Talk:Roche limit/Archive 2

Old discussions

This talk page got pretty long with old stale discussions. I moved it here: Talk:Roche limit/archive1. — Preceding unsigned comment added by Doradus (talk • contribs) 22:10, 30 October 2004 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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. --Doradus 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)[reply]

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

M*

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

...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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

PS. I corrected the formula after a (very) long discussion on plwiki during FA nomination procedure.

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)[reply]

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)[reply]

External Links

The Weisstein reference does not show me what it claims. 82.163.24.100 (talk) 18:16, 21 February 2009 (UTC)[reply]

Binary System

Can this article tell something about Roche limit in a binary System? --Gildos (talk) 01:27, 10 September 2009 (UTC)[reply]

Calculation of constant-density fluid case is erroneous

About the constant-density fluid case, the calculation in the article is likely erroneous for close-orbiting satellite. That is because many of them are likely to have tidally-locked rotation, meaning that one must add the rotation's centrifugal potential to the primary's tidal potential. Those potentials are approximately comparable in size:

V(tidal) = (1/2) * w² * (- x² + y² + z²)

V(centr) = (1/2) * w² * (- x² - y²)

V(total) = (1/2) * w² * (- 3x² + z²)

where w is the angular velocity in the orbit, x is the distance along the direction to the primary, y is the distance along the orbit-velocity direction, and z is the distance along the orbit-pole direction. This is in the small-satellite limit.

Chandrasekhar and others have been careful to include the centrifugal potential, though doing so means that the satellite's shape becomes a triaxial ellipsoid with all axes different, a "Roche ellipsoid". That makes the equations for the satellite's gravitational potential much more complicated than in the spheroidal case, it must be pointed out. But when one includes tidally-locked rotation, the Roche-distance factor goes from 2.423 up to 2.455. That factor multiplies (planet radius) * (planet density / satellite density)^1/3.

I'm thinking of adding a discussion of the effects of rotation.

Lpetrich (talk) 18:52, 9 July 2013 (UTC)[reply]

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. --Doradus 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)

It should be primary and secondary, since the Roche calculation applies also to an unbound secondary object. 94.30.84.71 (talk) 19:06, 17 September 2013 (UTC)[reply]

Reference to Original Paper

http://www.merlyn.demon.co.uk/gravity5.htm#Roche now has links leading to images of the original material in three parts, poorly focused. If nothing better can be found, those should be added. Note though that some sources show more to some readers than to others - links should be checked at least from Europe. 82.163.24.100 (talk) 20:30, 1 November 2009 (UTC)[reply]

Nice! I went ahead and added direct links to the three parts to the source section. It would be unfortunate if they do not work for everyone, but then again it would be better than the existing no links at all. --Noren (talk) 01:11, 22 July 2010 (UTC)[reply]

Now http://www.merlyn.demon.co.uk/gravity6.htm#Roche. 82.163.24.100 (talk) 21:10, 18 August 2010 (UTC)[reply]

Now http://www.merlyn.demon.co.uk/gravity8.htm#Roche. 94.30.84.71 (talk) 19:17, 17 September 2013 (UTC)[reply]

Roche Period

Web page http://www.merlyn.demon.co.uk/gravity8.htm#RP shows that, for a secondary in a circular orbit, the period at the Roche Limit is inversely proportional to the square root of the product of Big G and the density of the secondary, with a numerical constant factor of, IIRC, the order of ten. That result appears to me to be certain, at best not well known, and worthy of inclusion after checking. 94.30.84.71 (talk) 19:33, 17 September 2013 (UTC)[reply]

Rigid-satellite calculation

Firefox 24.0 on Mac OS X 10.8.5 shows:

> Failed to parse (unknown error): d = 2.44\; R_M\left( \frac {M_M} {M_m} \right)^{\frac{1}{3}}

which isn’t ideal. JDAWiseman (talk) 23:14, 23 September 2013 (UTC)[reply]

Do you get the error for this?

d=2.44\;R_{M}\left({\frac {M_{M}}{M_{m}}}\right)^{\frac {1}{3}}

If so, report it to Template talk:Math. If not, then bypass your cache and reload the article. — Reatlas (talk) 09:39, 24 September 2013 (UTC)[reply]

Unclear article

The § Roche limits for selected examples states the assumed density of comets but doesn't clearly state that the first table is talking about the roche limit for comets. Blackbombchu (talk) 21:37, 20 March 2014 (UTC)[reply]

Rigid satellites

I do not recall reading, in any professionally-edited medium, any discussion of the ""Roche Effect"" in a body of significant rigidity.

If reputable sources for such can be found, one or more of them should be specifically cited; they may justify use of "Roche" in a rigid-body situation, even though Roche's papers were about liquid bodies.

Otherwise, mention of rigidity should be substantially eliminated.

The "Pebble on Surface" calculation can happily be retained, with some rewording.

However, a better approximation to the right number is obtained by the more realistic approximation which considers the secondary of consisting of two equal rigid spheres in contact - see Gravity 6, already cited.

My recollection of Roche's result (it is too late to re-read the papers tonight) is that the secondary fissions into two halves rather more readily than it spews grits to zenith and oppositely.

82.163.24.100 (talk) 21:39, 18 August 2010 (UTC)[reply]

This article may be of interest to you; Holsapple, K. A. determines a limit of distance beyond which the shape of a body is "unstable and globally catostrophic" using a Mohr-Coloumb "rubble pile" model. I'm uncertain whether he calls this distance the "Roche Limit" even though the analogy is obvious - he does not do so in the abstract. --Noren (talk) 02:31, 8 September 2010 (UTC)[reply]

The case listed as "rigid" seems to be the simplified calculation where the deformation of the body is neglected, that is, the case of the Roche limit for a satellite assumed to be spherical. This is not a real-world case, but it is one simplified enough to solve exactly as an exercise. This should be made more clear. I'm not sure why viscosity or friction are mentioned; neither one plays any part in the calculation-- they only slow down the break-up, but don't change the limit.

I've made a stab at making this more clear, and added a reference that goes through this simplified calculation (although it does not actually call it "rigid satellite"). See if this helps. Geoffrey.landis (talk) 21:39, 23 March 2011 (UTC)[reply]

There is currently following statement included: "If the satellite is more than twice as dense as the primary, as can easily be the case for a rocky moon orbiting a gas giant, then the Roche limit will be inside the primary and hence not relevant." This is not supported by anything, and it disagrees with given references : http://www.asterism.org/tutorials/tut25-1.htm That article gives Roche limits for asteroids having density of 3 with primaries as Jupiter Saturn and Sun, all of them with densities below 1.5, yet having Roche limits well outside the radius of their primaries. Given formula for roche limit in this wiki article disagrees such a claim also. In order to roche limit equal radius of the primary, ratio of densities must be 2.44^3:1 = 15.527:1, not 2:1 as claimed. Unless somebody can explain that such a false claim should be removed from both places it is made. 178.55.5.19 (talk) 14:55, 17 February 2013 (UTC)[reply]

The definition given in the lead paragraph makes it clear that the Roche limit refers to a secondary held together by purely gravitational forces. Since such a body cannot be said to be rigid, "rigidity" does not apply, and the word should be removed unless, that is, an alternative definition is covertly implied, in which case the alternative definition should be given. Similarly, references to "fluid" (as in the caption to the illustrations) is inappropriate, and should be removed. Discussion of the minimum orbital radius for tidal stability of rigid or fluid bodies should be given under a separate heading which makes the distinction clear. A further contribution I would like to make is that I came to this page after Googling "tidal radius". It seems that Google brought me here because the two words "tidal" and "radius" both appear (albeit separately) in the lead paragraph. This is no fault of the article, but, given that Google operates in this way, it would be useful if a comment were included in the lead paragraph advising the reader that although both concepts refer to gravitating bodies, they are otherwise unrelated, and that the present article does not refer to "tidal radius". An appropriate link to "tidal radius" would also be useful to anyone similarly misdirected by Google. g4oep — Preceding unsigned comment added by 77.96.60.31 (talk) 11:48, 23 October 2014 (UTC)[reply]

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)[reply]

I believe the French would pronounce it like the English word "rush". 67.232.193.190 (talk) 05:18, 13 December 2008 (UTC)[reply]

The correct pronumciation must be that which the Roche family themselves used. Only a French person, preferably from the Languedoc-Roussillon region, can indicate the proper pronunciation. 82.163.24.100 (talk) 21:45, 18 August 2010 (UTC)[reply]

Anyone who knows a bit about French can tell you that it has to be /ʁoʃ/ – Up to the point where the typical French r might not be used in the Langue d'Oc. But then again the Roche family might be Oil-speaking northerners who had moved into the Languedoc region ...

Until someone finds out more about this, I am putting in what I believe would be a "modern standard french" pronunciation, mainly just to replace the American pronunciation.

Sorry I couldn't make that fancy curly-bracket-IPA-link work for me.

--BjKa (talk) 15:15, 21 December 2012 (UTC)[reply]

The current English pronounciation is given as /rɔːʃ/, added by Bringback2ndpersonverbs in March 2018. I think it's likely that this user has the cot-caught merger - the BrE pronounciation I'm familiar with is /rɒʃ/, to rhyme with "mosh"; I've also heard /roʊʃ/, to rhyme with the first syllable of "ocean". There are various non-authoratitive sources for the pronouncation available on-line - is there a generally-acceptable source we can use? If so, we should cite it for the pronounciation - if not, we should probably remove the English pronounciation altogether. Tevildo (talk) 01:05, 13 January 2019 (UTC)[reply]

Well, my USE anecdatum would agree with your second pronunciation, with the o from ocean, but I'm not an astrophysicist. A quick search found this with that pronunciation, too, but that's not exactly definitive. Possibly it's a better source than the no source in the current page...--Noren (talk) 16:26, 13 January 2019 (UTC)[reply]

The New Shorter OED has /rəʊʃ/ (no). Webster's Third has "rōsh" (snow) or with a dot over the "o" (saw) (no IPA). Random House also has "rōsh" (over). Most prefer a long "o". Random House also has the French pronunciation as Rôsh (R in French rouge) (ô in raw). — Joe Kress (talk) 18:50, 13 January 2019 (UTC)[reply]

The OED pronunciation was under the headword "Roche", Webster's was under "roche limit", and Random House was under "Roche limit". By the way, "Shorter" (two volumes) means words obsolete before 1700 and example sentences (90% of the original OED text) were removed, the remainder is virtually identical to the 20 volume OED. — Joe Kress (talk) 22:36, 13 January 2019 (UTC)[reply]

Thanks! It seems the "ocean" pronounciation is the correct one, and the SOED or Webster's are both impeccable sources. It's not obvious what the respelling should be - H:PRK would suggest "rohsh", but I'd be tempted to use the incorrect "mosh" pronounciation on seeing that. "Roash", perhaps? Tevildo (talk) 20:33, 14 January 2019 (UTC)[reply]

Roche limit, Hill sphere, and radius of the planet.

This sentence is in the section mentioned above. "Celestial body cannot absorb any little thing or further more, lose its material." I'm sure it's not a correct sentence, but not sure what the intent is. Can someone please fix?

Rick Norwood (talk) 13:57, 30 September 2015 (UTC)[reply]

And what about the Lagrangian points? Do L1 and L2 are lying on the Roche Limit? --178.11.68.222 (talk) 12:31, 8 January 2018 (UTC)[reply]

I removed the section because it was completely unsourced and incoherent. It was also extensively damaged by an attempted copyeditor in January 2021. –LaundryPizza03 (d c̄) 08:01, 19 November 2022 (UTC)[reply]

What is the distance d?

What is the distance $d$ for the Roche limit? Is it between mass centres of bodies? It is not written clearly in the text. Voproshatel (talk) 16:24, 13 January 2023 (UTC)[reply]

It is between the centers of mass of the bodies. This is generally understood to be the case when talking about the distance between two bodies. Fermiboson (talk) 13:55, 29 January 2023 (UTC)[reply]

Exception?

In 2023, two rings were discovered around the minor planet 50000 Quaoar, both well outside what is thought to be its Roche limit. The main message about these rings is that they are not yet understood. And until astronomers have come up with an explanation for these rings, it is highly inappropriate to call them exceptions, and certainly not exceptions to the Roche limit (The Roche limit is not defined as the limit within which planetary rings can occur; so far it was presumed that planetary rings can only occur within the Roche limit, but if they are found outside it, it doesn't change the definition of the Roche limit). Three references are given: The Guardian, Nature and The New York Times. The first reference is merely a news item, based on the journalist's understanding of the original Nature-paper. And The NYT-item is also not an original article, and thus prone to interpretation errors by its author (Kenneth Chang, the journalist, calls the minor planet a 'world'). The source for the NYT-item can be found in Astronomy & Astrophysics Letters April 21, 2023. The only references that really matter are the latter one and the Nature-paper itself. Reading that last paper, it is clear that the authors themselves still have more questions than answers about these rings. Also take into account that the rings so far have only been detected indirectly: no one has yet seen or photographed them. My opinion is that it is highly premature to discuss this discovery here as an 'exception'. With what we currently know, it might be appropriate to take it up as a footnote after the line "almost all known planetary rings are located within their Roche limit" in the paragraph 'Explanation', where also two rings of Saturn are mentioned as 'exceptional'. Wikiklaas 23:06, 2 May 2023 (UTC)[reply]

Yeah, it all seems too early and speculative. I rm'd the section William M. Connolley (talk) 08:22, 3 May 2023 (UTC)[reply]

What is 'R'?

The rigid-body formulas explain all terms.

The fluid formulas contain a term 'R'. What is 'R'? Is it the radius of the primary? The radius of the satellite? Something else?

--Cowlinator (talk) 21:24, 6 December 2023 (UTC)[reply]