Talk:Tidal locking

From Wikipedia, the free encyclopedia

Contents

[edit] Final Configuration

"The simple picture of the moon stabilising with its heavy side towards the Earth is incorrect, however, because the tidal locking occurred over a very short timescale of a thousand years or less, while the maria formed much later."

Why does it matter when the maria formed? If the density of the moon shifts to a different side following tidal locking, wouldn't the moon re-orient itself subsequently? Afterall, it is a similar adjustment that causes the tidal locking in the first place, following perturbations in the shape of the moon. Could someone address this? —Preceding unsigned comment added by Blueil77 (talk • contribs) 00:21, 13 January 2008 (UTC)

[edit] Misleading Sentence: Both bodies sychronize

It results in the orbiting bodies synchronizing their rotation so that one side always faces its partner

The above sentence implies that both bodies face their partner. But when one body is tremndously more massive than the other, isn't it only the tiny one whose rotation gets synchronized:

  • moon and earth - moon faces earth, but earth rotates 28 extra times per lunar rotation/revolution
  • mercury and sun - mercury's rotation synchronized in a 2:3 harmonic to its orbit period, while suns rotates considerably faster (?)

--Ed Poor

I think the fact that it says, "one side", implies that only one of the pair will show the same face at all times. Maybe this could be changed so it's more explicit?

--Don

It is mentioned later-on that Earth is still slowing down its rotation due to its interaction with the moon. —Preceding unsigned comment added by Blueil77 (talk • contribs) 00:25, 13 January 2008 (UTC)

[edit] Earth/Sun

Am I correct in assuming that at some point (however far off) the Earth will become tidally locked with the Sun and we'll have a cold and likely uninhabitable side of the planet?

You are probably correct. It will most certainly happen three years from now, just after I am graduated from college. (Sorry, but I had to.) If we assume that the moon used to not be locked to us, then we can assume that the earth will become locked to the sun, thereby making the entire planet (save maybe the twilight zones) uninhabitable, not just the dark side. -- D. F. Schmidt (talk) 16:43, 23 August 2005 (UTC)

I think the Earth would be locked to the Moon and thus it couldn't be locked to the Sun at the same time.  Grue  20:03, 25 September 2005 (UTC)
The articles says that tidal locking applies to bodies that closely orbit each other, and it's stated above that if there's a huge difference in mass, only the one with less mass will be tidally locked. So, the earth's rotation won't slow down due to tidal locking in any significant way at any time in the forseeable future. MarXidad 20:59, 27 February 2006 (UTC)
"Closely orbit each other"? If I have it correct, the earth has slowed down in its daily rotation (I'm not sure where I got this from, except I know that it was somewhere on Wikipedia). Why would the earth slow down, but its proximity to the sun (which has significantly more mass)? I don't think it would lock or be so affected by Venus. D. F. Schmidt 16:58, 25 April 2006 (UTC)
The Earth has slowed down due to interaction with the Moon, especially due to friction caused by the tides. The Venus thing is almost certainly a coincidence. Deuar 18:36, 25 April 2006 (UTC)
If the Earth becomes tidally locked with the Sun will the Earth be able to support an atmosphere?
No reason why not, but any atmosphere will probably be blasted away much sooner (in only 5 billion years or so) when the Sun goes into red giant phase. Deuar 10:53, 31 March 2006 (UTC)
Thank you kind sir :)
If the Earth had no moon, the timescale for the Earth to be tidally locked to the Sun would be in the order of tens of billions of years.
The presence of the moon complicates things. The Earth would tidally lock to the moon first. After that, the Sun continues to act on both bodies, slowing the rotation of both. The interaction is complex but I think it works like this. Both bodies rotate slower, cauing the tidal bulges of both bodies to lag behind the centre line connecting the centre of gravity of both bodies. This causes the orbit of the moon to shrink. Eventually the moon will be orbiting the Earth so closely that it breaks up and accretes onto the Earth. Then the Sun drags what is left to a tidal lock many billions of years from now.
Of course, the Sun will expand into a red giant long before this happens so the point is moot anyway. --B.d.mills 10:31, 3 July 2006 (UTC)

[edit] Please Clarify: Self-gravitation

Larger astronomical bodies which are near-spherical due to self-gravitation, become slightly prolate (ovoid).

Could someone elaborate on this sentence? Why only large bodies? Does tidal locking not apply to small bodies? Define self-gravitation. (I assume it means the forces pulling it into its own center, but why would this only apply to large bodies?) Does the material of the body (gaseous vs. rock) make a significant difference? It took me a couple read-throughs to realize the the orientation of the major axis of the prolate spheroid is towards the primary. (I saw the diagram at spheroid and assumed that the major axis was also the axis of rotation. Took me a while to break out of that mindset.) Could someone who knows what they're talking about (i.e. not me) spell out this part of the article a bit more? Thanks! --Nmagedman 22:21, 20 May 2006 (UTC)

[note: in response to this, Deuar did significant work rewriting the page. --Nmagedman]

Since I've restored the old one, I guess this now needs an answer. Which is, in small bodies the internal forces (structrural integrity) can be large enough to oppose gravity. Larger bodies (earth, moon) are essentially fluid. William M. Connolley 13:39, 24 May 2006 (UTC)
The tidal bulges are there in small bodies as well, but are tiny in comparison to their structural irregular shape. Example: all the small and irregularly shaped inner moons of Jupiter are tidally locked. Deuar 18:17, 26 May 2006 (UTC)


I thought the old explanation was clearer, myself William M. Connolley 18:45, 21 May 2006 (UTC)

Overall, I agree, but they both have their strengths and weaknesses. The new explanation clarified many points that I didn't get from the first one. Thank you, Deuar. I greatly appreciate your effort! I need to review both versions and then can try merging the best of both of them. --Nmagedman 17:58, 22 May 2006 (UTC)

Hmm, well, maybe. How about:
Rock (ice, gas, water, whatever) is not elastic, and when it is continually being reshaped by two bulges passing by during every rotation, friction is created. The planet is always pulling the bulge somewhat backwards, so the effect is that the moon is being gradually slowed down by this friction. Over time the rotational speed slows down. The rotational energy of the moon is being lost into space as heat.
I don't believe this (that the planet is slowed by friction). The planet is slowed because there is a gravitational torque on the bulge (as the previous version said). Friction is just a manifestation of this. William M. Connolley 19:13, 22 May 2006 (UTC)
Hi again. Yeah, well, re-reading what I wrote (this time in the light of day), I can see it could use improvement. The list I made is too choppy, for one. I'm a bit pressed for time for a couple of days, so i'll just be lurking for a while.
Regarding friction/torque, you need more than just a constant torque or a constant force to slow things down. What is also needed is a way to lose energy as time goes by (the friction). Suppose you had a body making a rotation in 10 hours. Now you start applying a small constant torque on it. It will indeed change its rotation rate so that now it takes, say, 10 hours plus one minute. But this rotation rate will not change any further if you have no friction. Deuar 08:57, 23 May 2006 (UTC)
Not at all sure I understand you. From 2 viewpoints: (1) postulate, if you will, a frictionless fluid body (either a real superfluid helium one, or an abstract mathematical one). Are you asserting that tidal forces will not cause it to slow and lock? (2) as you've just said, the torque on the bulge slows the thing down a bit; but that doesn't get rid of the bulge, or the torque, so it will slow further. Further, what you need is a way to lose angular momentum. Friction doesn't do that; the torque does. William M. Connolley 15:20, 23 May 2006 (UTC)
Hi William, yeah well you're posing pretty good questions. It seemed simple at first but it looks more involved now. I'm still convinced that you need friction because energy must be permanently lost somehow but there's also angular momentum to lose. Thinking... Deuar 21:11, 23 May 2006 (UTC)
Hmm - I think i get it now. Consider this analogy: we've got a pendulum made from a weight hanging on a spring (so the spring is the arm of the pendulum), and the pendulum can go all the way around the central attachment. It goes around the attachment in a vertical plane. Furthermore there's no friction in the attachment, or with the air. The weight is like a rock on the moon's surface, the spring is like the moon's gravity, and the downward earth gravity is like the planet's. By the way, the weight is a rock on the surface, not the bulge! It becomes a part of the bulge only for a time when the spring is stretched more than average.(I was confused about this for a while) Anyway, we give the pendulum a big push, so that the weight easily flies all the way around the attachment - this is the moon's rotation. So as our "rock" approaches the bottom of a swing, it is like approaching the planet-facing point. Because the earth's gravity is acting against the spring here, it streches a bit, and the rock is further down than if the spring was stiff, and it's like being part of the "bulge". It continues to go around, and as it begins to rise up from the bottom of its rotation, it experiences a torque from the earth's gravity, and begins to slow down. As you rightly pointed out! No friction. Now, it's genrally going pretty fast, so soon it passes the height of the attachment, and later gets up to its maximum height, exactly above the attacment point. It's slower now then whan it was at the bottom. However, now, once we're over the top, the torque from the earth's gravity starts applying torque in the opposite way and the weight begins to speed up as it travels around the attachment and down. In fact, by the time it gets down to the bottom, it's travelling at the same speed it was before. All assuming no friction. This is why we can have torque periodically slowing it down, but by itself it just causes an oscillation that repeats on and on without locking. Now if we also add friction, say in the spring, analogously to friction between rocks in the moon, it's clear what happens. It will fly around, slowly losing a tiny piece of speed with each revolution, until at last it's lost so much that it can't crest the top of the rotaton, and falls back down on the same side, swings around on the bottom a few times, and eventually stops, hanging straight down. It's locked.
So, regarding your points, (1) The superfluid will slow from the torque, but later speed up, and won't lock. (2) Yeah I kind of agree. The subtle thing is that a rock on the surface is not the same as the bulge (the bulge hardly moves at all, in fact). Hopefully the analogy above clears up this difference. What I wrote in the article needs fixing. Deuar 22:15, 23 May 2006 (UTC)
I disagree. You're missing the fact that the bluge isn't fixed on the sfc - it moves. But because of inertia its always a bit behind the rotation. Your analogy does not work. Furthermore, I'm a bit disturbed that you are essentially working this out in arrears... you should know this already, if you're editing it against objections William M. Connolley 22:32, 23 May 2006 (UTC)
Perhaps you misunderstand the analogy. The weight is a point on the surface, the surface is rotating. That's why the weight goes around. The bulge is always pointing roughly towards the planet, which is "down" in the analogy, but of course the surface rotates past the bulge all the time. The bulge occurs in the place where the extension of the spring is greatest, not necessarily where our test weight is at any time. And, the analogy does allow for a bulge that is slightly towards the direction of rotation because of inertia. If the weight goes around clockwise, the greatest extension of the spring will be at around 7 o'clock (not 6) because of the weight's inertia. It takes a little while for the spring to start significantly pulling it back up.
I'd also like to point out that my editing was before any of your objections, and in fact inspired by complaints about the previous version! (see first post in this thread). I have not edited the explanation since you raised doubts. Deuar 09:49, 24 May 2006 (UTC)
Sure. What I'm pointing out is that the previous explanation was correct, even if some people found it hard to follow. The current one isn't. Friction doesn't get rid of angular momentum. Now, you seem to be saying, lets leave your version in place while you try to work out how to fix it. Thats not good. William M. Connolley 10:47, 24 May 2006 (UTC)
I don't know why you're trying to put words in my mouth now. That's not good :-) I believe we were merely discussing the physics. Deuar 16:50, 26 May 2006 (UTC)
Before we invest time & energy into a rewrite, could we reach consensus on what the scientific views are? Does anyone have a WP:RS astrophysics textbook or whatnot? --Nmagedman 11:31, 23 May 2006 (UTC)
Time for a brain strain indeed. Deuar 21:35, 23 May 2006 (UTC)

I decided that I preferred the old version of the mechanism section, so I've restored it. William M. Connolley 13:39, 24 May 2006 (UTC)

I had hoped you could provide a more objective reason than your personal dislike, but hey, the old version is mostly correct so it can be worked with equally well. Friction is needed to tidally lock and should be mentioned. A frictionless superfluid body would rotate around, with a bulge, but without ever slowing. Flowing around without slowing is the defining feature of a superfluid after all. Deuar 16:50, 26 May 2006 (UTC)

[edit] Prolate

I was just curious about the use of the term "prolate". I had understood that tidal forces deform a body so that it is oblate. Please correct me if I am wrong. --CKozeluh 15:52, 9 June 2006 (UTC)

Check out spheroid. Deuar 21:38, 9 June 2006 (UTC)
That's just it; spheroid, oblate, equatorial bulge, and moon all suggest that the planets and moons are oblate. Do tidal forces act counter to rotational forces, bringing an orbiting body closer to perfect sphericity? CKozeluh 19:22, 15 June 2006 (UTC)
Usually the body is flattened (in the "north-south" direction) by centrifugal force due to its rotation in the "north-south" direction, and then also becomes prolate in one of the perpendicular directions (as in "towards the planet") from tidal forces. The end result is an ellipsoid. Deuar 19:35, 15 June 2006 (UTC)
The same question occurred to me while reading the article. Now I've read this exchange, and I still don't see that the question was settled. I'm further puzzled by Deuar's reference to "rotation in the north-south direction". I'm out of my element here, but isn't north-south defined as the poles of the rotational axis? So rotation in a north-south direction seems to be an oxymoron. Educate me, please. 24.58.11.50 03:38, 5 February 2007 (UTC)
Umm, sorry about the confusion caused by my sloppy writing above; I meant flattened in the north-south direction (wording now fixed). We have a combination of two basic deformations:
1) Oblate flattening at the poles due to rotation.
2) Prolate elongation of the tidal bulge at the equator.
On the scale of things, this prolate part is negligible unless teh body in question is orbiting close to another large mass. Also, this "prolate" part of the deformation travels around the equator with every revolution, unless our body is tidally locked. Deuar 14:05, 5 February 2007 (UTC)

[edit] Timescale for tidal locking

The following sentence is misleading:

The moon has oriented itself so that its heavy side with much more extensive maria, which we know as the near side, is oriented towards the Earth.

This is misleading. The moon would have tidally locked to the Earth fairly quickly (probably less than a thousand years). However, the maria would likely have formed over a longer period. In other words, by the time the maria formed, the moon would already have been tidally locked. To imply that the maria formed before the moon tidally locked is misleading at best and factually inaccurate at worst.

A reference to the time to tidal locking is given by a formula here: http://groups.google.com/group/rec.arts.sf.science/msg/e05283a619187a8f?dmode=source&hl=en --B.d.mills 10:20, 3 July 2006 (UTC)

That's a really good point. I've now found that kind of formula elsewhere, e.g. this paper, so it's not a discussion group fluke. I get 90,000 years at current earth-moon distance, but this must have been much quicker in reality since they were once closer, and the rate decreases as orbital radius to the sixth.
I find that short timescale totally amazing. It may itself be worth remarking in the article.
Regarding the maria, it's obviously as you say, but I would find it very odd if they just happenned to form on the earth-facing side. My best guess is that the equilibrium locked position must have migrated at some stage either gradually or in a jump to orient the heavy hemisphere towards Earth. This could use some kind of literature search... Deuar 20:34, 3 July 2006 (UTC)
The crust of the moon is thicker on the far side than the near side. (Ref: http://ottawa.rasc.ca/articles/hanmer_simon/geology/moon3/index.html) The moon is also believed to have had a magma ocean immediately after it formed. Evidence for this comes from the presence of certain minerals found by the Apollo and Luna missions. If the moon did tidally lock in under a thousand years, it would likely have locked with the crust still molten. This explains why the crust is thicker on one side than the other; when it solidified the moon was already tidally locked. It also explains the presence of the maria on the near side. With the thinner crust there, it would have been easier for basalt to flow on the surface after an impact. So the presence of the maria on the near side are actually a consequence of tidal locking and not a cause.
By the way, my calculations suggest that the moon tidally locked a lot quicker than a thousand years, I simply used that as a safe figure in case my rough calculations were flawed. The moon is believed to have formed very close to the Earth, maybe about 20,000 km from the centre of the Earth. At that distance, my calculations suggest that the tidal locking could take only a few months. --B.d.mills 11:22, 4 July 2006 (UTC)
I follow the argument up to and including "when it solidified the moon was already tidally locked", but I don't see what being Earth-facing has to do with crust thickness. The difference in gravity due to the Earth is completely miniscule. Deuar 14:07, 4 July 2006 (UTC)
The moon's crust is thicker on the far side, but an explanation of the mechanisms that may cause it will require further research. --B.d.mills 12:26, 5 July 2006 (UTC)

I find In fact the tidal locking was so rapid, that the frictional energy produced must have melted the moon's surface, and so must have occured even before the highland terrain solidified. somewhat doubtful. It has no source. And according to the discussion above, it was molten at the time anyway William M. Connolley 19:54, 4 July 2006 (UTC)

That makes sense. It's gone. Deuar 20:15, 4 July 2006 (UTC)
Let's better say, that there could have been maria everywhere on the Moon surface, but since one face points to the Earth all the time, the other face is much more exposed to asteroid impacts and the maria disappeared between asteroid impact craters during bilions of years, making the outer side even more light-weight and the tidal lock even more strong... Does this sound more probable?
It is just not possible to guess now, whether the Moon has been beside the Earth from the early times, when its surface was molten...
~ Semi, July 5, 2006
See the discussion above on the thickness of the lunar crust. The thin crust on the near side allowed magma to flow onto the surface, whereas the thicker crust on the far side blocked the flow of magma. Your conjecture about not being able to guess the age of the surface is incorrect, because the lunar rocks brought back by Apollo and Luna missions have been dated accurately by the decay of the radioactive isotopes that they contain. --B.d.mills 02:34, 8 July 2006 (UTC)

[edit] Pluto's moons: Nix and Hydra

Would the recently-discovered moons of Pluto (Nix and Hydra) be tidally locked? There's no mention of them in the list. These moons would have a unique tidal lock, because they would be locked to two bodies (Pluto and Charon) and this is worth mentioning in the article. --B.d.mills 12:56, 5 July 2006 (UTC)

Well from the locking time formula in this paper (which I should get around to putting in the article, I guess), I get the following time constants for the bodies in the Pluto system to lock:
Charon: 300,000 years; Pluto: 4 Myr; Nix: 50 Gyr; Hydra: 300 Gyr.
These are necessarily extremely rough, I made the same assumptions as in the paper, plus an initial rotation period of 12 hours for each body. I wouldn't trust these more than being within a factor of 10, and they are more likely over-estimates than under-estimates because of the gradual increase in orbital radii that must have occured in the meantime. So Nix looks like it might or might not be locked, while Hydra is significantly less likely to be locked. Maybe there's a paper somewhere that considers this more rigorously for Nix and Hydra. Deuar 18:23, 5 July 2006 (UTC)

[edit] Venus corrected

I have corrected the Venus paragraph.

Could you, please, recalculate the Venus apparent solar day and sidereal day, before reverting my change?

The more propper value seems, that after 13 Venus years and 8 Earth years, the both planets meet on an almost same place, and it takes 12 Venus sidereal days, with a difference, that is negligible, when compared to a relatively short time of radar insight.

The apparent solar day on Venus, as stated elsewhere here, copies an older bug in NASA planet factsheets, where the ea-day value in that cell is mistaken by hours in that column heading...

The apparent Solar day on Venus calculation: if it orbits arround the Sun in 224.70069 ea-days, which is 1.60° ea-daily, and meanwhile spins in counter direction after 243.0185 ea-days, which is 1.48° ea-daily, relative to star background, then the Sun moves apparently only 0.12076° ea-daily. From this, the apparent travel of Sun one "day" arround, as seen from the Venus surface, takes 8.28 Earth-years!

So how is that?

~ Semi, July 5, 2006

These calculations are flawed because they do not allow for the retrograde rotation of Venus. With the rotation and the revolution going in opposite directions, the mean solar day is about half a Venus orbit long. --B.d.mills 02:01, 6 July 2006 (UTC)
Aha, thank you. I was probably misleaded by some earlier version of Venus article here, that readed such, that the yearly and daily move of Sun are contrary and should subtract. It is no more on the Venus article... So they should actually add (both apparently counter-clock-wise from the surface) and make the value of 3.08° ea-daily, or 116.75 ea-days per whole rotation... Sorry for my previous error above.
Then, after 583.9 days, the planets meet on an almost-opposite side of the Sun (ie. after 1.60 earth-years, which is 0.6 of a circle arround from the previous meeting), and if it shows same face to Sun, it shows the opposite side (same as in previous meeting) to the Earth...
.
So I have corrected the article yesterday to say an almost same thing in other words, but related to the sidereal day in ratio 8:13:12... Could this be yet reformulated to say both?
.
Actually, these planets meet on an almost-same place (relative to Sun) just after 5 consecutive meetings, after 8 Earth-years and 13 Venus-years, which is after 12 Venus-sidereal-days (relative to star background) and after 25 Venus apparent-solar-days...
.
During the times, when these planets meet, the Earth moves retrograde from the Venus perspective. Also, when Venus meets Mercury or Jupiter (the only other tidally-important planets from Venus perspective, beside the Sun, and only tidally-important during the meetings), they also move retrograde from the Venus perspective, which all together is the most probable cause of the retrograde spin of the Venus...
Also, the Venus is one of few planets, that has got no moon to induce/support the prograde spin. The only other such planet (without a moon) is Mercury, whose rotation is also "relatively strange"... Could this be connected?
.
~ Semi, July 6, 2006
It is pretty doubtful that any of the planets including Earth have significant tidal effect on Venus. That's why these matching periods puzzle people. Earth's tidal effect appears on the face of it to be completely miniscule, not to mention the other planets. The tidal force falls off as radius to the third power, while the rate of tidal locking falls off as radius to the sixth. There were some papers published suggesting that the tidal slowing may be bigger than normal on Venus because of its large atmosphere, but this is rather speculative and is not an accepted explanation by any means. Deuar 18:57, 6 July 2006 (UTC)
Venus and Mercury have no moons and rotate slowly. There is a connection - moons that orbit faster than the planet rotates will be accelerated in their orbit by tidal forces, eventually causing the moon to crash into the planet. This is happening with Phobos. So moons have to orbit Mercury and Venus a long way out to avoid being deorbited into the planet. When moons orbit so far out, the Sun can perturb them out of orbit - see Hill sphere. So Venus and Mercury cannot have satellites with stable orbits. --B.d.mills 23:52, 6 July 2006 (UTC)

[edit] locked satellites

An application of the rough formula for locking time (6×1010 Rμ/msmp² years) to the satellites of the outer planets indicates that the current status of most of them can be confidently estimated. Apart from a couple of exceptions, all the inner regular satellites have a locking time of less than several million years, which is barely an instant in astronomical time. Conversely, all of the irregular outer satellites except Triton have locking times in excess of 1×1012 years, and obviously are extremely unlikely to have been locked so far. The exceptions to this clean division are:

  • Polydeuces: 100 Myr. Still only a fraction of solar system age, and even allowing for a factor of 10 uncertainty, so it is fair to assume that it is probably locked by now.
  • Hyperion: Known from observations not to be locked
  • Iapetus: 500 Myr, known from observations to be locked
  • Nix: 350 Myr to 7 Gyr depending on its size: faster locking if it has low albedo
  • Hydra: 1.4 Gyr to 25 Gyr depending on its size like for Nix

So it turns out that even allowing for a factor of 10 uncertainty in the above estimates, Pluto's new moons are the only ones for which the locking situation is strongly ambiguous. Accordingly, I have standardised the "(presumably)" tag to "(assumed)" for bodies on the "locked" list in the article, since their lock is pretty certain. Also added Oberon which was missing for some reason. Deuar 20:04, 16 July 2006 (UTC)

[edit] Mechanism section

Please could someone clear up the mechanism section - it's a little unclear. A gif animation would make this much clearer - I'd do it myself but I don't quite understand it! Thanks --El Pollo Diablo (Talk) 10:41, 9 October 2006 (UTC)

You're too right, it was unclear. I have gingerly made a rewrite which imho improves the situation − gingerly, because this has been tried before, not necessarily with good results. Diagrams would indeed be very useful... Deuar 17:37, 22 October 2006 (UTC)


Trivia to be added - Edmund Halley was the first to note the effect. He found that the revolution period of the moon is getting shorter and concluded that that moon is accelerating. 202.156.12.12 04:15, 17 December 2006 (UTC)Rajesh

[edit] Tau Boötis is known to be locked to the close-orbiting giant planet Tau Boötis Ab

This sounds like BS. Could someone please provide a reference to this statement, or at least describe how one could obseverve that an extra-solar planet is indeed tidally locked? Lunokhod 23:52, 24 January 2007 (UTC)

Presumably a tidal locking timescale has been estimated, and it's much much less than the known age of the system? By the way, what's BS stand for (sorry)? Deuar 15:32, 25 January 2007 (UTC)
Then perhaps we should move this entry from "List of known tidally locked bodies" to "Bodies likely to be locked". Either there has been some amazing advancement in imaging extra solar planets that I am ignorant of, or this is only an inference! Also, if the planet is a gas giant, then the Q and k2 will be much different than for a solid object. Perhaps this could be discussed in the article? A quick google search suggests that the Q of jupiter is 1 billion, as opposed to ~100 for the Earth. And this would increase the tidal locking timescale accordingly. Lunokhod 15:54, 25 January 2007 (UTC)
Go for it, I reckon. Deuar 16:21, 25 January 2007 (UTC)

[edit] Mercury

So...Mercury isn't tidal locked to the Sun? Then why is it on the list? --MPD T / C 02:43, 1 March 2007 (UTC)

I think that tidal locking is not the same as "synchronous rotation", even though the intro seems to say so. Perhaps it is better to say that tidal locking is a process where tidal torques leave one body on a spin-orbit resonance. Synchronous (1:1) is the lowest energy configuration. Lunokhod 13:04, 1 March 2007 (UTC)

[edit] Tidal braking

In the UTC article, there's a [[tidal locking|tidal braking]] link. Can someone please provide a definition for tidal braking in this article, even if they are the same? The "Locking of the larger body" paragraph in the "Mechanism" section seems an appropriate place to do this. Thanks. Xiner (talk, email) 01:22, 11 March 2007 (UTC)

Actually, I think the tidal acceleration article seems to discuss this in more detail. I've linked it to there instead. Deuar 15:55, 19 March 2007 (UTC)

[edit] Unclear Description of Orbital Resonance

Under "Mechanics," the description of orbital resonance is rather unclear. This is what it says:

Rotation-Orbit resonance: Finally, in some cases where the orbit is eccentric and the tidal effect is relatively weak, the smaller body may end up in an orbital resonance, rather than tidally locked. Here the ratio of rotation period to orbital period is some well-defined fraction different from 1:1. A well known case is the rotation of Mercury—locked to its orbit around the Sun in a 3:2 resonance.

It does not specify why or how this happens. I don't know myself, but I'm guessing it happens because the smaller body's rotation does not change when it reaches the aphelion of its orbit, thus causing it to skip ahead. However, my guess does not explain

A. why the opposite does not happen at the perihelion, thereby nullifying the effect, or

B. why this would cause resonance to occur in well-defined ratios such as 3:2 (in the case of Mercury).

I would find it very helpful if a little more research were done and this paragraph were revised. I myself would not know where to look, and the page on orbital resonance does not seem to describe rotational resonance at all.

[edit] Tidal locking and developing life

I removed the following from the "planets" section:

Tidally locked planets may present problems for developing life, as one side of the planet will always be facing away from the star and the other side will always face toward it; in the absence of significant heat redistribution by atmospheric winds or hydrospheric currents, this would result in constant temperature extremes. [citation needed] On the other hand, tidally locked large satellites of gas giants rotate with respect to the central star, providing places for developing life that avoid extremes in temperature.[citation needed]

Besides that being mere speculation, and being without any references, it also mentions "large satellites of gas giants" which are clearly not planets. If this text should be on this page at all, then please in a separate section, and with references. Jalwikip (talk) 14:10, 19 November 2007 (UTC)