View Full Version : Satelite's Precessions

mike42

2008-Feb-26, 04:05 PM

Hi Folks!

I have a question about precession, especially concerning moons (satelites):

1) Nodal precession:

As I understood, the nodes of the (earth-)moon rotate backwards, i.e. their longitudes getting smaller. Is that generally true for prograde orbits and is it generally the opposite for retrograde orbits? I found a few papers involving some serious maths (don't have the refs at hand) seeming to state that everything depends on inclination and eccentricity, but I am not sure, if I got that right.

2) Apsidal precession:

Are the apsides usually going forward (in direction of the orbit), i.e. their argument getting larger for both prograde and retrograde orbit? Or is it rather their longitude getting always larger?

Anyway, I found this table of satelite orbits on the JPL-page (http://ssd.jpl.nasa.gov/?sat_elem) also giving precession periods. But they don't specify the direction of either of them. Does anyone know a place where I can find a more detailed list?

thanks for any help!

Mike

grant hutchison

2008-Feb-28, 12:01 AM

The nodes regress: they always move in the opposite sense to the satellite motion. The rate of regression depends on the inclination, the eccentricity, the oblateness of the parent body, and the ratio of orbital semiaxis to parent radius.

The apsides rotate in the orbital plane, which is a change in argument. The rate of rotation likewise depends on the inclination, the eccentricity, the oblateness of the parent body, and the ratio of orbital semiaxis to parent radius. Orbits rotate within the plane in the same sense as satellite motion when inclination is less than ~63.4 degrees, and in the opposite sense at higher inclinations.

Edit: In exact terms, the cutoff occurs when inclination = arcsin(sqrt(4/5)).

Grant Hutchison

mike42

2008-Feb-28, 08:52 AM

The nodes regress: they always move in the opposite sense to the satellite motion.

So, an orbit with inclination=90 deg (wrt its laplace plane) does not precess at all in terms of nodes?

Btw. how is pro/retrograde defined exactly? Wrt the laplace plane, ie. i>90 meaning retrograde? Or is it wrt to the orbital plane of the parent body, or even its equatorial plane? (I found various, mostly inexact definitions.)

The apsides rotate in the orbital plane, which is a change in argument. [......] Orbits rotate within the plane in the same sense as satellite motion when inclination is less than ~63.4 degrees, and in the opposite sense at higher inclinations.

...or bigger than 116.6 degrees, I suppose?

So, if an orbit with i=10 degrees has a nodal regression of 2 deg/cy and its (argument of) apsides rotate forward (say with 1 deg/cy), it still may well be that the longitude of the apsides regress? In other words, whether the apsides move clockwise or counter-clockwise (in terms of ecliptic longitude) one can not judge from the direction of motion alone. Is that correct?

Mike

grant hutchison

2008-Feb-28, 11:04 AM

So, an orbit with inclination=90 deg (wrt its laplace plane) does not precess at all in terms of nodes?Well, what I've written so far refers to bodies in a low enough orbit that their Laplacian plane is the equatorial plane. (Your mention of "satellites" made me think of artificial satellites in LEO.) It should also apply, with slightly different variables, to objects that are so far out they precess in the ecliptic plane, effectively uninfluenced by the parent planet's oblateness. In between (as I think you know), there's a zone where the Laplacian plane lies between ecliptic and equatorial: I'm slightly nervous of making strong claims about that zone, given that the satellite is precessing under two different influences of similar magnitude. I think that it's still true that a 90-degree inclination to the Laplacian gives you stationary nodes on the Laplacian, but I actually don't know.

Btw. how is pro/retrograde defined exactly? Wrt the laplace plane, ie. i>90 meaning retrograde? Or is it wrt to the orbital plane of the parent body, or even its equatorial plane? (I found various, mostly inexact definitions.)I think you can define the inclination relative to any plane you care to use. There are satellites of Uranus, for instance, which actually flip from direct to retrograde and back again (relative to Uranus' equator) as their orbits precess in the ecliptic plane.

...or bigger than 116.6 degrees, I suppose?Yes, that's why I came back and edited in the trig definition, so you could see the behaviour at inclinations > 90 degrees.

...So, if an orbit with i=10 degrees has a nodal regression of 2 deg/cy and its (argument of) apsides rotate forward (say with 1 deg/cy), it still may well be that the longitude of the apsides regress? In other words, whether the apsides move clockwise or counter-clockwise (in terms of ecliptic longitude) one can not judge from the direction of motion alone. Is that correct?Yep, I think that's correct. If you find an inclination close to the critical value, the apsides will rotate very slowly, but the nodes will still regress. So you'll always be able to find a family of orbits in the vicinity of the critical inclination for which the nodal regression dominates: while the argument of the periapse increases, its longitude decreases.

Grant Hutchison

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