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I'm not sure for how many planetary systems we have a fairly good idea of the inclination and "argument of periapsis" of the orbital plane. I'm assuming there are a number of them that we can estimate due to co-planar assumptions in a multi-planet system or some other means.

I am asking if there are enough "known" orbital planes that we can have some statistical idea whether or not the orbital planes are somewhat aligned with the galactic plane or whether local forces would dominate galactic orbit angular momentum. (I'm thinking the latter is true.)

We would have to keep in mind of course that a system "directly above" us and parallel to the galactic plane would be "face on".

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    $\begingroup$ I think that this problem would be very difficult to answer because almost all of our detection methods rely on us being able to see the system edge-on. Since our system is aligned with the galactic disk, most of the systems we can find are also aligned with the disk. I suppose you could make an assumption that every star has a planetary system, then take a volume of space and count how many systems we know of. The remaining stars would have systems that aren't as aligned with the galactic disk. $\endgroup$
    – Phiteros
    Commented Mar 26, 2017 at 23:54
  • $\begingroup$ I just reviewed orbital elements and found out that the "argument of periapsis" can be known as the "argument of periastron" and a third element is known as the "longitude of the ascending node". I think I've got it now. $\endgroup$ Commented Mar 27, 2017 at 1:39
  • $\begingroup$ Somewhat confused as to how that answers your question, but okay. $\endgroup$
    – Phiteros
    Commented Mar 27, 2017 at 1:40
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    $\begingroup$ @Phiteros I don't follow your argument. A transiting/doppler-detected planet can be rotated by any angle around our line of sight and yield the same observational result. The orbital plane is undefined. $\endgroup$
    – ProfRob
    Commented Mar 27, 2017 at 6:44
  • $\begingroup$ I don't see how the argument of periapsis applies to the rest of this question. $\endgroup$
    – zephyr
    Commented Mar 27, 2017 at 12:46

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Unless I'm missing something we know very little about the orbital planes of individual systems.

Most exoplanetary systems (all of the transiting ones, whereas it is just a strong bias for the doppler-discovered planets) are discovered because they have high orbital inclination. This means that the planet's orbit takes it across the stellar disc as we view it. However, the entire arrangement can be rotated around the line of sight by any angle and this would give exactly the same observational signature.

That being the case, the only test I can immediately suggest is that if the planes of planetary systems were preferentially aligned with the galactic plane (ours isn't), then it would result in a lower planetary detection rate for stars in directions perpendicular to the Galactic plane in comparison with stars lying in the Galactic plane.

There may be some mileage in pursuing this, though I don't know of any results. For example the main Kepler field was directed just out of the Galactic plane, but the K2 fields are at a variety of Galactic latitudes along the ecliptic plane. If one could normalise somehow for the types of stars observed in each field and the differing data qualities then it might be possible to look at the "planetary occurrence rate" as a function of Galactic latitude.

The "all-sky" ground-based surveys are less suitable for this because they tend to avoid the Galactic plane because it is too crowded with targets.

The question can be more broadly addressed by looking at visual binary systems - i.e. those where you can resolve the components and study the orbit. There, it is possible to say something about the orbital planes of individual systems. A paper by Agati et al. (2015) has looked at this. They collected 51 systems with the requisite information, that are close to the Sun. They found only weak evidence for any anisotropy of the orbital poles and this weak alignment was closer to the ecliptic pole than the Galactic pole.

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