Hello, I am a Master's student studying Astrodynamics.
I am using Machine Learning to estimate future Orbital Element of satellites.
Recently, I found that True Longitude, Mean Longitude, and Argument of Latitude show almost linear form, and I would like to find information about this.
Is there a paper or journal related to this result?
And can I consider these as linear?
Hi, true anomaly and argument of latitude do not vary linearly with time even in the case of a Keplerian motion. At the opposite, the mean anomaly is a linear function of time for a Keplerian motion:
M=M0 + n(t-t0) with n=sqrt(mu/a^3)
where mu is the Earth's gravitational constant and a is the semimajor axis of the orbit.
As soon as you consider orbital perturbations (atmospheric drag, harmonics of the Earth's potential, lunisolar perturbation, etc..), this changes the picture.
Best
@Richard Epenoy
Thank you for your reply.
I am trying to estimate future state of satellites with only past orbital elements, using Machine Learning.
I am using GMAT as a propagator to make my data.
I had good result on other elements, but had hard time fitting with mean anomaly(M) and argument of periapsis(ω).
Trying different settings with GMAT, I found that when eccentricity is low ( e < 0.001), M and ω show non-linear movement whereas for high eccentricity they are linear. This is what you mentioned, considering perturbations.
In this case, I couldn't find a way to fit the data with machine learning.
After that, in the same condition, I found that True Longitude( = Ω+ ω + ν) , Mean Longitude( = Ω + ω + M ) , Argument of Latitude ( = ν + ω) moves linearly.
This is just one case, so I am trying to get more data to see if this really works.
So, I would like to know if there is any information on these three elements, True Longitude, Mean Longitude, and Argument of Latitude. Thank you very much.
Regards.
Short answer is, for a perfectly circular Keplerian orbit with no perturbations, the three quantities should vary linearly with time. Mean longitude does vary linearly with time even if the orbit is not perfectly circular. Let's elaborate, shall we!
First let me introduce my nomenclature here:
ML: Mean Longitude
TL: True Longitude
MA: Mean Anomaly
TA: True Anomaly
EA: Eccentric Anomaly
AL: argument of latitude
omega: argument of periapsis
Omega: right ascension of the ascending node
As illustrated in the answer of Dr. Richard Epenoy on this same question, MA does change linearly with time for any elliptical Keplarian orbit with no perturbations.
Defining the ML as:
ML = omega + Omega + MA
For elliptical Keplarian orbit with no perturbations (omega & Omega are constants), ML should vary linearly as MA does vary linearly.
Let's talk about TL, but first let me talk about TA. For a perfectly circular orbit (i.e. e=0), MA = TA = EA
As EA changes linearly with time for circular orbits, then TA does as well.
Please consult this answer for illustration:
https://space.stackexchange.com/questions/23988/how-to-get-true-anomaly-from-time
We define the TL as follows:
TL = omega + Omega + TA
Again, if TA is linear, then TL has to be linear as well. This is only true for circular Keplarian orbits with no perturbations. In fact, TL in this case is exactly equal to ML as TA = MA, which is what your code is resulting.
As for the argument of latitude (AL), you can write the famous equation for it as follows:
AL = TA + omega
As TA is established to be linear, AL should vary linearly as well (Again, for a circular orbit with no perturbations).
If you would like to read more, you can give a look at Vallado's Fundamentals of Astrodynamics and Applications.
Dear Kim, there is no reason to explain that the three parameters above move linearly with time. With kind of orbit are you considering (Low Earth Orbit, GEO orbite, ...)? By propagating with GMAT you should clearly find that v for e.G; moves quickly near the perigee of the orbit (cas e is great enough). Best
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