Talk:Horseshoe orbit
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editI think this article needs better images. Since this is the English Wikipedia, labels in images should be in English unless the labels are historic in nature. Will (Talk - contribs) 06:24, 14 February 2007 (UTC)
This is to advise that the graphic "Horseshoe thin" is showing the L4 and L5 points on the wrong side of the planet (assuming counterclockwise orbital movement). L4 should be leading the planet.Nfr-Maat (talk) 23:42, 12 July 2014 (UTC)
images removed - misleading
editI've asked the authors of http://www.scielo.br/pdf/cam/v24n1/06v24n1.pdf for permission to publish their image on p105. —Preceding unsigned comment added by Lfast321 (talk • contribs) 17:54, 20 October 2007 (UTC)
Non-encyclopedic content?
editI just noticed that the second section of this article has the following text:
- My non-expert calculation say that for an asteroid like 3753 Cruithne the inner and outer rings are less than a million km apart. So the horseshoe is very thin, more like the next diagram.
To my knowledge, Wikipedia isn't supposed to be written in first person. However, I'm not sure of any good way to replace this text. Perhaps someone with a little more experience should look into it and figure something out? - green_meklar 00:53, 20 December 2007 (UTC)
- I took it out, but have little to replace it with. :) 81.93.102.185 (talk) 11:51, 6 January 2008 (UTC)
Suspected nonsense
edit- "gravitational exchange of energy changes the object's apparent direction"
What does this mean? A gravitational exchange would change an object's actual direction, not just the apparent. OrangeDog (talk • edits) 02:59, 2 April 2009 (UTC)
- It changes the apparent direction of an asteroid as observed from the Earth. The asteroid changes between falling behind and catching up. However, relative to the Sun, both the Earth and the asteroid continue to orbit in the same direction as before. Is that clear? - Fayenatic (talk) 13:36, 2 April 2009 (UTC)
What has that to do with a "gravitational exchange of energy"? Surely it is just down to relative position and viewing angles. OrangeDog (talk • edits) 18:03, 2 April 2009 (UTC)
No OrangeDog, it is a real change of energy. Conceptually the same as skaters (ice or rollerblade) do. One slows down, the other speeds up. The asteroid gains energy from earth, and thus orbits slightly faster. It then loses energy back to the earth, and then orbits slightly slower. So it is more than just position and angles. — Preceding unsigned comment added by 58.164.242.37 (talk) 00:57, 22 March 2018 (UTC)
Possible clarifications/removals?
editThere are a few statements that seem (to this layperson) unclear:
- "But Earth's gravity exerts an outward accelerating force, pulling the satellite into a higher orbit and – counter-intuitively – slowing it down."
First, this and the other use of "counter-intuitively" probably just need to be excised as POV. But to clarify the meaning - by "slowing it down", does the writer mean: 1- an actual decrease in velocity (contrary to the acceleration mentioned just before), or 2- change to a higher orbit, which lengthens the orbit period? Both? Neither? - Special-T (talk) 19:16, 3 May 2009 (UTC)
- The acceleration is a change in velocity, as it alters the direction of movement. I believe the speed decrease is due to increased orbital period, though I don't have the sources to check. OrangeDog (talk • edits) 04:09, 10 July 2009 (UTC)
- In this context, a "speed decrease" must be talking about the object's relative ground speed (the amount of distance that it covers across the face of the Earth), which is basically meaningless to objects in orbit. In order to actually be in a "higher orbit" any object in orbit would need to increase it's velocity. Note however that "higher orbit" is an extremely imprecise term... higher how? Greater perigee? apogee? what? I'm honestly not sure what the source of that sentence is, but it's basically nonsense in it's current form.
— V = I * R (talk) 05:10, 10 July 2009 (UTC)
- In this context, a "speed decrease" must be talking about the object's relative ground speed (the amount of distance that it covers across the face of the Earth), which is basically meaningless to objects in orbit. In order to actually be in a "higher orbit" any object in orbit would need to increase it's velocity. Note however that "higher orbit" is an extremely imprecise term... higher how? Greater perigee? apogee? what? I'm honestly not sure what the source of that sentence is, but it's basically nonsense in it's current form.
- I'm leaning on the side of "if it looks like nonsense it probably is" with this article. OrangeDog (talk • edits) 05:16, 10 July 2009 (UTC)
I just looked at an old version of this article and, while the information wasn't presented that clearly, there was an explanation of this phenomenon that seems simpler to me:
- The asteroid and the earth are orbiting the sun in the same direction.
- Assume the asteroid is in a higher (i.e, farther from the sun, therefore slower) orbit than earth
- The earth will catch up to the asteroid, at which point the earth's gravity pulls the asteroid into a lower (therefore faster) orbit.
- The asteroid is now orbiting faster than the earth, and eventually catches up to the earth from "behind"
- A similar action occurs, but the earth's gravity now pulls the asteroid into a slower, higher orbit
- Asteroid orbit slows and earth eventually catches up to it from "behind" the asteroid
- Back to the first step. From earth's perspective the asteroid appears to approach from earth's "front", move across (into lower, faster orbit), recede (as it's moving faster), eventually approach from "behind", cross into a higher, slower orbit, and recede.
If someone with knowledge of this area (that wouldn't be me) thinks this is useful and can present it more coherently, please pick up the ball and run with it. Oh yeah, assuming it's reasonably correct! - Special-T (talk) 13:50, 31 July 2010 (UTC)
Orbital period
editThe 2010 TK7 article contains such statements as "the asteroid orbited the Sun with a period of 365.389 days" and "its average period over long time intervals will exactly equal that of Earth", but neither that nor this article do not define what "period" and "average period" (relatively to Sun) mean for an aperiodic orbit. Possibly it can be defined in terms of longitude, but it should be defined before usage in articles. If someone is willing to rectify this, then look on 3753 Cruithne also – it is a good example how such complex things can be explained. Incnis Mrsi (talk) 14:21, 8 January 2013 (UTC)
Question moved from article
editSurely relative to the Sun, the asteroid rotates around the sun, in about a year; slight wobble every hundred (?) years. If we want to use weird frames of reference, lets choose the asteroid as centre. Now the earth is on a horseshoe orbit, but what about the tadpole asteroid? — Preceding unsigned comment added by Bparslow (talk • contribs) 10:04, 22 February 2014
Tadpole orbit
editThis section, Horseshoe orbit#Tadpole orbit, has a link to the page Tadpole orbit, which redirects back to this section. Either a separate page should be created, or the link should be removed. JMtB03 (talk) 00:18, 14 August 2014 (UTC)
- Thanks! It works best as a sub-section of this larger article, so I have removed the circular link. – Fayenatic London 18:47, 16 August 2014 (UTC)
More elaboration needed.
editThe statement: "The horseshoe becomes apparent only when mapping the movement of the asteroid relative to both the Sun and the Earth. " is completely true, as is the rest of the paragraph, but it is not enough. And that appears to be the source of confusion for several of the comments here. It needs to be elaborated that the horseshoe (and for that matter the tadpole) shape only occur if viewed in a frame of reference rotating with respect to the second body. If someone better than I at creating (for example) animated gifs can do one, showing both views - that is, a rotating and a non-rotating one - then this can be further discussed around that animation. — Preceding unsigned comment added by 58.164.242.37 (talk) 00:52, 22 March 2018 (UTC)