omega$)?
• Or will it be $3\pi\cdot \cfrac {J_2} {G\cdot M_e \cdot a^2}$ behind (West) of $P$ in longitude (considering $\Delta\Omega$ only)?
• Anything else?

Cheers!

Note: $a$ = semi-major axis, $e$ = eccentricity of the orbit, $i$ = inclination of the orbit, $G$ = gravitational constant, $J_2$ = a coefficient in the spherical harmonic expansion of the spheroid's gravitational potential field

Answer 1

The formulas you mention are obtained from the Lagrange Planetary Equations, that describe the perturbed central motion, with some averaging along the orbit.

For context and notation, in the case of central motion, the set of orbital elements $\{a,e,i,\omega,\Omega,t_p \}$ are constant and equivalent to know the initial conditions, the solution being (for $e <1$) an ellipse. $t_p$ is the time at periapsis; knowing the current time and the period (your formula) $T=2\pi/n$ is enough to determine the mean anomaly $M=n(t-t_p)$ from which we can determine the true anomaly, the angle from the periapsis to the current position, using Kepler's equation; $n$ is the orbital frequency.

In the case of motion perturbed by the oblateness, the orbital elements become time dependent. The intuitive view is that at each instant of time $t_s$ we have a different orbit with elements $a(t_s),e(t_s),\ldots$. In the present case, it happens that the 3 elements $a,e,i$ are periodic with small amplitude, so their average variation is zero and can be ignored except for very high fidelity calculations. Because of that, we can imagine the motion as if the ellipse that is the trajectory (that is the same because $a,e$ are constant) rotates in space: the orbital plane rotating around the $z$-axis (determined by $\Omega$) and the apsidal line rotating in the orbital plane (determined by $\omega$).

So, for an orbit with inclination $i$, if the central body does not rotate, the orbit rotates towards East for $i<\pi/2$, and so the satellite will not pass over $P$. When it passes at the same latitude depends on the period and the change in $\omega$ (apsidal rotation). The satellite moves in this rotating orbit.

What is the period and how much is it is a little ore complicated. We can have two views: when it completes a turn relative to the inertial reference frame (stars) or relative to the periapsis, which is rotating. We can use both. In the case of the Earth, we call sidereal year to the first and anomalistic year to the second (check Sidereal, tropical, and anomalistic years). What about the period? Lagrange Planetary Equations determine the change in $M=M_0+n_M(t-t_0)$ with a changed orbital frequency $$ n_s=n+\frac{3nJ_2(3\cos^2 i-1)}{4(1-e^2)^{3/2}}(R_e/a)^2; $$ at each instant $t$, we can determine the current $M$ using the current osculating orbit for the true anomaly.

I suggest the book of Sidi, "Spacecraft Dynamics and Control: a practical engineering approach", it's the only reference that comes to mind that explains the change in $M$.

Answer 2

Since the period $T$ doesn't appear in either of those formulas of precession angle, I don't understand why you need a definition of $T$ (the formulas are self-consistent without $T$). In fact, we can not define period very well in some cases, since the orbit may not close on itself due to precession, which makes the movement aperiodic.