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I am developing a fictional system which I want to calculate to as much detail as I can. Thus I have a question about astrophysics and astrophysical formulae.

I know that Apsidal Precession is effected by Relativity, Tides, Quadrupole, and perturbations from other planets. Which, when added together, gets the total apsidal precession per unit time.

My question here is, how would you find how much other planets perturb the apses of a moon around a planet.

I have the masses and semi-major axis and eccentricity and polar-equatorial radii difference, and inclination to the ecliptic for each planet in the system.

How, from this, can I find the Apsidal perturbation in degrees per year, if I can't from this information, what other information do I need about the system.

Your help would be much appreciated. Please and thank you.

PS: if it makes you feel more comfortable, you can explain the physics in terms of calculating it for Earth, as long as you still give the formulae for finding that value.

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  • $\begingroup$ You might start with the Wikipedia entry and its references which seems to provide the detail you're looking for. en.wikipedia.org/wiki/Apsidal_precession $\endgroup$ Commented Apr 19 at 15:11
  • $\begingroup$ I looked at the references and they don't contain that answer to my question. Sorry. $\endgroup$ Commented Apr 19 at 15:21

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I don't think that perturbations by other planets will lead to an apsidal precession as such. Apsidal precession is due to the gravitational potential having a consistently stronger or weaker radial gradient along the complete orbit as illustrated in the following diagram Apsidal Precession

In the image above, the planet orbiting in the red circle will pull the planet in the blue orbit outwards everywhere (depicted by the green arrows) i.e. slightly weakening the attractive force of the Sun and thus increasing its gradient outwards. Only the fact that the perturbing force is always directed outwards from the orbit is responsible for the apsidal precession (forwards in this case)

Now if you have a moon orbiting a planet, the situation will look schematically like this

Perturbed Moon

In this case the perturbing force due the planet in the red orbit will pull the orbiting Moon outwards in one half of its orbit but inwards (towards its own planet) in the other half. This won't lead to a consistent precession but rather some periodical perturbation (the direction of the precession depends on the sign of the radial component of the perturbing force). I think trying to model this theoretically by some simple algorithm is a futile attempt which would probably result in larger errors than ignoring the perturbation altogether. You may only be able to correctly calculate the effect of other planets on the Moon's orbit by large scale numerical computations.

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  • $\begingroup$ This is an interesting answer. The apsidal precession of Earth's Moon is attributed primarily to the gravity gradient from the Sun (if I understand correctly) so this argument must allow for that effect while ruling out any affect from other planets. (Kibble, T. and Berkshire, F. H. (2011) Classical mechanics. Imperial College Press, 5th edition is cited in The Earths long-term climate changes and ice ages: a derivation of Milankovitch cycles from first principles, Section II. Precession of the Orbit of the Moon) $\endgroup$
    – uhoh
    Commented Apr 20 at 1:47
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    $\begingroup$ @uhoh The paper you linked to deals with the nodal precession (rotation of the orbital plane) not the apsidal precession. The latter exists as well though for the Moon (see en.wikipedia.org/wiki/Lunar_precession ) and is certainly caused by the Sun as well, but in this case the top picture I showed in my answer would apply if you consider the situation in the geocentric system. The red circle would then be the Sun. And due to its mass, the precession is therefore quite massive (360 degrees in 8.85 years). I don't see how any planet could contribute to the Moon's precession. $\endgroup$
    – Thomas
    Commented Apr 20 at 7:54
  • $\begingroup$ Okay thanks for the reply. I will strap-in and read further. :-) $\endgroup$
    – uhoh
    Commented Apr 20 at 15:13

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