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Could someone address the formula e = SQRT(1+2*energy*h^2/u^2) which seems to imply e>=1 always, since h^2 and u^2 are always positive. To match the planetary data, it seems like the equation should be e = 2-SQRT(1+2eh^2/u^2). Perhaps we could plug in numbers for one of the examples? 2607:F140:6000:1C:551C:8D2:AC56:37E5 ( talk) 16:41, 29 October 2018 (UTC)Ralph Berger
what about Eccentricity (mathematics) ? there's other equation, which is wrong ?
The effect of excentricity on insolation is missing. —Preceding unsigned comment added by 130.184.251.50 ( talk) 22:28, 26 October 2009 (UTC)
I just reverted the article to an earlier version because of inserted vandalism. The value of the eccentricity of Earth's orbit was also changed in the same vandalism. I assume the previous value was correct but I'm uncertain so it should be checked by someone who knows better than I. -- Pigman ( talk • contribs) 21:17, 26 November 2006 (UTC)
Is that part about eccentricity and longer winters "favorable for triggering the next ice age" actually true? It set my bogometer off, so I thought I'd ask if anybody else has heard that theory. The only reference in the paragraph is that AOL page, which seemed a little un-scientific and POV. -- W0lfie 20:26, 8 December 2006 (UTC)
pet peeve: please don't use "theory" to mean "conjecture" in a science talk page. a theory doesn't just mean some random idea in science. —Preceding unsigned comment added by 99.233.20.151 ( talk) 05:05, 19 March 2008 (UTC)
I reckon that this formulation is somewhat wrong: "Planet Neptune's largest moon Triton is believed to be the only astronomical body that has a perfectly circular orbit with an eccentricity of 0." Is it not quite harsh to claim that Triton is the only body in the entire universe to have a perfectly circular orbit? Either it should be stated that it is in our Solar system or it should state that it's the only know body... or both. -- Bilgrau 11:46, 23 June 2007 (UTC)
Is it even possible for something in nature to be perfectly anything? I suspect that the O.E. of Triton is close to but not exactly equal to 0. 65.167.146.130 ( talk) 15:46, 31 October 2008 (UTC)
Am I wrong, or is there no actual definition of eccentricity given in this article? I read the whole thing, but I still have no real idea how it is defined. --01:13, 16 December 2009 (UTC) —Preceding unsigned comment added by 90.162.97.176 ( talk)
Graph of the variation of eccentricity of rocky planets (Mercury, Venus, Earth, Mars) does not agree with upper value for Earth in the text beside the graph, assuming the heavy curve is for the Earth. (I do not know if there is a problem with the other three planets, obviously Earth is most important from human perspective.) If the upper value in the text occurs outside the time span of the graph, an explanation would be in order. G Warren Coleman wd4nit@yahoo.com — Preceding unsigned comment added by 76.105.116.92 ( talk) 05:45, 16 November 2011 (UTC)
The graph is only plotted for the next 50,000 years, the text in the article refers to the last 750,000 years. -- Kheider ( talk) 15:58, 16 November 2011 (UTC)
Regarding the definition in the lead, the orbit of stars and clusters through the galaxy follow a non-Keplerian rosette path. This orbit can still be assigned an orbital ellipticity, even though it isn't used in the same mathematical sense. [1] [2] [3] Regards, RJH ( talk) 17:02, 11 August 2012 (UTC)
The graph is hard to fathom. The vertical scale at the origin disagrees with the vertical scale on the right by X10. Is this an error?. If not, an explanation is needed on what the two scales mean. Also, what is the meaning of the four arrows on the diagram? Spinning Spark 10:12, 1 November 2013 (UTC)
Under climate it states "Apsidal precession slowly changes the place in the Earth's orbit where the solstices and equinoxes occur (this is not the precession of the axis). Over the next 10,000 years, northern hemisphere winters will become gradually longer and summers will become shorter. Any cooling effect in one hemisphere is balanced by warming in the other—and any overall change will, however, be counteracted by the fact that the eccentricity of Earth's orbit will be almost halved[citation needed], reducing the mean orbital radius and raising temperatures in both hemispheres closer to the mid-interglacial peak.". This seems presumptuous and remains uncited. While the total solar insolation may balance out, Many factors could influence the effects on the climate such as exposure to land vs water etc. Unless a counter argument can be posed i think it should be removed or reworded. Nickmista ( talk) 09:10, 14 February 2015 (UTC)
Currently, the text of the article says "The eccentricity of comets is most often close to 1. Periodic comets have highly eccentric elliptical orbits with eccentricities just below 1 ...". This is totally nonsense! If you go to the JPL Small-Body Database Search Engine, select "object kind: Comets", select "Limit by object characteristics: (select orbital parameter)" and choose "period (years) < 200 " for periodic comets, you will get 701 periodic comets. Then if you refine the search by adding the limit "eccentricity < 0.9", you will still get 659 periodic comets. And so on. What I want to say is, that only a small fraction of periodic comets have "eccentricities just below 1"! This would be true only for non-periodic comets - out of 343 non-periodic (period > 200 years) comets only 4 have e < 0.9. The eccentricity of periodic comets is most often below 0.7, namely, 566 out of 701 comets, that is more than three fourths. 467 periodic comets of 701 have eccentricities: 0.3 < e < 0.7 .
Will anyone object if I correct that part in the article's text? Sergei Schmalz ( talk) 15:16, 29 March 2015 (UTC)
References
If an orbit is one dimensional (like an object in vertical freefall), its semiminor axis zero, so the eccentricity=1, while I'd still call it an ellipse. So it seems like we have to say eccentricity=1 AND semimajor axis=infinity as both needed to define a parabola. Tom Ruen ( talk) 04:48, 21 January 2017 (UTC)
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The claim that...
"Comet McNaught has a hyperbolic orbit while within the influence of the planets..."
... is not supported by the citation for that sentence. The cited article makes no mention of "hyperbolic" — Preceding unsigned comment added by 23.119.204.117 ( talk) 21:10, 28 April 2018 (UTC)
This article lists the eccentricity of Neptune as 0.0086, while the Neptune page has 0.009456. I don't know which is correct. — Preceding unsigned comment added by Probablyarobot ( talk • contribs) 08:56, 3 November 2018 (UTC)
The animated gifs, https://upload.wikimedia.org/wikipedia/commons/a/ae/Binary_system_orbit_q%3D3_e%3D0.gif, do not represent realistic motion because the periods of both orbits are shown as the same, whereas in reality the one with a longer major axis has longer period, according to Keplers laws. IMO, these gifs are likely to cause more confusion than any benefit because orbital eccentricity is a property of the whole orbit and animation is really unnecessary, but a wrong animation is worse than unnecessary. I am deleting them. Kotika98 ( talk) 15:49, 12 August 2021 (UTC)
I'm having difficulty reconciling the two equations for e in the Definitions section. They differ where one has EL²/mα² and the other has εh²/μ². I'm reading ε = E/m, h = L/m, and μ = Gm. When I plug these three values into the second definition I get EL²/m³α² instead of EL²/mα². Why am I getting this extra m² in the denominator? Should μ = G instead of μ = Gm or h = L instead of h = L/m or what? Vaughan Pratt ( talk) 07:22, 2 February 2022 (UTC)
On further consideration, for e = 0 (a circular orbit), both equations would need EL²/mα² and εh²/μ² to be -½. But none of E, m, or ε = E/m can be negative. So now I'm even more mystified. Vaughan Pratt ( talk) 07:46, 2 February 2022 (UTC)
To be more precise, μ = GM, the mass of the body being orbited (the center of mass), and hence ε h² / μ² = ( E / m ) * ( L / m )² / G² M² = E L² / G² M² m³ = E L² / ( G² M² m² m ) = E L² / α² m. where I have used M as the mass of the body being orbited (the center of mass), and m as the reduced mass of the orbiter, mred , as it appears in the article. In addition, please keep in mind that in all of physics, bound states (such as closed orbits) have total negative energy. Therefore, the total energy, E, as a quantity is a negative number with the following bounds - G M m / r < E < 0. It never gets more negative than zero kinetic energy ( v = 0 ), and is never less negative than approaching zero, which at E = 0 the orbit stops being bound and becomes an escape parabola, consistent with e = 1. MMmpds ( talk) 15:09, 2 February 2022 (UTC)