Talk:Hilda asteroid

Latest comment: 1 year ago by Tamfang in topic special eccentricities

Family

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What's this bit about not forming a true asteroid family because they don't "descend from a common parent body"? As I understand, an asteroid family is defined by having common orbital elements--anyway, is it even commonly accepted that any of the asteroids have a common parent body? I'm going to comment it out as OR. --Aelffin 17:06, 30 August 2006 (UTC)Reply

  • I'm pretty sure that info is right, but I can't seem to find exactly the right reference - all the papers I've looked at call the Hildas a "group" not a "family". Hirayama proposed that groups came from a common parent body - and the Murra & Dermott book identifies Koronis, Eos and Termis directly as being descended from a common body and recommends Low et al 1984 and Neugebauer et al 1984 for confirmation that families are descended from a common group. So it seems like all the info you edited out is correct and can be included. Asteroid families are definitely found through common orbital elements, but it doesn't seem to be how they're defined. (Just as a galaxy cluster is found through it's infrared or X-Ray emission, but that's not how it's defined.) I don't believe it to be OR. WilyD 17:28, 30 August 2006 (UTC)Reply
Okay, I read about it in Asteroid family and I stand corrected. It just struck me as wrong at first. Referencing would certainly help though. --Aelffin 14:17, 31 August 2006 (UTC)Reply
I can accept that, but I haven't been able to find "just the right reference", you know? It's all a few references plus a little inference plus general OR hanging out with other astronomers. WilyD 14:19, 31 August 2006 (UTC)Reply
A little bit of OR never hurt anybody; at least not when it comes straight from the proverbial horses' mouths. --Aelffin 15:00, 31 August 2006 (UTC)Reply

Elliptic or triangular orbits

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I think the article should point out that the orbit of each Hilda asteroid is generally elliptic, and the apparent triangle is just a dynamic sum of all Hildas at each point in time with a reference rotating along Jupiter's orbit. A figure should be included to illustrate a single Hilda's orbit demonstrating that it travels through all of Jupiter's L3, L4 and L5 through 3 cycles (2 Jupiter cycles).

Background: As an amateur astronomer I am well aware of Kepler's laws, and so it astonished me to read this article about a special group of asteroids apparently moving in triangular orbits. This misunderstanding was based on the following text and also the graphs:

Hildas move in their orbits so that their aphelia put them opposite Jupiter, or 60 degrees ahead of or behind Jupiter at the L4 and L5 Lagrangian points. Over three successive orbits each Hilda asteroid passes through all of these three points in sequence.

Not really believing it I sketched possible orbits by hand and realised: These asteroids, as most celestial bodies, generally demonstrate elliptic orbits where the speed is slower at aphelion (at a Hilda triangular vertex, near Jupiter's L3, L4 or L5) and faster at perihelion (at the middle of a Hilda triangular edge). Indeed each object passes through L3, L5 and L4 in that order in 3 orbits, but each elliptic orbit remains relatively constant.

I believe the concentration of Hildas at the L3, L4 and L5 points is explained partly by the generally slower speed of objects at their respective aphelia and partly by the special gravitation potentials at the Lagrange points. Assuming this is substantiated by literature it should also be included in the article.

--Eddi (Talk) 02:49, 24 November 2007 (UTC)Reply

The article needs a lot of clearing up, but the grey area in fig. 2 is the orbits of the Hildas. WilyD 03:28, 24 November 2007 (UTC)Reply
I have now anglified some japanese and tried to clarify a few issues. Kindly review and correct.
Fig. 2 shows the sum of all Hildas' orbits over time. I think it would be useful to include a graph of a single Hilda's orbit – albeit generally elliptic – and its constellations with Jupiter. And to emphasise that there are as many orbits as Hildas, but the sum portrays a triangle at each time. --Eddi (Talk) 04:53, 24 November 2007 (UTC)Reply
 

Such a diagram is easy enough to make, but is style consistancy important? I usually make plots with GNUplot, which means they look vaguely like the one on the right (which I made). What's that plot made with? Super Mongo? WilyD 14:47, 24 November 2007 (UTC)Reply

I don't think the style of illustrations needs to be consistant, for example black/white or colour is not important. It would be great if you could make a plot of a Hilda orbit, Jupiter's orbit and their relations through 3 (2) cycles. --Eddi (Talk) 03:06, 26 November 2007 (UTC)Reply
LIke an animated gif? That'd take longer to produce. WilyD 04:43, 26 November 2007 (UTC)Reply
I thought of a diagram with various positions of Jupiter and a Hilda marked through time, but not animated. The time points can be the Hilda's aphelia and perihelia through 3 cycles and some intermediate points. The positions could be numbered e.g. J1, J2, ... and H1, H2, ... to be the positions at time 1, 2, ..., or the colours of the planet and asteroid could be varied to illustrate the time sequence. The diagram might involve just a model asteroid and not necessarily a true Hilda object. --Eddi (Talk) 23:22, 26 November 2007 (UTC)Reply
Well, it's easy enough to use a true Hilda object (probably Hilda) but I'm not sure how to make such a diagram without it being confusing and too busy. WilyD 04:00, 27 November 2007 (UTC)Reply
I'd vote for an animation, if someone can do one.--agr 13:37, 30 November 2007 (UTC)Reply
It's not actually terribly hard, although it "might take some time". WilyD 14:19, 30 November 2007 (UTC)Reply
 
Please complain about what's wrong with this image (I realise the restart isn't perfectly synched. I mean apart from that)

Very well! (And the restart is just fine.) I suppose the sun, planet and asteroid can be shown as dots or something instead of crosses. It would also be neat if the Trojan, Greek and Hilda regions could be indicated on the outer circle orbiting along with Jupiter, so that the asteroid's passing through or near all those regions can be observed. --Eddi (Talk) 23:58, 4 December 2007 (UTC)Reply

Err, I redid it, but all the timesteps are 36.525 days, except the last one which is 29.22 days - the difference maybe too small to notice by eye, but I'll fix that. The symbol representing the Sun, Jupiter and 153 Hilda? Yeah, converting those to any of the other GNUplot shapes is easy enough. I can, of course, draw on the orbits of Mars or Earth or whatever, but animating them is problematic ... this animation runs 8668? .. 8669? days and then restarts, which works fine because it's two orbits of Jupiter and 3 of 153 Hilda, but anything else will jump (well, except Trojans or other Hilda asteroids, obviously). I'm not totally clear on what you're looking for with other bodies, I guess. WilyD 03:15, 5 December 2007 (UTC)Reply
I like it. I'd just change the symbols and not clutter it up. We can explain things in the caption, or maybe exhibit three stills showing the aphelia in relation to Jupiter.--agr (talk) 04:33, 5 December 2007 (UTC)Reply
Is it possible to add 3 objects with orbits and periods overlapping exactly that of Jupiter, only at −60, +60 and 180 degrees angles to Jupiter? They could be small circles to show the approximate regions of the Trojans (L5), Greeks (L4) and Hildas (L3). Then it could be seen directly that the model Hilda moves through or near those 3 circles or objects. --Eddi (Talk) 08:49, 5 December 2007 (UTC)Reply
Circles instead of stars are easy enough (as evidenced). I'm unsure of how to process this other request - the reality of L4,5,3 are not well illustrated with crappy animated plots in some what that's obvious to me. WilyD 15:43, 5 December 2007 (UTC)Reply
Could you make the Hilde circle small and the Sun yellow? I think we can do the rest with an explanation in the caption. Something like: "An animation of the orbits of Jupiter (red) and Hilde (green) about the Sun. If you watch Hilde for three orbits, you will see that whenever Hilde is furthest from the Sun (far left), Jupiter is alternately 60 degrees ahead of Hilde, 60 degrees behind it, or on the opposite side of the Sun."--agr (talk) 22:10, 5 December 2007 (UTC)Reply
Making the Hilda circle small is easy enough ... making the sun yellow makes it hard to see. I'm not sure you want me to do that. WilyD 22:31, 5 December 2007 (UTC)Reply
I want the sun to look like the sun. Orange would do just as well.--agr (talk) 02:28, 14 December 2007 (UTC)Reply
 
At the perihelion of Hilda (pink) it is in conjunction with Jupiter (blue); click here for a series of constellations.

Since I don't know how to make animations I made a series of 12 drawings that span 2 orbits of Jupiter and 3 orbits of Hilda. One constellation is shown to the right; click here for all 12. With this series I try to show that Jupiter's Lagrange points move in orbits that overlap exactly with Jupiter's orbit, only at −60, +60 and 180 degrees angles to the planet, and that successive aphelia of Hilda coincide with L5, L4 and L3. My 12 steps are of course very crude, but I think this principle can be animated rather smoothly. The plot would include 5 objects travelling around the sun, of which 4 with identical orbits (Jupiter, L3, L4 and L5) and one with a smaller orbit (Hilda). If there is no limit to the number of objects in a plot, they could just be represented by different symbols, for example big dots for Jupiter and Hilda, and circles for the Lagrange points. --Eddi (Talk) 00:31, 6 December 2007 (UTC)Reply

 

So I animated that to poor result - I think to do anything like this effectively, we'd need to be in jupiter's frame, and I'm not sure how well that'd go - it'd probably also need to be less jumpy. WilyD 19:40, 13 December 2007 (UTC)Reply

Oh, I didn't mean that my 12 drawings would make a nice animation, only that the principle of these drawings could be animated smoothly. Double the number of drawings might improve the result, but a good animation would require computed plots instead of hand drawings. As these drawings show, we have 4 objects in identical orbits and 1 object in another orbit, all of which are probably computable. --Eddi (Talk) 21:58, 13 December 2007 (UTC)Reply
Err, I meant it only as a test of concept. See how busy and readable it was. WilyD 22:14, 13 December 2007 (UTC)Reply
 
Approximate orbits of Hilda (pink) and Jupiter (blue) with L3, L4 and L5 as blue, green and red rings; click here for a series of 24 constellations.
Here is a new version with 24 drawings, which also span 2 orbits of Jupiter and 3 of Hilda. The three "Lagrange rings" have got different colours, while the labels L3, L4 and L5 are deleted. I think this will be less busy and more readable. It should be noted, however, that Hilda in this example does not follow the laws of Kepler but moves with constant speed, so computed plotting is still necessary to make it right. --Eddi (Talk) 02:06, 14 December 2007 (UTC)Reply
 
Hilda herself is a bit off course
So Hilda herself isn't quite lined up to show what you want - I suppose I could sim a "hypothetical" Hilda object but I'm not super-thrilled with that idea - I could try to find an example object that's better lined up, I suppose. What else? WilyD 16:29, 14 December 2007 (UTC)Reply
The restart isn't perfect either, if you watch - I think I dropped a frame or two. WilyD 18:44, 14 December 2007 (UTC)Reply
 
Hey, these plotting routines I've had to write are great!!!
OGV version converted from the GIF
I noticed the GIF was not animating (the number of frames may be over some wikimedia limit), so I converted it to the Theora video at the left, and replaced it in the articles Hilda family and 153 Hilda. -84user (talk) 11:21, 30 April 2010 (UTC)Reply

Which resonance?

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In one place this article says the Hilda asteroids are in 2:3 resonance with Jupiter; in another it is 3:2. Which is it? -- 23:26, 5 August 2009 (UTC)~ltaylor@csub.edu Larry Taylor —Preceding unsigned comment added by 66.75.198.179 (talk)

Sorry for the late reply Larry. Using asteroid 153 Hilda as a generic example, it goes around the Sun 3 times (7.92 year orbit) for every 2 times that Jupiter does (11.8 yer orbit). -- Kheider (talk) 15:43, 30 April 2010 (UTC)Reply
The article is still inconsistent though, saying in the intro that they are in 2:3 resonance and another time saying they are in 3:2. Based on the dicussion and the article Orbital resonance, we may use the following convention: "saying that B has an n:m resonance with A is a shorthand for saying that (secondary body) B completes n orbits in the same time as (primary body) A completes m orbits.". Using this convention, Hildas have a 3:2 resonance with Jupiter. We can also say that there is a 3:2 Hildas-Jupiter resonance. I'll amend the article in this way.--Joancharmant (talk) 10:09, 10 December 2011 (UTC)Reply

Etymology

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The eytomology needs to be included.174.3.123.220 (talk) 04:26, 20 May 2010 (UTC)Reply

Hildas and the Kirkwood Gaps

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Most of the orbital resonance points in the astroid belt are devoid of asteroids. These are called the Kirkwood gaps. But the Hildas are apparently the exceptions to this rule. Does anybody know why? If so, it should be in the article, if not, it should be mentioned as an unanswered question. —MiguelMunoz (talk) 06:25, 15 June 2012 (UTC)Reply

The 2:1 resonance is empty because most of its otherwise stable regions are destabilized due to Jupiter and Saturn being near their mutual 5:2 resonance.
http://iopscience.iop.org/1538-3881/116/3/1491 Agmartin (talk) 17:53, 25 September 2013 (UTC)Reply
The 4:1, 3:1, 5:2, 7:3 resonances are empty because interacting secular resonances inside the MMR resonances cause the orbits to become chaotic resulting in large jumps in eccentricity and planet crossing orbits.
https://www.oca.eu/morby/papers/1041a.pdf Agmartin (talk) 20:06, 25 September 2013 (UTC)Reply

Similar families?

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The Article should say whether similar families are known to exist, for other planets and/or other resonance ratios. 94.30.84.71 (talk) 10:10, 16 June 2012 (UTC)Reply

No it shouldn't, that's the job of asteroid family. SpinningSpark 15:59, 16 June 2012 (UTC)Reply

Please improve sentence

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Dear main authors; thank you for the good work. I am not happy with the sentence: "Consequently, a Hilda's orbit has a semi-major axis between 3.7 AU and 4.2 AU, an eccentricity less than 0.3, and an inclination less than 20°."

  • I deleted the word "consequently", because this sentence doesn't follow from the previous text.
  • The 3:2 resonance corresponds to a semi-major-axis of 3.97 AU (~4, as stated below). Can asteroids with semimajor axis 3.7 and 4.2 AU be called in orbital resonance with Jupiter? The corresponding orbital period ratios are 1.67 and 1.37. This needs clarification.

Regards, Herbmuell (talk) 03:15, 27 January 2016 (UTC)Reply

Resonances have a finite width, due to the planetary perterbation. The semimajor axes change with time. The *average* value must be ~3.97, but the instantaneous value can be farther away, as it librates around. Kepler47 (talk) 20:13, 10 November 2016 (UTC)Reply
I see. I added this information in the article. --Herbmuell (talk) 16:05, 19 November 2016 (UTC)Reply

special eccentricities

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I made a plot of two 3:2 orbits in the rotating frame. (I disregarded effects of Jupiter's gravity, though that's kinda contrary to the spirit.) The two differ in eccentricity: the black one touches the Trojan points, and the red one has a cusp where it "stands still" at aphelion. I'd like to add one for the eccentricity below which the curve is everywhere convex, but don't know how to find it. —Tamfang (talk) 08:22, 9 February 2016 (UTC)Reply

If you are still here .... I dont understand your difficulty. If you simply change the eccentricity to Hilda itself (about 0.14), you will get an orbit that is convex everywhere. Then simply add that to your two other orbits. The lower the eccentricty, the more the trilobe becomes like a circle.
Obviously I could draw a circle, which is convex everywhere, but I want the eccentricity that is the boundary between all-convex and otherwise. —Tamfang (talk) 00:54, 9 June 2023 (UTC)Reply

Suggest renaming and further editing

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Ref. 4 of the article, Brož, M.; Vokrouhlický, D. (2008), can be downloaded from arxiv. These authors call the Hilda asteroids a group, and so does the beginning of this article. The group consists of two families, the Hilda family and the Schubart family; the article also says so. Clearly, this article is about the Hilda group, and not the family. I suggest therefore to rename it "Hilda group". If we can agree on this terminology, the articles 153 Hilda and 1911 Schubart would then also have to be corrected similarly (distinction between group and family). See also the articles Asteroid family and List of minor-planet groups on this topic. --Herbmuell (talk) 15:07, 19 November 2016 (UTC)Reply

I agree, there are other articles on wikipedia with the same error, for example the Hungaria_family Agmartin (talk) 20:59, 19 November 2016 (UTC)Reply
The papers discussing them typically refer to them as Hildas or Hilda asteroids. Agmartin (talk) 21:41, 20 November 2016 (UTC)Reply

Long term stability?

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How stable are the orbits of these asteroids? The main article sort of SUGGESTS, without actually explicitly saying, that they are stable for a long time. That is, that their Keplerian elements change only very "slowly". But how "long" is "long? Hundreds of years? Thousands? Millions? And how do their orbits "decay"?

On a related topic ... "The Hildas Triangle has proven to be dynamically stable over a long time span". What exactly does this mean? Is this the same thing as the previous question? If not, then what is it? And again, what is a "long time" in this context? So many questions... — Preceding unsigned comment added by 2001:8003:E422:3C01:BCD6:9922:BB24:C3D0 (talk) 09:22, 30 December 2021 (UTC)Reply

One phrase missing from all of this.

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It needs to be explicitly stated that these asteroids orbit around THE SUN, not Jupiter. Even though the animations and the diagrams make this clear, well at least to me. I say this because in the Talk section for the Trojan asteroids, which whilst of course different from the Hildas, do arguably share some common factors with Hildas, there is quite a bit of confusion about this. — Preceding unsigned comment added by 2001:8003:E422:3C01:BCD6:9922:BB24:C3D0 (talk) 09:27, 30 December 2021 (UTC)Reply

PErturbation of their orbits by Jupiter.

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There should be a bit of a discussion of this. In this discussion shoudl be included something like the following:

From the diagram with the red and green orbits, it can be seen that the closest approach between asteroid and Jupiter - that is, when Jupiter is in opposition - occurs when the asteroid is at perihelion, and thus at its furthest from Jupiter. This would presumeably minimise the gravitational perturbation of Jupiter.

Also .... is this the case for ALL Hilda asteroids, or only Hilda itself? — Preceding unsigned comment added by 2001:8003:E422:3C01:C2D:60A8:DA74:74FD (talk) 22:33, 30 December 2021 (UTC)Reply

Two nurmber either incorrect; or need more explanation.

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You have: "The Hildas "rest" at their aphelia in the apexes for an average of 5.0–5.5 years, whereas they move along the sides more quickly, averaging 2.5 to 3.0 years. "

This seems very incorrect. If you look at the animated gif, the aphelion "rest" is more like 2-3 years, and the "rapid siding" is more like 5-6 years.

Also, it is perhaps way too simplistic to describe this with one set of values. Orbits with low eccentricities, such as Hilda herself, have "sharp" apices, like the red one, and the "rest" period is relatively short. Orbits with larger eccentricities have "lobes" at their apices, like the black one, and the "rest" period is much longer. — Preceding unsigned comment added by 2001:8003:E422:3C01:B5C0:7521:E3E5:C5E7 (talk) 03:29, 1 January 2022 (UTC)Reply