ISS nodal precession

Question

I was confused by JPL Horizons data showing time variations in the ISS longitude of ascending node, until I realized maybe the orbital plane itself rotates around earth...

A quick web search then turned up this thing called nodal precession, in which the orbital plane rotates around the polar axis of the host planet.

Can someone confirm if this is a thing with the ISS? Does its orbital plane suffer from nodal precession around the earth's polar axis? I'd always imagined its orientation would be fixed in inertial space, but now I'm inclined to say no?

Answer 1

I'd always imagined its orientation would be fixed in inertial space, but now I'm inclined to say no?

Trust your inclination!

If Earth had perfect spherical symmetry then the smaller effects like gravity from the Sun and Moon and some other smaller effect would still perturb the ISS' orbital parameters (in addition to the big one - drag - which will pull it back to Earth if it doesn't keep lifting itself back up)

Have a look at answers to

• How often does the ISS orbit align with the day/night terminator?
• Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?

In this answer I show how to calculate the rate of precession based on $J_2$ which is some measure of how oblate the Earth's gravity field is. The effect is about 1 part per thousand in LEO, which is pretty big!

A circular orbit with roughly the same altitude and inclination as the ISS will precess 360 ° in roughly 60 days.

Interestingly the inclination of the orbit is so high that when the nodes are over the morning and evening terminator (i.e. perpendicular to the direction of the Sun) it spends only a small fraction of one orbit in Earth's shadow. But 15 days later when the nodes fall closer to the Sun-Earth line the duration is much closer to half an orbital period.

• Would a lower LEO ISS orbit really have a shorter eclipse duration than a higher one?
• Does the ISS still "need" to be at around 400 km?
• How do I determine the ground-track period of a LEO satellite?

"But wait, there's more!" Sun-synchronous orbits

You can play orbital mechanical games with $J_2$. From that linked answer:

Let's see if we can calculate it. From this answer:

The first equation in Wikipedia's Nodal precession for the rate of precession $\omega_p$:

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

depends on the parameters of the orbit ($a, \epsilon, \omega$, i) and the Earth's equatorial radius $R_E$ and its $J_2$ term.

If you think about it, from an "inertial Earth frame"'s perspective the Sun revolves apparently around Earth every 365 days. If you play with $a$, $\epsilon$ and $i$ you can get the same rate.

Then with a high inclination orbit your satellite will always be in sunshine, it will remain at a more constant temperature (important for some space telescopes or Earth observation telescopes in LEO) and and won't need huge batteries to power its power-hungry synthetic aperture radar transmitter.

Answer 2

All orbits about the Earth except for polar orbits will see their line of nodes precess due to the Earth's equatorial bulge. The effect is strongest for near-equatorial orbits and for objects in low Earth orbit. In the case of the International Space Station, this nodal precession causes the ISS's right ascension of ascending node to decrease by about 5° per day.