# Equation for orbital period around oblate bodies, based on J2?

In this answer I point out that the period of items (ring particles, moons, spacecraft, etc.) around an oblate body will not scale exactly as $a^{3/2}$ because the closer you are to the planet, the stronger the perturbing effects are as a result of being much closer to the near-side of the oblate "ring" than the far side of it. Mathematically that turns out to be $1/r^4$ vs $1/r^2$.

I can mindlessly calculate orbits including the $J_2$ term as shown in this answer using these radial acceleration terms assuming an equatorial orbit:

$$a_0 = -\frac{GM}{r^2},$$

$$a_2 = -\frac{3}{2} J_2 \frac{GM R^2}{r^4},$$

where $a_0$ is the radial acceleration due to the monopole term and $a_2$ is the radial acceleration due to the quadrupole term — that part of the oblateness captured within the $J_2$ coefficient, and $R$ is the normalizing radius of the body used to keep $J_2$ dimensionless.

I can rewrite this as

$$a_{tot} = -\frac{GM}{r^2} \left( 1+\frac{3}{2} J_2 \frac{R^2}{r^2} \right)$$

and just decide that for the circular equatorial case I can set $r$ equal to the semi-major axis and the "effective mass" of the central body is increased by the factor in the parenthesis, but I am not sure if I've done this right, and certainly don't know what to do if the orbit is elliptical and/or inclined.

Question: What would an equation for the period of a circular orbit taking into account $J_2$ look like? Is there something that would include either elliptical and/or inclined orbits as well?

I'm also a bit confused about the mass and its distribution. I'd like to double check that the standard gravitational parameter $GM$ represents all of the mass including that in the equatorial bulge, and that we're not somehow double-counting that by using $J_2$.

A related and (still) unanswered question is For the mathematical relationship between J2 (km^5/s^2) and dimensionless J2 - which one is derived from the other?.

• @ChrisB.Behrens thanks for your several title rewrites, but after asking over 1,000 Stack Exchange questions, I've sort-of developed a sense for the way I'd like to write titles.
– uhoh
Mar 7, 2018 at 17:08

If you consider the orbital period to be defined as successive node crossings, that's known as the nodal period. For an orbit with semimajor axis $a$ around a spherical body with gravitational parameter $\mu$, the nodal period is equal to the Keplerian period: $T_0=2\pi \sqrt\frac{a^3}{\mu}$, however, as you point out, this changes when oblateness is taken into account. Wikipedia has one form for the expression taking the $J_2$ budge into account: $$T = T_0\left[1 - \frac{3J_2(4-5\sin^2 i)}{4\left(\frac{a}{R}\right)^2\sqrt{1-e^2}(1+e\cos\omega)^2} - \frac{3J_2(1-e\cos\omega)^3}{2\left(\frac{a}{R}\right)^2(1-e^2)^3}\right]$$

As you can see, it depends on the eccentricity $e$, argument of perigee $\omega$, and inclination $i$ of the orbit, as opposed to $T_0$ which is only a function of semimajor axis. $R$ is the equatorial radius of the body.

As an example, using this equation, an orbit around Earth with $a=6778~\textrm{km}$, $e=1\times10^{-3}$, $i=20^\circ$, and $\omega=0^\circ$ has a Keplerian period of about 92.56 minutes vs. a nodal period incorporating $J_2$ of about 92.20 minutes, the latter being a little under 22 seconds shorter.

• Yikes! That's more complicated than I expected, but then again I didn't really know what to expect. I'll take it for a spin. I'm never sure what qualifies as a period for an orbit that doesn't exactly repeat itself, but it looks like the nodal period is a pretty easy one to understand, and test for. Thank you for both the equation and the explanation!
– uhoh
Mar 8, 2018 at 4:13
• I'd planned on writing a short script to test a few oblique orbits to numerically to verify a few cases before adding a reward bounty, but I never managed to get "a round tuit". Thanks for the laborious MathJaxing :-)
– uhoh
Mar 4, 2020 at 4:30
• Note that the Wikipedia article says this only applies to "near circular orbits". I'm curious about how well it performs for more eccentric cases. Mar 4, 2020 at 9:46