Talk:Roche limit/Archive 1

Pronunciation

How is "Roche" pronounced? With a glottal "och" like the Scottish "loch", a hard "ock" as in "rock", or like the "tch" sound in "choice"? Mr. Brownstone 19:26, 18 Oct 2004 (UTC)

As Edouarde Roche was French, I assume it is a long "O" and a "sh", and that the only pronunciation I have heard. I've never heard it pronounced as in "loch". -- ALoan (Talk) 11:01, 19 Oct 2004 (UTC)

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Great article, I am very impressed with all of the stuff on the planets and tidal forces. I esp. liked the part about some moons being within the Roche limit but one one hand they are held together by tensil strength so they do not break apart but on the other any object on the surface not tied down might leave the surface.--ShaunMacPherson 23:02, 18 Mar 2004 (UTC)

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Yes, I have to say this is one of the coolest articles I have ever read here. I came to the discussion page just to say that. --Doradus 13:51, 18 Aug 2004 (UTC)

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As the current article states, "Since tidal forces overwhelm gravity within the Roche limit, no large body can coalesce out of smaller particles within that limit." Also, I note that the "Roche limits for selected examples" table has alot of moons orbiting approximately at their 100% (+-15%) non-rigid Roche limit. Seems worthwhile to discuss why so many moons are close to the Roche limits (and not further out). Wendell 03:27, 17 Oct 2004 (UTC)

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A very good article, properly organized to avoid bothering the non-techies with the math. But at the end it reminds me of an old Harris cartoon: Two mathematicians in front of a blackboard full of equations, one of them saying, "I follow your derivation down to the point where it says 'Here a miracle happens.'"

I got rid of one miracle by fixing the typo in an exponent. Then I set to work finishing the derivation: expand terms, simplify, forget s where it's added to d, simplify again. With great embarrassment I confess that every time I did it, I got the wrong answer, by a factor of 2.

The Wolfram site's derivation is much easier, and comes out right, even though one has to supply a derivation for the tidal force formula. It uses a mass element μ on the orbiting object, at distance r from the center. That difference should all cancel out, duhh.

It would be a favor, and would improve the article, if someone who can fill out this derivation would put up a couple of intermediate results to replace the arm-waving. Or I could replace the derivation with one using the other approach, which is not a copyvio, but we'd need a different diagram. Dandrake 17:52, Aug 24, 2004 (UTC)

I got the start of the derivation from here and I didn't check it :-( I've started working it out in the article and it looks to me like I'm going to be out by a factor of 8! It's very late I cannot see what I'm doing wrong. Check my working please. If we can't sort it out then it's very easy to change the diagram, if going with the Wolfram site's derivation makes things easier. Theresa Knott 00:46, 25 Aug 2004 (UTC)

Yes, that factor of 8 is the factor of 2 I was talking about (it emds up under the cube root, right?) You have no idea how much better this makes me feel. Dandrake 07:35, Aug 25, 2004 (UTC)

I think the problem is the part where you define what the tidal force is. Instead of using the difference between the gravity of the two bodies, you should start off with the formula from Tidal_force (and possibly link to that page) $F_{t}={\frac {2GMm}{r^{3}}}\,dr<<r$ Wuzzeb 01:59, 25 Aug 2004 (UTC)

Ok I fixed it, but it is still technically wrong. The tidal force is from a gradient of the force across one body, and splitting the body up into two (equal) pieces is the wrong way to aproach the problem. Wuzzeb 02:43, 25 Aug 2004 (UTC)

Looks good now. Should I fix up the diagram and reupload it? Wuzzeb 02:49, 25 Aug 2004 (UTC)

New diagram uploaded Wuzzeb 03:37, 25 Aug 2004 (UTC)

Thanks for the image update. -- Chris 73 Talk 03:44, Aug 25, 2004 (UTC)

Thank you so much everyone. I feel pretty daft putting in a deivation that I hadn't checked beforehand. Theresa Knott 08:35, 25 Aug 2004 (UTC)

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(William M. Connolley 09:27, 27 Aug 2004 (UTC)) This is a very nice article. I think it is true that only *small* satellites can orbit within their Roche limit... anthing the size of, say, the earth is close to fluid on these scales. I might be wrong though (both of jupiters moons cited on the page say on their own pages that they are small enought to avoid Roche).

Also, the article doesn't point out that the *fluid* roche limit is larger than the rigid one, ie deforming satellites will break up further out: presumably because once they start to deform a bit their self-gravity weakens.

Figure trouble

While both sets of figures are pretty, they're incorrect in a misleading way. The shape a liquid moon distorts to is a prolate spheroid, which means that when viewed from the side as depicted here the cross-section should be elliptical. Unfortunately, both Roche limit (ripped sphere).PNG and RocheLimitAnimation.gif both show, deceptively, shapes which are asymmetric. When the center of mass reaches the Roche limit the object would come apart on both sides, not just the closer side. Basically, the outer part is orbitting too quickly for the lesser gravity on it to the same degree as the nearer part is orbitting too slowly for the greater gravity on it, so the net outward force on each is equal. (This is true to a first order approximation, and I mean that in a scientific way... it's close enough that you shouldn't see a difference on these pictures. If the second derivative of the gravity is large enough there will be asymmetric distortion but this should be very small in most situations... in fact, it's small enough that the equations used here (taken from Wolfram) neglect this and assume a prolate spheroid. Noren 22:17, 27 Aug 2004 (UTC)

I'll change the png image accordingly - probably tomorrow so please check back. Thanks for your input. Theresa Knott (The token star) 22:30, 27 Aug 2004 (UTC)

Actually it was easier than I thought so I've done it this evening. (you may need to refresh your browser). I've added your explanation to the caption. Theresa Knott (The token star) 22:48, 27 Aug 2004 (UTC)

I've been bold and edited the caption heavily. I hope you like it. --Doradus 03:16, 28 Aug 2004 (UTC)

Much better! I like the new figure! I also edited the captions heavily in addition to adding a paragraph. I'm somewhat worried I may be overexplaining it, but the figure troubles seem to indicate that the earlier explanation wasn't clear enough. Noren 06:17, 28 Aug 2004 (UTC)

No offence intended, but I reverted the captions. I think your edits left them incorrect: it said "the near end is pulled toward the center" which is not true. Perhaps you meant that it is pulled toward the star, but then so is the rest of the body (such is the nature of gravity) so that doesn't really explain anything. The tidal effects are due to the differences in gravity, and the effect is that the near end of the body is pulled harder than the center of the body, which in turn is harder than the far end. That's what my caption said originally, so being unable to reconcile our two versions, I just reverted to the last correct explanation.

However, as you say, apparently this explanation is still not clear enough. --Doradus 12:56, 28 Aug 2004 (UTC)

The net force on the near end is toward the center [of mass of the system], while the net force on the far side is away [from the center of mass of the system.] I think your captions as they stand miss this latter point, and in doing so miss the whole point how tidal forces work. Noren 15:00, 28 Aug 2004 (UTC)

No, in an inertial reference frame, the net force everywhere in the system is toward the center of gravity. Even at the far side of the orbiting body. Tidal forces arise from the difference between the pull at different places on the orbiting body. The far end is sill pulled toward the center of gravity of the system, but just not as strongly as the center of the object is pulled.

If you want to consider a reference frame rotating with the orbiting body, then yes, the centrifugal force on the far end would be greater than the inward gravitational force. However, the explanation I gave doesn't depend on any particular reference frame, and I think that makes it simpler. --Doradus 15:59, 28 Aug 2004 (UTC)

Reading through the rest of the article, I see that it refers to rotating reference frames repeatedly, so my objection is moot. I'd say go ahead and reword the captions the way you like them. --Doradus 16:44, 28 Aug 2004 (UTC)

I may also change my animation accordingly, but currently i am on a trip and can´t get to my regular computer. In any case, an animation is probably better as a link only, not the image directly. Chris 73 Talk 07:40, Aug 28, 2004 (UTC)

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"As all parts must orbit at the same velocity, the centrifugal force is constant throughout the satellite" ... This is false. All circular motion satisfies this formula:

$a={\frac {v^{2}}{r}}$

This means the centrifugal force falls off linearly with distance for a given velocity. --Doradus 16:44, 28 Aug 2004 (UTC)

Hurm. Yeah, thinking about it some more I think you're right, the tidally locked satellite would experience different centrifugal force- my explanation is in error. I'm going to rewrite it in terms of differing orbits(which was how I initially worded it before I tried to 'simplify' it incorrectly.)Noren 22:50, 28 Aug 2004 (UTC)

On second thought, your analysis is incorrect, though my original logic is still wrong too. The anglular velocity is the same throughout the satellite- each part of the satellite rotates the same angle per unit time. Each part of the satellite travels in a circular path. The velocity of each part of the satellite isn't constant- the outer parts travel faster than the inner ones, linearly dependant on the radius. Since the velocity scales linearly to the radius, the force scales as the radius squared divided by the radius, and therefore increases linearly with the radius. Noren 20:45, 29 Aug 2004 (UTC)

You're right, it's the angular velocity that is constant here, not linear velocity. That would correspond to $\omega =v/r$ in the above equation. Substituting, we get this:

$a=\omega ^{2}\times r$

...which is what you just said.  :-) --Doradus 03:39, 31 Aug 2004 (UTC)

Figure viewpoint

The third of the series of images might be more instructive if the view was down the normal to the orbial plane, rather than in the orbital plane. That way in the destruction shot pieces in inner orbits can be seen to move more rapidly than pieces in outer orbits. This might (possibly) help with people's mis-perceptions regarding centrifugal force, etc. -- Kop 21:57, 2 Sep 2004 (UTC) (I posted this before in it'sown section but somehow the edits got lost.

One way to depict the satellite's oribal path without confusing it with the curve representing the Roche Limit would be to overlay a thin (red?) curved arrow centered on the orbiting body. The 3rd image of the object in the Roche Limit the curved arrow would overlay the Roche Limit curve. The length of the arrow could vary to represent orbital velocity. That way the first two images could be used in an orbital mechanics article. Besides, the arrow would have to have a length anyhow, so the length may as well convey information. Obviously, the curve radius of the arrow would vary depending on the size of the orbit. -- Kop 00:51, 3 Sep 2004 (UTC)

Figure trouble, part deux

Ok Noren, I have reworded the captions in a way I think you will like. What do you think? --Doradus 20:01, 29 Aug 2004 (UTC)

(William M. Connolley 23:04, 29 Aug 2004 (UTC)) You have made it worse! Your caption now implies that one end is pulled by gravity, whilst the other end is pulled by CF. A lot of what is there shouldn't be on the Roche page at all - it should be on the tidal forces page, and is, but better. Remember - the tidal forces exist independent on the centrifugal forces. Tidal forces are really change-of-coordinates forces.

Good point! A body not orbiting a planet, but falling straight toward it, still experiences tidal forces. I have simplified it further and just linked to tidal forces.

Objects on the planets surface don't necessarily fly away

I reverted this sentence: "However, an object resting on the surface of such a body and held in place only by that body's gravity could find itself pulled away by tidal forces." (emphasis mine). It had been changed to say "would" which is incorrect. If the object is on the limb of the body as viewed from the planet, then tidal forces will actually tend to hold it on the body's surface. --Doradus 18:18, 30 Aug 2004 (UTC)

I'm not exactly sure what you mean by a limb, but you're right that the position on the moon is relevant. On a rigid, tidally locked prolate spheroid moon in a circular orbit within the Roche limit, an object placed on the axis would "find itself pulled away" but one placed on that moon's equator would not. --Noren 22:05, 30 Aug 2004 (UTC)

The limb of a celestial body is the edge of its apparent disk. For a spherical body, this is (roughly) the great circle normal to the line of sight passing through the body's center. This is not the same as the equator, which is the great circle normal to the axis of rotation. --Doradus 03:23, 31 Aug 2004 (UTC)

Urhixidur is taking this article in a very verbose direction. This "sub-primary pole" part is tangential and distracting, and still doesn't warrant changing the "could" to a "would". I'm sure if I try I can think of other reasons that an object on the surface would not be pulled away, even at the sub-primary pole (say, local gravitational anomalies). I'd rather just keep it simple and leave it as "could". --Doradus 16:42, 31 Aug 2004 (UTC)

Inexact but precise

It seems a tad silly, Noren, to use \approx when you've got 10 significant digits. Mathematically, you're right. Physically...

Urhixidur 23:39, 2004 Aug 30 (UTC)

The use the equals sign to signify 'very close' is simply incorrect. The equals sign has a defined meaning, and this use doesn't fit its definition.

What is Urhixidur's objection here? That there are too many sig figs, or that we should use a symbol other than \approx? --Doradus 03:32, 31 Aug 2004 (UTC)

I made an edit (21:34, 30 Aug 2004) which changed a particular = symbol to \approx, I believe Urhixidur was responding to that specific edit. At least, I was defending that specific edit... --Noren 22:54, 31 Aug 2004 (UTC)

As a somewhat separate argument, I doubt the assumptions made to generate that number are actually valid to 10 sig figs... in particular, I don't think the shape will be precisely that of a prolate spheroid. Away from the long axis, the net force is asymmetric with respect to the 'near' and 'far' ends. This would cause distortions dependent on the size of the satellite with respect to the rest of the system, which the formulae actually used neglect. --Noren 00:31, 31 Aug 2004 (UTC)

That's more or less what I said.  ;-) Maybe we should use the ≊ symbol (which means "almost equal or equal"). Mathematically speaking, ≈ is "almost equal" whereas ≅ is "approximately equal". Those are the Unicode labels, anyway.

As for the derivation, Wolfram's seems rigorous, but it does start from the assumption of an oblate spheroid gravitational potential, which, as you say, is wrong (because the orbit isn't straight). The difference is likely to be very small, however, unless one is dealing with a very tight orbit (compared to the moon's size).

Urhixidur 03:28, 2004 Aug 31 (UTC)

Orbit to Roche limit ratios

I have changed the last table to make it (what I consider) more legible. If people don't like it, I'll revert it (or you can :-). --Doradus 18:23, 1 Sep 2004 (UTC)

Roche limit dependancy on density discussion?

We don't discuss at all in the non-heavy-math sections that the radius changes linearly based on the density. Perhaps a line somewhere stating that the denser the moon, the greater the self-attraction is, resulting in a smaller Roche limit? --Noren 23:52, 1 Sep 2004 (UTC)

Explaination of tidal force needed for complete article

The second sentence of the introductory paragraph has gone back and forth as a short explanation of tidal force. It has finally been deleted due to inaccuracies introduced. It is important that there be an explanation of tidal force somewhere in the introduction or the article will not stand on it's own, the lay reader won't 'get it'. -- Kop 22:01, 2 Sep 2004 (UTC)

It's true that the centrifugal force isn't a tidal force, however the Roche Limit explicitly applies to orbiting bodies and the centripital force (which increases linearly with distance at constant angular velocity) is relevant. I undeleted that sentence. Part of the reason I want something to that effect in goes back to when everyone trying to make a figure misunderstood the old explanation enough to think that the forces applied were asymmetric. --Noren 22:12, 2 Sep 2004 (UTC)

(William M. Connolley 22:16, 2 Sep 2004 (UTC)) Noren has reverted several of my edits as "vandalism". I am insulted, and in due course I expect an apology. For now, let me point out that, given the *definition* of roche limit on the page, centrifugal forces and orbit (or not) are *totally irrelevant*. Perhaps Noren might try reading the tidal force page.

You (Noren) appear to assume (for your physics) that the moon is tide-locked (think about it).

Indeed I do make that assumption, just as the Wolfram site does, and just as the explanation of Tidal forces in the liquid limit does. Perhaps you might try reading that Wolfram page. If you wish to back up your claim that centrifugal forces are "totally irrelevant", please explain the formula as listed on the page:

$d\approx 2.423R\left({\frac {\rho _{M}}{\rho _{m}}}\right)^{\frac {1}{3}}\left({\frac {(1+{\frac {m}{3M}})+{\frac {c}{3R}}(1+{\frac {m}{M}})}{1-{\frac {c}{R}}}}\right)^{\frac {1}{3}}$

without reference to centrifugal forces. --Noren 22:37, 2 Sep 2004 (UTC)

(William M. Connolley 17:03, 3 Sep 2004 (UTC)) The point about the wolfram site is that it makes the assumption *and says so*. Not silently. And only on the liquid bit: confusingly, the Wolfram site *doesn't* use spin for equation (6) (for a rigid body) but *does* factor in spin for the liquid case.

A perfectly spherical moon, as in Wolfram's equation 6, has no asymmetric force applied to make it spin lock. Of course, real moons which are even slightly non-spherical do feel asymmetric tidal forces and therefore typically spin lock. The non-spin-locked case is actaully the unusual case rather than the reverse. The very simplest case is fully derived on the page, and even this simplest case that is borderline being too much math; the fact that the most simple case is the one fully derived on the page does not imply that the simplifying assumptions used in it are universally valid. The full formula including orbital parameters is and was listed on the page does take into account a spinlocked moon and the results thereof- and is more generally applicable. I'm still waiting for you to defend your claim that orbit is totally irrelevant to that equation listed on the page and reproduced above. --Noren 16:40, 4 Sep 2004 (UTC)

(William M. Connolley 18:57, 4 Sep 2004 (UTC)) I made no such claim.

If we can explain tidal forces in a single short phrase (like "tidal forces--due to the differences in gravity felt by different parts of the body") then I'm all for it. However, this may be hard to do, because tidal forces are not a simple concept, so linking might be sufficient. --Doradus 02:49, 3 Sep 2004 (UTC)

Yeah, no phrasing tried so far (including those I've written) quite works, alas. I've been toying with a phrasing involving differing orbital periods... something along the lines of orbits nearer the center have shorter periods than those farther away, thus the near side is orbiting too slowly to be stable while the far side is orbiting too quickly to be in a stable orbit, but I haven't had any luck getting it concise and clear enough to work for the low-math section. I do think that reference to the different periods at differing orbital radii might be a comprehensible tack to try though. --Noren 05:29, 3 Sep 2004 (UTC)

I believe the problem is that writing clear sentences is hard, not that tidal force is complicated. After all, tidal force is just "gravity pulls more on the close side of a satellite than the far side, and closer orbits have shorter periods", and both these clauses are just different ways of saying the same thing, as is "the far side of the satellite is subject to centrifugal force as it's orbit spins it about it's primary." It would be nice to sumarize tidal force in a single sentence, but that's not a requirement. Regardless of how much verbage it takes, the article's not self contained without a tidal force explaination. And I just don't believe that a short explaination is so hard we should give up. I think several of the versions no longer in the article would do. If nothing else, recall that education is all about lying so as to simplify the real world enough to make it understandable. -- Kop 16:55, 3 Sep 2004 (UTC)

(William M. Connolley 17:03, 3 Sep 2004 (UTC)) Centrifugal forces, just like tidal ones, are symmetric.

Also, I'm sorry to say I think Mr. Connolley might be right, even if he is coming across as a smug dork. For instance, Shoemaker-Levy 9, mentioned in this very article, was not in orbit around Jupiter in any meaningful sense, yet it disintegrated because it passed inside its Roche limit with Jupiter. --Doradus 02:49, 3 Sep 2004 (UTC)

Comet Shoemaker-Levy 9 was indeed in orbit around Jupiter- a highly eccentric orbit which decayed. In 1992 its periapsis was only 1.3 R (Jupiter radii) away from the center of Jupiter, within its Roche limit, but its (highly elliptical) orbit decayed enough so that its periapsis was less than 1 R on the subsequent approach in 1994. This fact is absent from the Wikipedia Shoemaker-Levy 9 page, but is not contracticted by that page. I'm not seeing a contradiction. If you'd like an outside reference that the Roche limit is defined to specifically apply to orbiting bodies, see the the Encyclopedia Britannica entry. --Noren 05:29, 3 Sep 2004 (UTC)

Aha, I guess I missed your point there... it's the Roche Limit page which is in error about Shoemaker-Levy 9, not the Shoemaker-Levy 9 page. I'll edit. --Noren 05:44, 3 Sep 2004 (UTC)

Allright, let's get constructive. What's wrong with this sentence? (from the last revision in which it appeared):

"Within the Roche limit the net forces experienced by opposite ends of the satellite, gravity acting more strongly on the side closest to the body orbited and centrifugal force acting more strongly on the far side, are stronger than the forces holding the satellite together."

Yes, tidal force is symmertic. So is centrifugal force. But this sentence does not say that they are not, just that there is more of each force on on each side. After all, the differential of each force across the body is accounts for each force's net destructive tendency. --Kop 19:40, 3 Sep 2004 (UTC)

(William M. Connolley 20:27, 3 Sep 2004 (UTC)) It can be construed correctly, but it is easy (and possible) to misconstrue it as sayin ravity pulls one end and centrifual forces te oter. (BTW, I ave lost my hs and gs (I ad to paste tose)). *Also* (and I keep comin back to tis) te *defn* of Roce limit is balance of self-rav and tidal *not* centrifual.

Ok, tidal. So:

"Within the Roche limit the net forces experienced by opposite ends of the satellite, gravity acting more strongly on the side closest to the body orbited and less strongly on the far side, are stronger than the force holding the satellite together, the satellite's own gravitational attraction."

I wouldn't think we have to write something that cannot be mis-construed. That's why we link to the tidal forces article.

--Kop 21:44, 3 Sep 2004 (UTC)

(William M. Connolley 22:02, 3 Sep 2004 (UTC)) Yes, I agree with that, and link to tidal forces for the detail.

Ok. I've gone ahead and added this sentence to the article. -- kop 16:57, 5 Sep 2004 (UTC)

Definition

(William M. Connolley 08:44, 3 Sep 2004 (UTC)) The definition of roche limit is:

The Roche limit is the distance within which a satellite ... will start to disintegrate due to tidal forces exceeding the satellite's gravitational self-attraction.

I've stripped out the deliberate insertion of "orbiting", of course.

The definition of the Roche limit is that it applies to orbiting bodies. See the Encyclopedia Brittanica entry. You are welcome to discuss tidal forces acting on non-orbiting bodies on some other wikipedia page, but please leave this page alone- by the definition of Roche limit this page should be a description of tidal forces on orbiting bodies. --Noren 17:12, 3 Sep 2004 (UTC)

(William M. Connolley 19:22, 3 Sep 2004 (UTC)) That isn't reason, it is appeal-to-authority. The Roche limit applies, irrespective of whether the body is orbiting or not. You could, if you wished, write a page artificially restricted to the case of an orbiting body; but its not at all clear why you would wish to do so.

Well, on Wikipedia, appeals to authority are relevant, because Wikipedia is not meant to be a primary source. It is supposed to be a collection of knowledge from other sources.

(William M. Connolley 18:57, 4 Sep 2004 (UTC)) Subtlety and humour is clearly a bad idea. I'll make one more try though: go check the edit comment Noren made when he inserted his bit.

I interpret that to mean this: if you want to remove the references to orbits in this article, you need to find an existing definition of Roche Limit elsewhere that does not refer to orbits. So far, in my searches, I have been unable to do so; so for Wikipedia to discuss the Roche limit without reference to orbits would apparently be novel and unprecedented, and that's undesirable. --Doradus 17:08, 4 Sep 2004 (UTC)

(William M. Connolley 18:57, 4 Sep 2004 (UTC)) You need to *read the defnintion* currently used on the page, which (with omission of the word "orbit") is apparently uncontroversial. Do that, and you notice that *nothing other than grav forces enter into it*. Ergo, it is orbit-independent. You coud define another limit, one that includes orbtial (centrifugal) forces, if you like.

Um, using the definition on the Wikipedia page to justify itself is absurd. It doesn't matter whether anyone's definition of Roche limit (including ours) is "uncontrovercial" when the word "orbit" is removed; if we are breaking new ground by removing that word, then we are acting like a primary source, which is bad. --Doradus 02:27, 5 Sep 2004 (UTC)

Anyway, this may be moot, because I think it is impossible for an object within its Roche limit of a planet not to be in some kind of orbit about that planet. The orbit may be hyperbolic, or even degenerate (ie. a straight line toward the planet's center), but arguably those are also orbits. (I'm prepared for the "everything is in orbit around everything, then" debate if you want to go there. :-) --Doradus 17:08, 4 Sep 2004 (UTC)

If you took a perfectly written page about gyroscopic forces, one which explained that subject clearly and accurately, replaced the word 'gyroscopic' with 'tidal' throughout and put it up as the tidal forces wiki page- it would be entirely self consistent and would be an interesting page about an interesting topic, but it would still be wrong. This objection is to the definition you wish to use for the phrase "Roche Limit", not an argument about the internal self-consistency if your definition of the term is used. On the subject of what the definition of a phrase is, a citation of an encyclopedia is indeed an Appeal to authority but is not a logical fallacy, as the Encyclopedia Brittanica is a legitimate authority on the definition of terms. You have still not presented evidence in support of your contention that the formal definition of "Roche Limit" includes non-orbiting bodies. --Noren 18:33, 4 Sep 2004 (UTC)

(William M. Connolley 18:57, 4 Sep 2004 (UTC)) This is going round in circles, which is dull. The Roche limit applies, irrespective of whether the bodies are orbiting. You *could* restrict it to orbiting bodies only, but its rather hard to see the point.

OK, now lets look at this defn: it *defines* the Roche limit in terms of TWO forces: tidal and self-gravity. Tidal forces are due entirely to the grav field of the larger object *and do not depend on the motion of the smaller object*. Self-grav is due entirely to the mass of the smaller *and does not depend on the motion either*. The orbit is irrelevant, in this definition. This is also consistent with the wolfram defn.

This is all true, but irrelevant. If no other source defines Roche limit without orbits, then we can't do it either. If our definition of Roche limit is the only one that does not refer to orbits, then it's wrong. --Doradus 02:27, 5 Sep 2004 (UTC)

Now, as to the centrifugal forces: the problem here is that you *don't know what the spin of the moon is*. Wolfram *assumes* that the moon is spin-locked. But it needn't be. It could just as easily be spinning much faster (most satellites are). Or, in an extreme case, it might be spinning once-per-revolution (if it was orbiting) in which case there would be no centrifugal forces at all, since it wouldn't be spinning wrt to fixed stars.

But most moons in the solar system are in synchronous rotation- including, of course, the moon. Also, consider that the closer the orbit of a satellite gets to its Roche limit the more force is exerted to tidally lock it. --Noren 17:12, 3 Sep 2004 (UTC) _.

(William M. Connolley 18:25, 3 Sep 2004 (UTC)) According to what I have recently read, you are correct & I was wrong: most moons *are* tidally locked. However, that doesn't affect the definition of Roche limit, both here and on wolfram, which is written purely in the terms I give above. I notice that you haven't disputed that.

The first part of the article is for non-technical readers. Bluntly, if the satellite is not orbiting, there are no tidal forces.

(William M. Connolley 16:47, 3 Sep 2004 (UTC)) No, this is false. The tidal forces are totally unaffected by the motion of the two bodies (we are in newtonian physics here).

While Kop's statement is incorrect, so is yours. Had you bothered to actually think through Newtonian physics you might have realized that two separated points at the same radius at different points in a circular orbit feel no tidal forces with respect to one another, either attractive or repulsive (though they may be attracted to one another by simple gravitation, but I hope you'd agree that that is not a tidal force.) An erroneous phrase which you are repeatedly inserting on the page asserts in part that:

"tidal forces are repulsive along the axis to the central mass and attractive along the "equator" perpendicular to this: see tidal force."

These attractive forces are tidal restoring force, are part of the tidal force tensor but are distinct from the (purely repulsive) tidal force. They apply to and are a function of a free-fall reference frame.

In the case of a circular orbit, let the z axis be the direction toward the center of mass, the x axis be the (perpendicular) direction the satellite is traveling. The attractive and repulsive forces balance at a constant R from the center of mass. You may wish to get pedantic here and claim that there are equal and counterbalancing tidal attraction and centrifugal repulsion forces, but that would just be a game with frames of reference obscuring the fact that in basic Newtonian physics there's an arc of points in a stable circular orbit experiencing no tidal force in any direction.

It is true that along the y axis the attractive tidal restorative forces do apply and will be attractive in the orbital reference frame, but your claim that they apply to the entire circle perpendicular to the z axis is false. Your assumption of a freefall reference frame on the tidal force calculation is incompatable with an orbiting body reference frame. --Noren 18:33, 4 Sep 2004 (UTC)

It will stay a sphere until it finishes falling and smacks into the other object. The non-technical reader may not understand this, or even know that if it's not orbiting it's not a satellite. I think the word orbiting should be in the first paragraph and the first caption.

(Doradus 16:49, 3 Sep 2004 (UTC)) That is false. Even a non-rotating object falling straight toward a non-rotating planet still experiences todal forces. They occur because the near end of the object experiences a greater gravtiational exceleration than the far end.

Yes, and this phenomenon an interesting one but is offtopic on a page about the Roche limit. --Noren 17:20, 3 Sep 2004 (UTC)

Agreed. --Doradus 02:27, 5 Sep 2004 (UTC)

Because centrifugal forces must be present because the object must be orbiting? This implies that a fluid body in free fall would not disintegrate until it's inside the Roche Limit. -- Kop 18:39, 3 Sep 2004 (UTC)

(William M. Connolley 19:22, 3 Sep 2004 (UTC)) No, its perfectly on-topic, because a non-orbiting body who trajectory takes it inside the Roche limit will happily break up from tidal forces.

Why is it off-topic? Where did you get the idea that the Roche Limit deals only with orbiting bodies? --Doradus 20:05, 3 Sep 2004 (UTC)

That the object is in orbit is part of the definition of the term 'Roche Limit' itself. A lengthy discussion of pears on a page about apples would be similarly offtopic, regardless of the accuracy and quality of the writing. Non-orbiting bodies may or may not break up at the Roche limit due to tidal forces, but the definitions I've found for the term include either orbit or refer specifically to a satellite. It's my understanding is that orbit is part of the definition of the term 'Roche Limit' is- how this effects the math and physics has to come after the definition of what the words actaully mean. --Noren 18:33, 4 Sep 2004 (UTC)

On a separate note, I thought the centrifugal forces in question were those on the far side of the orbiting body, which, being farther out, is orbiting more rapidly than that it would if it were not part of the satellite and is hence subject to centrifugal force.  ? Nothing to do with spinning, which seems outside the scope of the article.

-- Kop 16:20, 3 Sep 2004 (UTC)

(William M. Connolley 16:47, 3 Sep 2004 (UTC)) Firstly, you are assuming that the body is tidally-locked - this is OK, providing you know you are assuming it. But take for example the earth, which isn't (wrt the sun; or even to the moon if you want). The centrifugal forces on the earth are far larger (because the spin rate is higher) than those you are assuming.

Secondly, just like the tidal forces, the centrifugal forces are symmetric about the center of the moon: they are repulsive on the near and far sides.

Ok. So, we're talking tidally locked. Even suppose we are talking about zero rotation about it's own axis, which is close enough to tidally locked for purposes of the non-technical explaination section. I don't understand how the centrifugal forces from spinning in orbit are replusive on the near side. On reflection, I'm not sure this is worth explanation as it seems to me that tidal forces would dominate the situation by far. Yes? -- Kop 18:39, 3 Sep 2004 (UTC)

(William M. Connolley 19:22, 3 Sep 2004 (UTC)) I haven't done the math as to which force is larger: at a rough guess, they should be similar (err, but don't quote me on that). As to why the centrifugal forces are repulsive on both sides: they are (roughly) just like the tidal forces: the centrifugal force *that is tending to break the satellite up* is the difference of the force at a given point, and the force at the center. Since (cf=centrifugal force) cf_near < cf_center < cf_far (assuming tidal locking), then (cfb=centrifugal_breakup, = cf - cf_center) cfb_near < 0 == cfb_center < cfb_far. Taking into account the directions, you have symmetry. I think this is right...

If centrifugal force is ~ half of what's pulling the satellite apart, and you need an orbit to get centrifugal force, then shouldn't an orbit be part of the definition? I'd think a circular orbit. The Roche Limit is a fixed sphere, but if centrifugal forces are so important a fluid object dropping straight down (the most degraded of the non-circular-orbital cases) would not break apart until well inside the Roche Limit. So in order to talk about the Roche Limit being where things break apart, then you also have to have those things in orbit.  ? -- Kop 19:54, 3 Sep 2004 (UTC) Are there two Roche Limits, one for bodies that relates to satellites in circular orbit, and one for orbits (including degenerate cases)? If so, then we just need to decide which kind is discussed where in the article. -- Kop 20:43, 3 Sep 2004 (UTC)

\*sigh* I guess I should read the math before commenting further. -- Kop 18:39, 3 Sep 2004 (UTC)

(William M. Connolley 19:22, 3 Sep 2004 (UTC)) Wisdom!

The "orbit" edit war

Please, Noren and William, quit reverting each other. You clearly disagree on which version is best, but until we resolve that question, neither version is so superior as to justify the continual reversions. Both of you just please leave the article alone until we reach a concensus. --Doradus 02:27, 5 Sep 2004 (UTC)

Well said! I been away from this article for a little while, and when i come back there's an edit war on. I'm sure we can come to a consensus if the reverting stops though. Theresa Knott (Nate the Stork) 13:11, 5 Sep 2004 (UTC)

(William M. Connolley 13:24, 5 Sep 2004 (UTC)) Hello Theresa. Yes we are having an unseemly war over the article you improved so nicely. I won't edit it further until you have had a chance to look at it and comment substantively.

Well it's not really my place comment extensively since you both know far more about the subject than I do. However I did not assume the body was orbiting when I drew the first set of images, nor when i started the derivation. The web sites I looked at all said that to calculate the limit, balance the tidal forces with the self gravity. There was no mention of any other forces. I suppose it depends on how you define ther Roche limit - is it the point at which tidal forces break up a sattalite, or is it the point at which a true sattelite would actually break up? How did Roche define it himself? Theresa Knott (Nate the Stork) 14:38, 5 Sep 2004 (UTC)

I agree with the jist of your point: that we should find some kind of existing authoratative definition. Often scientists don't define the terms that bear their names. For instance, you couldn't ask Newton what the term "newtonian physics" means; it's a term defined by later generations of scientists that differentiates Newton's views from those that came before and after. So don't be surprised if you can't find a good answer from Roche's own pen, and you must look elsewhere. Encyclopedia Britannica is the most authoratative citation I have seen so far, for what that is worth. --Doradus 03:09, 7 Sep 2004 (UTC)

(Later)

Quote from [1] explaing the now out of favour James Jeans theory of the formation of the solar system.

"Jeans theory doesn't involve the previously mentioned solar prominences. 
Jeans Theory instead involves the star passing within the Roche limit 
(The distance from the centre of a body, within which a second body would
 be broken up by gravitational distortion) to release solar material from the Sun.
This theory had trouble in explaining the angular momentum difference
 between the Sun and planets."

In this example the star is not orbiting the sun, yet is passing with the Roche limit of it and therefore breaking up. So I think it's reasonable to say that it's not necessary for a body to be in orbit for us to talk about a Roche limit. Theresa Knott (Nate the Stork) 15:11, 5 Sep 2004 (UTC)

Viewpoint of piccies

(William M. Connolley 17:31, 5 Sep 2004 (UTC)) The stuff about viewpoints of the piccies is a bit odd. All the (TK) pictures have the orbit drawn in, so I had assumed that all the pictures were looking down on the plane of the orbit. But now the last one says that we are looking from a different axis. But this is odd, because the tidal forces are (of course) symmetrical around the sun-moon axis. The *last* piccy has to be looking down on the orbit: but there is no reason why all the others shouldn't be too.

No it's not the orbit that's drawn in. The white line is supposed to represent the Roche radius. Having said that you are right about the viewpoint. I was thinking sideways on view but there is no reason it couldn't be looking down as long as it is explicitly stated that the the bodies are not orbiting. The advantage of a side view is it liiks correct even if the bodies are orbiting. Theresa Knott (Nate the Stork) 17:55, 5 Sep 2004 (UTC)

(William M. Connolley 18:20, 5 Sep 2004 (UTC)) Oops, of course not its the (sphere) of the Roche limit. The advantage of having them all "top down" is that there is no (potentially confusing) switch of viewpoints in the last piccy.

But the 3rd and 4th picture must be from different viewpoints if the mass is in orbit --elsewise you wouldn't see the 'tails' in the 4th picture, or you would see them in the 3rd. It's useful to know that either the mass in the third picture is not in orbit, or that it looks different depending upon viewpoint. --kop 18:53, 5 Sep 2004 (UTC)

Just saw the two new pics for the first time. Very cool. I'd also prefer if they could all be from the same POV (since the Roche limit lines are bound to be confused with orbit lines) but the captions are clear enough for me. --Doradus 03:00, 7 Sep 2004 (UTC)

The only way that pics 1-3 & 4-5 can be the same POV is if we explicitly state that Pics 1-3 are for a non orbiting mass. Which is probably not a bad idea anyway. Thoughts anyone? Theresa Knott (Nate the Stork) 07:48, 7 Sep 2004 (UTC)

The Roche limit applies to satellites. The most direct online reference I've found states that in 1850, the French astronomer E. A. Roche (1820 – 1883) stated “no satellite can exist closer to a planet than 2.44x its radius or 1.44x from its surface.” If a part of the Roche limit wikipedia page does not match this definition, the correct course of action is to change that wiki content to reflect the correct definition of Roche limit rather than to pretend that the phrase "Roche limit" has a different meaning. The fact that free falling bodies may also break up (approximately) at the Roche limit doesn't change its definition. Unlike satellites, free falling bodies will not have enough time to equilibrate with respect to changing forces making it a dynamic, and therefore much more complicated, problem to model.

All that being said, I don't see why 1-3 and 4-5 must be of different views- I've been visualizing 1-3 as depictions orbiting bodies viewed from above the plane of the orbit, with the white arc representing both the Roche limit and orbital path of the orbit at the Roche limit. The cross-section of a prolate spheroid is the same from either of those two directions, so the elliptical shape looks correct as viewed from either a 'top down' or 'from the side' perspective. Adding the arc (perhaps in a different color?) of the orbital path to 1-3 as well some caption modification would clarify this. I don't see an advantage in changing the perspective along the sequence. --Noren 16:56, 8 Sep 2004 (UTC)

(William M. Connolley 08:41, 7 Sep 2004 (UTC)) (Drop indent). It would make things simpler, which would be nice. Pics 1-3 pretty well imply that the object is falling in rather than orbiting anyway.

On vaguely similar lines, there ought to be something in the article about how long it takes an object to fall apart. If the object is orbiting (almost) but slowly spiralling in, then clearly it has time to fall apart noticeably. If its just falling straight in then I suspect you would barely notice its fragmentation before it went plop.

Having thought a bit further... pic 4 (& 5) may or may not be for the same viewpoint but it is important that the object now be orbiting (o.w. no rings). So some change-of-scene is necessary. In fact, might it be better (in order to emphasise this) to separate 4&5 to somewhere lower down, in order to avoid confusion? 4 & 5 are about something else - how different orbits smear the thing out into a ring - whereas 1-3 are about tidal forces.

Theresa Knott (Nate the Stork) 09:59, 7 Sep 2004 (UTC) (replying likes this saves indenting at all) Sperarting them out is a good idea IMO. As for how long it takes to fall apart, how long did it take Shoemaker-Levy 9? does anyone know? Theresa Knott (Nate the Stork) 09:59, 7 Sep 2004 (UTC)

Unfortunately, we didn't detect it until March 1993- it is believed to have broken up at perijove on or about July 7, 1992, a date determined from its orbit rather than direct observation. While it was separated at that time, I expect the distance between pieces increased during the two years of its final orbit. It passed 1.3R away, or only .3 R from Jupiter's 'surface'- so I suspect some atmospheric drag may have come into play in addition to the tidal forces... --Noren 16:56, 8 Sep 2004 (UTC)

Doradus 15:32, 7 Sep 2004 (UTC) -- Whether we emphasize or de-emphasize the fact that pics 1-3 are of a non-orbiting body depends on how we settle the question of whether the Roche limit is defined to include non-orbiting bodies. I still have not yet seen anyone post a reference to a definition that doesn't mention orbits, so I'm inclined to (for instance) remove picture 3 and let the other four represent the viewpoint normal to the ecliptic (if you see what I mean).

Debris a little dim?

I hate to critique such nice pictures, but I find the debris a little hard to see. Particularly in the 4th picture and the reddish debris in all the images. On a related note, I wonder whether the contrast in tone is enough that someone who's color blind could see the images. (I'd guess that converting them to black and white and seeing how they look would be the appropriate test.) Also, the rightmost arrow in the 4th picture does not seem to be pointing along a circular orbit. None of these are big problems, but it seemed worth pointing them out while the images are fresh-on-the-mind. -- kop 19:33, 5 Sep 2004 (UTC)

OK no problem I can sort that out. Theresa Knott (Nate the Stork) 22:11, 5 Sep 2004 (UTC)

Centrifugal force

There is no such force, is this in reference to the false sense of centrifugal force within the inertial reference frame? Otherwise, its a word that should be changed. The roche limit could be seen as the actions of a centrifugal force in that it spins things appart, but is it not just the change due to the interaction of the unever distribution of net forces. That is all I have to say, I knwo this has been discuessed before, but a featured article witha fictional force is not a tiny issue. Ctrl_build 10:04:19, 13 Oct 2004 (UTC)

(William M. Connolley 21:36, 13 Oct 2004 (UTC)) I'm not really sure what you are saying here. The *definition* of roche limit doesn't involve centrifugal forces. Are you complaining that CF force isn't a real force?

(Doradus) Regardless, centrifugal force is not fictional. In rotating reference frames, there is a very real acceleration (and therefore force) away from the axis of rotation. Likewise for coriolis forces. There are numerous references to it in the 1911 Britannica, so at least it wasn't considered fictional by the experts of the day. Any reference that claims the force is fictional is merely playing with semantics, ie. "what exactly is a force?". If we keep off the ivory tower and define a force as anything producing acceleration, then centrifugal force is real.

Sorry - I meant to say the other day that I added a link to centrifugal force, which I hope deals with the original point (this article is not the place for a discussion of whether centrifugal force is fictional or not. Having said that, I'm not sure I understand how centrifugal force can be considered a "real" force, as opposed to a fictional force - isn't the point that there is an observed acceleration in a co-rotating frame of reference, which would lead a co-rotating observer to infer the presence of a force, whereas, as is apparent to an observer in a non-co-rotating frame, the "acceleration" is due to the absence of a (real) centripetal force - no?)-- ALoan (Talk) 11:00, 18 Oct 2004 (UTC)

(Doradus) I don't see how what you said makes the force "fictional". Einstein's general theory of relativity tells us that rotating reference frames are as valid as any other. If a rotating reference frame exhibits centrifugal force, then that's no more fictional than any other force.

There is a nice discussion of fictional force / fictitious force / imaginary force (all redlinks - perhaps I should write an explanation) in Talk:Coriolis effect. I suppose I am guilty of thinking in terms of classical mechanics (or, at least, special relativity and inertial frames) rather rather than than general relativity. At a guess, it may come down to whether the coriolis force and centrifugal forces can do work (I hope I am right in saying that they cannot). Gravity is a bit of a special case, since no work is done in keeping a satellite in orbit, so in that sense gravity is fictional (for example, move to a co-rotating frame of reference and a satellite in a circular orbit stays exactly where it is) but work is done (or released) when the satellite is moved into or out of orbit. So perhaps gravity is, in some sense, both "real" and "fictional". I can live with that.  :) -- ALoan (Talk) 11:01, 19 Oct 2004 (UTC)

(Doradus) Lots of "real" forces do no work. When you are standing stationary on the ground, you are exerting a force on the ground, but are doing no work. As for whether a given force (such as centrifugal force) "can" do work is, I think, a circular argument (no pun intended) because centrifugal force will do work when viewed from those reference frames where it exists, and will do no work in frames where it does not exist.

(William M. Connolley 21:02, 21 Oct 2004 (UTC)) I had to think about this for a bit... the idea is (and I'm not saying I support it, I'm just trying to describe it) it that an unreal force is one that does no work *under any cicumstances*. Since coriolis is perpendicular to the velocity, it never does any work. Thats not true of gravity.

(Doradus) Hang on, just because coriolis forces act perpendicular to (inertial) velocity doesn't mean they don't do work. It just means they never change an object's (inertial) speed. Consider this thought experiment: imaging standing in the center of a large, light rotating disk with a tube running radially from the center. If you push a bowling ball down the tube, you will see it exert a coriolis force on the side of the tube, slowing down the disk, and thereby doing work.

(William M. Connolley 20:29, 23 Oct 2004 (UTC)) I disagree with your example. "Coriolis force" is the thing that you see in the equations of motion in a rotating frame of reference compared to an "inertial" frame. If we roll a ball across a frictionless rotating table, then from a not-rotating-with-the-table POV we see the ball move in a straight line. But from the table we see a curve, described by the equations with the C term. In your example, from the table we see the ball moving in a straight line; whilst from outside we see it moving in a curve. How do we explain the motion of the ball, from the POV of the table? After all, the equations (on the table) include the C term. Answer (of course... err, though it took a bit of thinking): there is *also* the force from the sidewall of the tube. This force exactly balances the C terms, and is why (from the POV of the table) the ball goes in a straight line. This force also exists when you transform back into the non-rotating frame, and this (not C) is the force that slows the table down.

Great work comments

Great work on this article to those who have contributed. I especially like how it explains everything thoughtfully, *BUT* also has advanced mathematical concepts at the end for more indepth look. Many of the mathematics articles *ONLY* have the advanced stuff since explaining it thoughtfully in layman's terms is the hardest part :). Good work everyone! --ShaunMacPherson 01:20, 17 Oct 2004 (UTC)

Viewpoint for drawings

All the drawings appear to make sense to me if we regard them as views from above the orbital plane. I think that choosing this single POV is less likely to confuse readers. Or am I missing something? -- Karada 12:15, 17 Oct 2004 (UTC)

(Doradus) Agreed. Someone has apparently already adjusted the captions.

Review needed for this and tidal force article

I removed the comment:

"(this is for a secondary body freefalling in a straight line to the primary body; in the case of an orbit the 2 becomes a 3)"

This makes no sense: the object must be in free-fall for this derivation: otherwise you don't have a tidal force. An orbit is one kind of free fall. -- The Anome 10:24, 19 Oct 2004 (UTC)

But it says freefalling in a straight line, with the intention of emphasis on straight line.--Patrick 23:03, 2004 Oct 21 (UTC)

(William M. Connolley 10:33, 19 Oct 2004 (UTC)) I don't understand you. The (gravitational) tidal force exists independently of the motion of the object: free-fall, rotation, whatever.

We need a review of the derivation in tidal force at the same time: these two articles appear to contradict one another at the moment. -- The Anome 10:29, 19 Oct 2004 (UTC)

(William M. Connolley 10:33, 19 Oct 2004 (UTC)) The mess arises because it depends. As far as I can see, the strictest defn of tidal force is just the gravitational one. However, if you're interested in the breakup of a body, you might as well add in the rotational one, because it has a similar form. At that point you need to assume something about the rotation, and assuming its tidally locked makes sense (apparently most bodies are) but this should be pointed out.

Assessment comment

The comment(s) below were originally left at Talk:Roche limit/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Please Change your vote beacuse this article is very good Amir (talk) 17:47, 29 August 2008 (UTC) Why is the formula for Roche's limit for rigid satellite stated as 2.44R((density of primary/density of satellite)^(1/3)) but the derivation yields R((2*density of primary/density of satellite)^(1/3))? 2^(1/3)=1.26 and not 2.44? Killer199208 (talk) 07:52, 17 December 2012 (UTC)

Last edited at 07:52, 17 December 2012 (UTC). Substituted at 15:41, 1 May 2016 (UTC)