Talk:Roche limit

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Contents

[edit] Old discussions

This talk page got pretty long with old stale discussions. I moved it here: Talk:Roche limit/archive1.

[edit] Consistent naming of the two bodies

I have changed all reference to the two bodies so they are consistently called "Primary" and "Satellite". If you disagree with this choice of names, please try to change them consistently throughout the article. --P3d0 22:12, Oct 30, 2004 (UTC)

I made it more consistent by changing the introductory sentence to also use the "Satellite" nomenclature, consistent with the rest of the article. --Noren 17:27, 15 Nov 2004 (UTC)

[edit] Oblateness term

I think that Weisstein got it wrong at Mathworld --- Let's say that the celestial body is perfectly spherical: then the semi-minor axis radius would equal the semi-major axis radius, so c=R, right? So, (1-c/R) would be zero, and the output of the formula would diverge :-(. I think Weisstein meant to say that c is the difference between the semi-major and semi-minor radii. This makes the formula work. We could define it that way, but it is quite a mouthful --- it turns out that the oblateness of a body is the ratio of the difference to the semi-major radius, so that seems more compact.

I tried to double check this against a non-Mathworld source, but could not. Can someone come up with a double check? Thanks! -- hike395 17:50, 15 Nov 2004 (UTC)

No, R is defined by both Weisstein and here as the radius of the primary, while c is the length of the semi-major axis of the satellite. So, while it's true that this is indirectly related to oblateness, the statement that c/R equals the oblateness of the satellite is incorrect- a perfectly spherical primary with a radius of 10 with a perfectly spherical satellite with radius of 1 would have c/R equal to .1, for example. We could reiterate the definition of R here if that would clarify the issue. --Noren 04:03, 16 Nov 2004 (UTC)

I'm sorry, that isn't clear to me at all. Weisstein defines R in the main body of the article [1]. c is not defined in the main article. The main article references an article about the oblate spheroidal gravitational potential[2]. There, c is defined as the semi-minor axis, not the semi-major axis of an oblate spheroid. It isn't clear to me (from the text description, at least) whether he is referring to the oblate potential of the primary body or of the satellite. It's quite confusing. Is there a different source that we can double check this with? It seems suspiciously mushy. -- hike395 07:42, 16 Nov 2004 (UTC)

I'd certainly like to read an alternative source, but I'm not very skeptical of the one we have, so my motivation to seek another one out is low. An important problem with your 'oblateness' definition of the c / R term, as you pointed out, is that it becomes very messy in the limits, for example the 'spherical' limit when c = R. on the other hand, the obvious limiting case if the term relates to the relative sizes of the two bodies is that R > > c and M > > m ... in this case the \left( \frac{(1+\frac{m}{3M})+\frac{c}{3R}(1+\frac{m}{M})}{1-\frac{c}{R}} \right)^{\frac{1}{3}} term goes to 1, which is the behavior we expect given that this term is ignored entirely in the cruder first approximation.
Throughout the Roche limit discussion, capital letter variable definitions refer to the rigid primary, while lowercase letters refer to the satellite. Weisstein makes the assumption for purposes of this derivation that the primary is spherical- that's why the letter used is R rather than an A,B,C,a,b, or c- it refers to an assumed spherical body which has a Radius. Finally, you're right that c refers to the semi-minor rather than the semi-major axis of the satellite, that was my mistake. --Noren 16:06, 16 Nov 2004 (UTC)
I'm confident one of the two following statements is true: 1) c refers to the difference in equatorial and polar radii of the primary, or 2) Weisstein's derivation is wrong. I believe this because a satellite deformed by tides is a prolate spheroid, while a fast rotating primary is an oblate spheroid. Weisstein specifically uses the oblate spheroidal potential: he must either be referring to the gravitational potential of the primary, or he is confused.
If c is the difference between the equatorial and polar radii, it goes to zero for a perfect sphere, so there is no problem with divergence at all. If c refers to the semi-minor axis of the primary (as stated by Weisstein), then the formula is easily shown to be incorrect. -- hike395 04:46, 17 Nov 2004 (UTC)
I found this paper [3], but is on only available as an on-line abstract, and I don't have a subscription to Icarus. Can anyone help out? -- hike395
This Roche limit stuff is really complicated. Check out [4][5]. There are many possible factors, including modeling the internal friction of the satellite and/or modeling the hydrodynamics of the satellite. The Weisstein derivation looks overly simple compared to what real astrophysicists use. -- hike395 06:58, 18 Nov 2004 (UTC)

[edit] Why clockwise?

Why are the illustrations (well, one of them anyway) of the disintegration process showing clockwise movement? The natural direction is ocunterclockwise, as we tend to look at things from the "north" pole...

Urhixidur 17:53, 2004 Nov 15 (UTC)

Okay, I fixed the image now.

Urhixidur 17:57, 2004 Nov 15 (UTC)

[edit] Mixing percentages and ratios

The table in the Roche limits for selected examples section showing where various solar system bodies are relative to their roche limits mixes ratios and percentages. The table header indicates that the data's in ratios, so I was thinking I'd change the percentages over to ratios to make it all consistent. However, I figured I'd check here to make sure there wasn't a reason why percentages were used in the first place. Bryan 09:23, 21 Nov 2004 (UTC)

Thanks for asking. I changed them to percentages in the first place because I think people are accustomed to using percentages for numbers between 0 and 2 that have two or more significant figures. It also highlights those bodies that are close to their Roche limits. However, go ahead and change to ratios if you like it better. Mine is just one man's opinion. --P3d0 17:52, Nov 21, 2004 (UTC)
Either ratios or percentages are OK with me -- hike395 18:01, 21 Nov 2004 (UTC)

Is the notation using a consistant separator? There seems to be both , and . being used in different ways... IE. the 0,6... in Saturn. 70.177.71.206 16:24, 29 August 2006 (UTC)

Good point! fixed. Deuar 16:38, 29 August 2006 (UTC)

[edit] Help needed on wikijunior solar system book

I'm working on a book for children over at wikibooks and someone has written the following:

Mercury does not have a moon. Mercury's rotation is so slow that if Mercury had a moon, it would crash into Mercury or get broken up. This would happen because the moon's gravity would cause tidal effects on Mercury. There would be two bulges called tidal bulges on Mercury. One would bulge toward the moon, with the other bulge being on the opposite side of Mercury. The moon's motion in its orbit would be faster than Mercury's rotation because Mercury's rotation is very slow. That would cause the moon to be ahead of the tidal bulge all the time. The gravity from the bulge would pull back on the moon. This would cause the moon to become closer to Mercury and Mercury's rotation to speed up. This would continue to happen over millions of years until the moon got broken up by Mercury's gravity or crashed onto Mercury. Mercury had existed for billions of years, so if it had any moon, it is long gone.

[6]

Now aside from the fact that this explanation is confusing to me (let alone a kid) I'm not at all convinced that the science is actually correct. So I was wondering if any of you fine people (flattery will get me everywhere I hope) could help me rewrite? Theresa Knott (a tenth stroke) 2 July 2005 19:41 (UTC)

Sounds confusing but basically right. Tidal forces make satellites tend toward the planet's rotation period, given enough time. If the planet's rotation is faster (like with the Earth/Moon system), the planet donates angular momentum to the moon, causing it to move slowly away from the planet and (paradoxically) slow down. If the moon's revolution is faster (like with the Mars/Phobos system), the planet takes angular momentum from the moon, causing it to move slowly toward the planet, and to speed up. The good news is that all orbits eventually tend toward a stable, tidally locked orbit over time. The bad news is that they often get inside the Roche limit or outside the Hill sphere before they reach stability, and therefore stop being a moon altogether. --P3d0 July 3, 2005 00:26 (UTC)
Having done the calculations, I think the remark is right. Mercury's Hill sphere radius is about 220 Mm, while its geosynchronous orbit radius is 244 Mm. That means a moon with enough angular momentum to sustain a geosynchronous orbit would not be within Mercury's sphere of influence, so it would be in orbit around the Sun. Likewise, a moon with an orbital radius within Mercury's Hill sphere would orbit faster than geosynchronous, and would therefore bleed angular momentum into Mercury until it found itself inside Mercury's Roche limit, and either broke apart or collided with Mercury. Perhaps someone can confirm my calculations? --P3d0 July 3, 2005 03:59 (UTC)

[edit] History section

This article lacks of a history section where it describes when the idea of roche limit was discovered, and who have developed the ideas and formulas. Could someone add it? CG 12:44, 10 January 2006 (UTC)

[edit] Shape of satellite

I notice that the rigid body calculation is for a spherical satellite. Presumably if it is elongated and tidally locked, the long axis will point towards the primary, and the tidal forces along this axis will be significantly greater than for a spherical body of the same size. Hence its Roche limit will be farther out from the primary. I think the effect can be quite significant, and could be calculated for particular bodies. E.g Amalthea is quite elongated and its elongation is known. Does anyone know any details of this? Deuar 21:34, 9 June 2006 (UTC)

It's a bit more complicated than that. Consider the ends of the long axis. You're right that there's more tidal force the more elongated the satellite is, as that farthest points are moving away from the center of mass. However, also of importance is the fact that at constant total mass, the gravity holding the satellite together decreases with elongation of the satellite, as the rest of the mass is moving farther away. So elongation both increases the tidal outward force and decreases the inward force of gravity.
For a formal physics treatment of this, check out [this icarus article]- added it to the Phobos page a while back when there was some dispute as to whether Phobos was within its Roche limit. --Noren 05:13, 10 June 2006 (UTC)

[edit] Bolding "Roche Radius" in intro sentence

As per the Manual of Style, I've bolded "Roche radius" in the intro sentence, as it's an alternative name for the same concept. I hope no one disagrees? T. S. Rice 22:21, 19 June 2006 (UTC)

[edit] M*

The formula containing the oblateness was corrected according this source (see M^*).--Beaumont (@) 15:52, 10 December 2006 (UTC)

...and it seems like the oblatness is NOT what is meant in the formula: c stands for the semi-major axis and R stands for the distance between two bodies --Beaumont (@) 18:50, 10 December 2006 (UTC)

...ooops the linked source has been removed. Can one verify the formula, please. --Beaumont (@) 07:58, 11 December 2006 (UTC)

Finaly, I have verified the formula. The oblateness of the primary has nothing to do with the Roche Limit. Actually, in the second part of the paper on mathworld, R denoted the distance between the two bodies. And our description of the formula was a misinterpretation (or hoax). Now all the calculations have been removed from the mathworld article. The best solution is to remove this "oblateness" correction (not only wrong but also unsourced). --Beaumont (@) 15:46, 12 December 2006 (UTC)

I'm not M, but I agree with you for the most part, and was in opposition to the use of "oblateness" two years ago when it was added(see the discussion between User:Hike395 and myself above about "Oblateness Term".) I was unable to convince others at that time. One minor correction: c was the semiminor axis of the satellite according to Wolfram's nomenclature, which in this prolate case was the longer axis. I found this counterintuitive as well, but it did make for a pun so I had to point it out here. See the subpage that still exists, unlike the Roche Limit one. --Noren 16:12, 15 December 2006 (UTC)
Thanks for your remark. Yes, on the subpage c stands for the semiminor axis - as you claim. However, I studied carefully the "main" page, and there it was semimajor... As for today, the deleted calculation can still be recovered from the google cache. You may observe that in (8) c is subtracted from R, which stands for the distance between the two bodies. The difference R-c gives the distance needed for centrifugal and gravitational potential (and semiminor axis makes no sense there). You may also notice that (10) uses c^2-a^2, which is impossible under the square sign. Indeed, (10) comes from the subpage with c and a exchanging the meaning (unfortunately, the subpage has been modified too; before it gave (10) explicitely). Further, it should be noticed that "R" in the "main" page changed the meaning within the very page. I think these notation conflicts gave rise to our problems. Nonetheless, the calculation was essentially correct. It is sufficient to observe c<<R to get rid of the problematic part of the resulting formula (R can not make part of it!) and to obtain eventually the reasonable term (1+m/(3M)) instead. A minor correction, since usually m<<M as well. I did not insert it as my OR ;-) I just deleted what was (IMHO) obviously incorrect. --Beaumont (@) 08:36, 18 December 2006 (UTC)
PS. I corrected the formula after a (very) long discussion on plwiki during FA nomination procedure.

[edit] Image Animation

Would it significantly help the article to have someone animate the first four images showing tidal forced on an orbiting mass into a single piece? It might be easier to understand then. Pharod42 02:35, 23 May 2007 (UTC)

[edit] Lack of Notability?

Why is this article flagged for lack of notability? It is a discussion of an important topic in celestial mechanics. Yes, it needs sources, but that does not diminish its notability in any way, shape, or form.--Popefelix 02:18, 27 June 2007 (UTC)

[edit] Pronunciation of "Roche"

The pronunciation is given as /roʊʃ/. The diphthong indicates that this is a US pronunciation, so I believe the page should say as much. I don't know how "Roche" is pronounced in British English, but given that American English tends to diphthongise "o" in non-English names where British English does not (compare the US and UK pronunciations of "Notre Dame", "Pinocchio" [first "o"] and "Pocahontas" [first "o"]) , I would hazard a guess that the UK pronunciation is /rɒʃ/. — Paul G (talk) 18:30, 4 March 2008 (UTC)