2
$\begingroup$

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?.

$\endgroup$
1
  • 1
    $\begingroup$ @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. $\endgroup$
    – uhoh
    Mar 7, 2018 at 17:08

1 Answer 1

5
+100
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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! $\endgroup$
    – uhoh
    Mar 8, 2018 at 4:13
  • $\begingroup$ 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 :-) $\endgroup$
    – uhoh
    Mar 4, 2020 at 4:30
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 4, 2020 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.