Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?

Question

I came across the orbital data for a low Earth orbit spacecraft and one thing which I am not able to understand is why does its Right Ascension of ascending node and Argument of Perigee keep changing every day.

Firstly, it seems extremely unrealistic to me, that they are being changed by manoeuvring the spacecraft because:

1. They are very costly affairs and for doing them over the period of months, one would need a mammoth amount of propellant.
2. I can't see what purpose it would serve.

So, to me it looks like they are either changing by themselves or the orbital data itself is wrong. I suspect the data being wrong as I haven't come across any phenomenon where orbital plane keeps changing and and also rotating by itself with time. It would require strong external disturbances for that to happen by itself and I doubt there are such strong and sustaining disturbances acting on the spacecraft. So, is the data incorrect?

Edit: I checked data for two more spacecrafts on the same website and they also seem to be behaving similarly. This is either something I don't know of, or the data is garbage.

Answer 1

You are correct to a point that the RA of the ascending node and argument of perigee won't change over time without some external force acting upon the satellite. In a simplified gravitational field, an object's orbital plane remains fixed.

Unfortunately, reality is a lot more messy.

Earth's gravitational field differs significantly from that created by a hypothetical point mass - this is known as the Geopotential Model. In particular, the Earth's oblateness (equatorial bulge), defined by the J2 coefficient, has a significant effect on orbits. It causes them to precess over time, changing the RA/longitude of their ascending nodes.

Furthermore, perturbing influences from other bodies in the solar system need to be accounted for when accurately plotting orbits - they can have small but non-negligible effects.

As to your second point, these effects can be very useful for satellites. For example, Sun Synchronous Orbits are precisely designed to precess at a rate of 360° per year, maintaining a constant angle between the orbital plane and the sun. This is ideal eg. for keeping a fixed illumination for surface observation.

See this related question on J2 pertubations for some more detail on the maths.

Answer 2

@Jack's answer is excellent, I'll just address @Niket's question in this comment a little further.

Just to be sure, is the data realistic? Like does the day to day variation seem realistic?

tl;dr: The drift makes sense and is within 1% of what we can easily calculate.

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There are some of the orbital parameters plotted at the bottom of the page. The RA of the ascending node cycles through 360 degrees in about 60 days. That's about right for a ~45 degree inclined orbit in LEO, the ISS does that.

It's a big deal for the ISS because it means it cycles through periods where it goes into Earth's shadow every 90 minutes and then periods where it's in constant daylight as its orbital plane rotates around the Earth.

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.

Let's use 6378137 meters for $R_E$ (from this answer) and 1.0826E-03 for $J_2$ (from this answer).

The satellite's period $T$ in your data table is 15.59029 revolutions per day, or about 5542 seconds. Then use:

$$\omega = \frac{2 \pi}{T} = 0.0011338 \ \text{sec}^{-1}.$$

From this answer get

$$a^3 = \frac{GM}{\omega^2}$$

where GM is Earth's standard gravitational parameter of about 3.986E+14 m^3/s^2. That makes $a=$ 6768601 meters, or an altitude of about 390 km.

Plug those all in to the first equation, and we get $\omega_p = -1.2149 \times 10^{-6} \ \text{sec}^{-1}$ If we multiply that by 60 days or 5184000 seconds, we get -6.298 which is almost exactly $-2 \pi$ or one complete cycle, just what the plot shows!

The argument of perihelion at first looks like it drifts steadily then flips by 180 degrees around day 85, but that's actually a smooth shape change since the eccentricity hits zero and bounces back. That looks like a natural precession as well, and not an orbital maneuver.

So @Jack's answer nailed it.

Python script for the plot is available here: https://pastebin.com/rx1np9Mv