Talk:Argument of periapsis

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Suggest Correction to Figure 1[edit]

Figure 1 is intended to show the argument of periapsis (omega). In the figure, omega is labeled 'argument of perihelion'. Perhaps it should be relabeled 'argument of periapsis' to agree with the title.

Mike Kestell 03:35, 23 August 2007 (UTC)

Perihelion is what the periapsis is called if the orbit is around the Sun. The image is a svg so it would be easy to change it. Bobby D. DS. 00:04, 25 September 2007 (UTC)

Applicability to exoplanets?[edit]

This article, and the related articles on longitude of the periapsis, longitude of the ascending node, and orbital elements as well as the related figures (File:Orbit1.svg and File:Angular_Parameters_of_Elliptical_Orbit.png) illustrating those elements, are confusing when applied to exoplanets and to their values for "ω" reported using the arg_peri parameter in the orbit planetbox. A reader seeking an understanding of how these concepts apply to exoplanet orbits, and in particular to the to the values of "the argument of periastron" given in the inboxes in exoplanet articles would only be confused by these materials.

I'm reluctant to try to correct this myself, since I'm not certain I understand the relevant definitions well enough, or how they are used in the exoplanet literature.

As near as I can tell, the bottom line is that, for exoplanets, Ω is treated as being zero, so that ϖ and ω are interchangeable, and ω is measured orbitwise from the point where the approaching planet passes through the plane perpendicular to the line-of-site.

This is difficult to map to the articles mentioned above. These basically say that the longitude of the ascending node, Ω, is measured in the reference plane, from the reference direction to the ascending node, and that the argument of periapsis, ω, is measured in the orbital plane, from the ascending node to the pericenter. But for exoplanets, we don't generally know the inclination, i, and we we treat these planes as the same (e.g. for curve fitting, dealing with the uncertainy for i by reporting a minimum mass). Moreover, since, for all exoplanets, the reference direction is the line-of-site, the (actual) nodes will be always be the points where the orbit passes through the plane perpendicular to the line-of-site, and Ω should thus either be 90° or 270°.

That's as far as the application of the definitions given in the articles cited above get us. Beyond that, I'm guessing. It appears that the convention (perhaps for all edge-on orbits?) is just to (a) forget about Ω and (b) treat the "approaching node" as the ascending node. I.e., we only measure "from" the that node to the pericenter (ω = ϖ) and don't bother measring "to" there from the reference direction (Ω).

If I have that right, I can fix these articles. If not, can someone please enlighten me (and fix the articles). AldaronT/C 03:25, 24 July 2009 (UTC)

  1. This issue has nothing to do with exoplanets per se. For any orbit determined spectroscopically, whether it's a spectroscopic binary orbit or an exoplanet orbit determined using the radial velocity method, the angle of inclination and the longitude of the node will both be unknown.
  2. The angle ω, the argument of periastron, is still defined in the normal way. Unfortunately, many sources (e.g. the Catalog of Nearby Exoplanets) call ω the longitude of periastron; strictly speaking, this is a misnomer. Whatever it is called, though, ω is still the angle from the node to the periastron, measured in the plane of the orbit.
  3. The inclination is unknown. But, since it's not necessarily 90°, the orbit is not necessarily edge-on.
  4. Ω is unknown and doesn't have to be 0°, 90° or 270°.
  5. Since Ω is unknown, ϖ is also unknown, so it's not tabulated.
  6. Also, remember that for exoplanets, M sin i and a are not fitted directly but are estimated using the estimated mass of the star and Kepler's Third Law. See equations (1) and (2) in [1].
In short, although we don't know two of the orbital elements, the other four are still defined in the normal way. Spacepotato (talk) 06:50, 24 July 2009 (UTC)
Thanks for the thorough reply. Maybe it's not clear from my question, but I see that points 1 and 2 are true. The crux of my question is about the contradiction between 1 ("the longitude of the node will ... be unknown") and 2 ("ω ... is still defined in the normal way"): Since the normal way of defining ω is to measure from the ascending node, and since the location of that node is unknown, there has to be some convention for specifying it. Moreover, that convention seems logically constrained by the fact that the (true) inclination is defined in such a way that the nodes must lie in the plane perpendicular to the line of site (and thus at either 90° or 270°). So it seems that the convention is (a) to take the "approaching" node at 270° as the starting point for measuring ω (in the normal way) and (b) to just forget about Ω since it's arbitrarily fixed by the relationship between inclination and reference direction.
Regardless of how all of this is arrived at, the relevant question is whether the convention for exoplanets is to treat Ω as "unknown", so that ϖ is also unknown, or whether the way inclination is defined makes Ω the same for all exoplanets so that it is just ignored (in fact treated as zero), so that ω and ϖ can be treated as equivalent (as many authors apparently explicitly do, as you point out in 2). AldaronT/C 16:36, 24 July 2009 (UTC)
  1. The convention is to treat Ω as unknown (which it is.)
  2. It's not that ω and ϖ are equivalent, but that ω, instead of ϖ, is being called the longitude of periastron. To quote our article: "Sometimes, the term longitude of periapsis may be used to refer to ω, the angle between the ascending node and the periapsis..." The quantity ϖ is not being used at all.
  3. In your first paragraph immediately above, I don't understand the reasoning you are using to conclude that the nodes have to lie at 90° or 270°. They don't have to; in fact, for the orbits of visual binary stars, we can measure Ω, and it can take on any value. To summarize the geometry of the situation:
    (a) The reference plane is the plane of the sky, which is perpendicular to the line-of-sight and tangent to the celestial sphere.
    (b) The plane of the orbit is inclined to this plane at the angle i. So, unless i=0, it will intersect this plane in a line, the line of nodes.
    (c) Draw the line of nodes on the celestial sphere.
    (d) At the location of the system on the celestial sphere, draw a line towards the North Celestial Pole.
    (e) Ω is the angle from the line drawn in (d) to the line drawn in (c).
Spacepotato (talk) 18:32, 24 July 2009 (UTC)
Ah: 3.b. is the critical point. It wasn't (and actually still isn't, quite) clear to me that that was what the convention for extrasolar planets. My reasoning was all based on the guess that the reference plane was coplanar with (rather than perpendicular to) the line of site. AldaronT/C 19:13, 24 July 2009 (UTC)

Formula for equatorial orbits incorrect?[edit]

The page states that for equatorial orbits, the argument of periapsis is assumed to be:

 \omega = \arccos { {e_x} \over { \mathbf{\left |e \right |} }}

Can this be right? Argument of periapsis can be in range [0, 360) degrees, while the arccos function has the range of [0, 180] degrees. Formula obviously won't work for periapsides in the III and IV quadrant.

Makes more sense if the formula is:

 \omega = \arctan2 ({e_y}, {e_x})

Because its range are all four quadrants. However, I have no reference for this...

The formula would be correct if the longitude of ascending node is defined differently (ie. 180 degrees for quadrant III, IV periapsides), but that's not what the wikipedia page on ascending node states. —Preceding unsigned comment added by 109.121.79.228 (talk) 16:01, 19 June 2010 (UTC)

Why do |n| and |e| appear in the denominator? If they're unit vectors then |n| and |e| are both 1 and are not needed. Should they be deleted, or should the requirement that n and e are unit vectors be dropped? Vaughan Pratt (talk) 21:22, 22 December 2013 (UTC)
Good point. Either way would work, but to avoid confusion with other Wikipedia articles, I removed the unit-vector requirement. --Lasunncty (talk) 06:35, 1 January 2014 (UTC)

Diagram could be improved[edit]

The diagram shows a perfectly circular orbit with periapsis exactly 90 degrees from the node. This is unfortunate becuase it masks many subtle details. For example, precession causes radial motion of the node. The diagram also implies that periapsis occurs at the highest point over the ecliptic. NOrbeck (talk) 10:36, 16 July 2010 (UTC)

I'll work on it. I didn't intentionally design it that way, but I admit it does appear as you say. I've actually been admiring this image: Orbital elements.svg. Would something like that be better? --Lasunncty (talk) 12:54, 16 July 2010 (UTC)
I agree. Also, the periapse should be labeled. Trying to convey 3D concepts in 2D is inherently confusing (like this figure and the cited alternative); I think two side-by-side images showing the orbit in two different planes would be better. Jdeast (talk) 15:41, 21 June 2011 (UTC)

arg of periapsis 90deg at northmost distance ?[edit]

Article says this: An argument of periapsis of 90° means that the orbiting body will reach periapsis at its northmost distance from the plane of reference. which is true, given diagram in Fig. 1. However, if the inclination of the orbit is between 180 and 360 degrees, in that case the article should specify that an argument of periapsis of 90° means that the orbiting body will reach periapsis at its southernmost distance from the plane of reference. Yes the orbit in the diagram should be a little elliptical so it is clear what periapsis is, not just some radius on a circle. — Preceding unsigned comment added by 91.201.80.240 (talk) 12:11, 5 April 2012 (UTC)

what?? what??[edit]

I'm smart... really, I usually am. But wow, I have no idea how to interpret that picture or this article. "plane of reference from North to South"? wait, what North? What South? The plane of reference in the picture does not intersect the planet at all! What on earth could North or South possibly mean in this context? Red Slash 02:42, 30 April 2014 (UTC)

As the body goes around its orbit, it will cross the plane of reference at two points, the ascending and descending nodes. Does that help? --Lasunncty (talk) 03:20, 1 May 2014 (UTC)
Hmm, let me be more specific.
"The argument of periapsis (also called argument of perifocus or argument of pericenter), symbolized as ω, is one of the orbital elements of an orbiting body. Specifically, ω is the angle between the orbit's periapsis (the point of closest approach to the central point) and the orbit's ascending node (the point where the body crosses the plane of reference from South to North)."
First, which plane of reference? The linked article lists no fewer than four possibilities. Once I figure out what the plane of reference is, maybe the concept of north/south will have some meaning. In fact, I can't see any possible meaning or reason behind this article without knowing that. Is it just some arbitrary plane somewhere in space?
Anyway, I think once I figure out (or you helpfully tell me/add to the article) what the plane of reference is, it'll make a lot more sense. I think the root of a lot of my confusion is that awful picture. Like, I only just figured out that the center of the diagram must be what the planet/moon/whatever in question is orbiting around. That's not labeled. And the label for the orbiting body is absurd. "Celestial object"? Seriously? The article text uses "orbiting body"--what the heck is this celestial object? I think I know now that it's the orbiter, not the orbitee, but it's really unclear. The plane of reference is not explained or defined, there are multiple unexplained Greek symbols, and the point on the orbit that the blue line leads to is not defined. Why does it point there? If I understand the article, that point is the unique point in the orbit where the orbiting body is closest to the orbited body... but why is that not labeled? And why does the plane of reference travel through the sun? (Or whatever the center of the body's orbit is.)
Thanks for all your help! Red Slash 03:44, 2 May 2014 (UTC)
The linked article answers your question: it depends on the celestial body in question. The only tricky case is when that body is the planet Earth, since the ecliptic is the plane of Earth's orbit. Since the Earth never ascends or descends (other than very slightly due to tugs from Jupiter, Venus, etc.) the ascending node is by convention taken to be zero. For all other planets the reference plane is still the ecliptic, i.e. the plane of Earth's orbit, and you should have no problem with that since no two planets in our system have the same orbital plane. Planetary systems of other stars currently don't have reference planes (that I'm aware of anyway), though the star's equatorial plane might be a reasonable choice in theory, though in practice determining it might be tricky. Binary stars, no idea. Vaughan Pratt (talk) 07:21, 2 May 2014 (UTC)