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This answer introduced me to the word Areostationary Orbit - (hint, comes from Ares), an orbit that would remain stationary above a point on the equator of Mars.

The article mentions that gravitational interaction with the two moons of Mars, Phobos in particular, would be a factor that would require station keeping.

This 2012 paper discusses locations in the gravity field of mars where a Areostationary orbit would be possible due to small variations in the gravity field of the planet. Certain points seem to be slightly stable, and satellites put in certain particular locations above the Martian equator would tend to stay there, although they would be orbiting these locations by a few degrees. If the wobble it too large, it sort-of defeats the "stationary" part. You'd have to do some steering, and at that point it would be better to just call it a synchronous orbit. (Fixed dishes will be superseded by steerable phased arrays well before then).

However, once I started seeing terms like monodromy matrix and heteroclinic orbits in that paper, I decided to ask here instead.

Question: What factors would make station-keeping of an Areostationary Mars satellite necessary? Would the pull of the Moons destabilize orbits at Mars' natural gravitational Areostationary points over the period of years, requiring station-keeping?

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What factors would make station-keeping of an Areostationary Mars satellite necessary?

Mars' rather lumpy gravitational field, The Sun's gravity, Mars' two moons, solid body tides, and solar radiation pressure.

That 2012 article addressed Mars' rather lumpy gravitational field (partly), but it didn't address any of those other perturbing factors. Those other factors will exist, even at the supposed stationary points investigated in that article. Those stationary points also exist for geosynchronous satellites. However, for a geosynchronous satellite, the fuel penalty for not being near one of those stationary points are small. The perturbations from the Moon and Sun dominate, requiring considerable fuel (about 40 m/s per year) for north-south stationkeeping. The total stationkeeping budget for a geostationary satellite is about 50 m/s per year.

That north-south stationkeeping will still be needed for an aerostationary satellite. Mars axial tilt is a bit more than is the Earth's. This tilt makes the orbits of equatorial orbits become inclined due to third body effects (e.g., the Sun, the Moon in the case of the Earth, and Mars' two moons in the case of Mars). Solar radiation pressure causes circular orbits to become non-circular. Stationkeeping will still be needed, even in the vicinity of those stationary points.

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  • $\begingroup$ OK, I think I understand - while Mars' lumpy field can tend to keep a satellite orbiting near a stationary point, Mars' moons can still excite north-south motion. The moons are closer and orbiting much faster, but also much lower mass than Earth's Moon. Is there any way to roughly estimate (or read about) if it's strong enough to have an effect over say 10 years? $\endgroup$ – uhoh Dec 17 '16 at 23:44
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    $\begingroup$ The Sun is also a big factor, @uhoh. One way to get a rough estimate is to model Phobos and Deimos as rings of mass and apply perturbation theory (e.g., planetary equations of motion) and average over time. I'm not going to do that; it's a lot of work. An alternative is to use a numerical simulation. You'll need a Mars rotation model, a Mars gravity model, ephemerides for the Sun, Mars, Phobos, and Deimos, a good numerical integrator (e.g., not RK4), and a controller (perfect control should do) that keeps inclination, drift, and eccentricity within limits. Also too much work. $\endgroup$ – David Hammen Dec 18 '16 at 0:24
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    $\begingroup$ Plucking a number out of the clear blue sky, I'd guess on the order of 10 m/s per year for stationkeeping for an aerostationary orbit near one of those stationary points. That "on the order of" gives my guess a lot of leeway. 3.4 m/s is within that range, as is 32 m/s. $\endgroup$ – David Hammen Dec 18 '16 at 0:26
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    $\begingroup$ I've stumbled across adsabs.harvard.edu/full/1977SvA....21..513Z clicking "print this article" yields the pdf. For Earth, $\lambda_{22} \approx -15°$ puts the $C_{22}$ maximum (okay, minimum potential) at 15° West longitude as well. This longitude would be an example of the stationary point you've mentioned, wouldn't it? $\endgroup$ – uhoh Jun 4 '18 at 5:21
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    $\begingroup$ @costrom - Not much. The closest Jupiter gets to Mars is 3.5 astronomical units, and those encounters are rare and short-lived. That's over twice the mean distance between Mars and the Sun. That alone makes Jupiter's contribution less than a tenth that of the Sun due to the inverse cube relation of third body effects. Divide by 1000 (the ratio of Jupiter's mass to that of the Sun) and you get a very tiny effect. $\endgroup$ – David Hammen Nov 6 '18 at 14:28

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