Talk:Tidal locking/Archive 2

Archive 1

Archive 2

Well done

To all persons responsible for this article - it explains this phenomenon very clearly and simply. Take a bow, the lot of you. ElectricRay (talk) 13:24, 6 January 2014 (UTC)

• Good start, but there is a lot that is still unexplained. Like how does a planet like Mercury obtain a "tidal lock" to the Sun, having no surface fluids? Why is not every orbital element ultimately in tidal lock (or some other form of resonance) with its larger, nearest neighbor? (For example, why does the Earth have day/nite, yet Mercury does not?) ALL orbits are elliptical so I remain skeptical. Need some expert to follow up on this. There is much more to address in this comments section. --2600:6C48:7006:200:D84D:5A80:173:901D (talk) 02:54, 22 February 2018 (UTC)
  • I think you're possibly mixing up the Earthy concept of ocean tides with tidal force in general. There's no mention of surface fluids in the article; just bulges and the forces acting on them. Ocean tides are just one example. The article does mention that tidal locking takes a long period of time to occur. Also, Mercury does experience day/night because it is rotating (slowly) with respect to the Sun. Praemonitus (talk) 03:49, 22 February 2018 (UTC)

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)

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?

The discussion, the way it is formed, is so unclear to me. The moon does rotate once everytime it circles the earth, correct? Doesn't that mean the same side of the moon always faces the sun, not the earth? One side of the moon is always in the dark to us and one side is always in the sun? Please help me here...

Cvgittings (talk) 05:03, 10 December 2010 (UTC)

--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)

The usage of the term "synchronized" implies that the moon just happens to spin at the right rate to always face its same side to the earth, which is not the case. It is tidally locked because one side is more dense, which is the side that always faces us. Would it be correct to describe an "orbiting" ball tethered by a string to a focal point, synchronized? In my opinion it wouldn't be.

All I am saying is that its misleading to what is actually* happening with tidal locking between the earth and moon. Its not a coincidence, like describing it as synchronized suggests.

No, "synchronize" in this case means that a body's the rotational period becomes equal its own orbital period. It does not imply a coincidence. --JorisvS (talk) 10:05, 1 May 2013 (UTC)

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)

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)

Lunokhod, thank you for explaining that. I was very confused. I don't want to edit the page because I am not an expert. If someone else could fix the definition in the first paragraph to clarify that synchronous rotation is only one potential orbital pattern resulting from tidal locking, that would be great. Pauldebarros (talk) 16:10, 22 September 2021 (UTC)

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)

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.

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. 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.

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)

It's back with a reference to a badly written anonymous paper on arxiv.org. Deleting. Oumot (talk) 00:17, 13 December 2017 (UTC)

Isn't the animation confusing?

The animation shows two bodies orbiting a central body at the same rate, yet they are at different distances from the central body...what is the point of this?, It might lead some people to think that this is a real orbital configuration...Jellyboots (talk) 20:55, 29 January 2009 (UTC)

Final configuration

I took this out. It has no sources and I don't think I believe it. Does anyone speak for it?

There is a tendency for a moon to orient itself in the lowest energy configuration, with the heavy side facing the planet. Irregularly shaped bodies will align their long axis to point towards the planet. Both cases are analogous to how a rounded floating object will orient itself with its heavy end downwards. In many cases this planet-facing hemisphere is visibly different from the rest of the moon's surface.

The orientation of the Earth's moon might be related to this process. The lunar maria are composed of basalt, which is heavier than the surrounding highland crust, and were formed on the side of the moon on which the crust is markedly thinner. The Earth-facing hemisphere contains all the large maria. 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. The maria are instead formed from heavier lunar magma that responded to the tidal lock by gravitating towards the Earth.

William M. Connolley (talk) 18:39, 9 March 2010 (UTC)

I speak for it and so does Gravity-gradient stabilization. Tidal locking forces are proportional to the spin angular velocity difference between the bodies orbital velocities. As the locking body slows down so the locking force reduces. This continues until when you take the limit no locking force exists. So, at the end there is no phase lock information remaining. Between the moon and Earth there is a fixed phase (empirical). This phase lock is occurring and it cannot be from tidal locking. As William describes above the moon cannot have symmetrical weight. Hence the heavy side faces the gravitational attractor i.e. the Earth.

We should add a section on phase locking which is as a result of Gravity-gradient stabilization.

User:pcrengnr —Preceding undated comment added 16:56, 24 March 2016 (UTC)

Side?

Since when does a spherical object have "sides"? --77.109.223.37 (talk) 07:56, 15 June 2010 (UTC)

Since there are defined features on that sphere, that's when. —Preceding unsigned comment added by 129.186.253.87 (talk) 19:45, 27 August 2010 (UTC)

Contradicting formulae in "Timescale"

In the section "Timescale", I think the second ("simplified") formula given for t_lock

$t_{\textrm {lock}}\quad \approx \quad 6\ {\frac {a^{6}R\mu }{m_{s}m_{p}^{2}}}\quad \times 10^{10}\ {\textrm {years}}$

is incorrect, if the first formula ($t_{\textrm {lock}}\approx {\frac {wa^{6}IQ}{3Gm_{p}^{2}k_{2}R^{5}}}$ ) is right. Namely, the following conclusion drawn from the second formula

One conclusion is that other things being equal (such as Q and μ), a large moon will lock faster than a smaller moon at the same orbital radius from the planet because $m_{s}\,$ grows much faster with satellite radius than $R$.

contradicts the first formula: There, the satellite mass only appears in the numerator of the formula given (via the "Inertia momentum" term). If we assume the satellite mass to roughly increase at the third power of its radius (i.e. assuming constant density, which seems a plausible assumption), we get five powers of R in both the numerator and the denominator of the fraction. Thus, tidal locking should essentially be independent of the satellite's mass, all other things (except the radius) being equal. Can someone clarify this for me? --Roentgenium111 (talk) 21:43, 23 August 2010 (UTC)

♦ I agree that there is something wrong with the ("simplified") formula. Earth's moon was tidally locked by the time of the Lunar Cataclysm, but plugging in the values for the Earth/Moon system in the ("simplified") formula gives a time to lock of 2.3 × 10³¹ years which we know cannot be true. The time to lock the Earth/Moon system must be less than the difference between the time of formation of the moon (~4,450 million years ago) and the time of the Lunar Cataclysm (~3,900 million years ago) which is approximately 550 million years. -az — Preceding unsigned comment added by Sciencebookworm (talk • contribs) 17:33, 9 March 2011 (UTC)

I don't know how you got that high a figure! I calculated 3.8 million years for the moon tidally locking to Earth. (using the same simplified formula). If anything the simplified formula seems to underestimate the tidal locking times, by a factor of 100, particularly for planets. When I multiply the results by 100, I get - 384 millions years for the moon - 5.1 billion years for a planet in the habitable zone of Epsilon Indi - 543 billion years for the Earth to the sun. The results I get after multiplying the simplifed formula in the article by 100 are very similar to the results from other formulae for calculating tidal locking time. For example, 'Tidal Locking Time in years = (ρ*((a/0.0483)^6))/(M^2)' - where a = distance in AU from primary, M - Mass of primary, as fraction of the sun (eg -Epsilon Indi M=0.762), ρ (density of satellite, Earth taken for Epsilon Indi= 5512 Kg/m^3 or for the moon= 3346 Kg/m^3). Using that formula I get - 4.1 billion years for Epsilon Indi- 433 billion years for the Earth - 8.4 million years for the Moon to tidally lock to Earth, almost the same values. Or, using the formula taken from 'Peale et al - 1977'. 'Tidal Locking Time in Seconds = (a/(0.027*(M^(1/3))))^6/486' - using CGS units where M= Stellar Mass (grams) a= orbital distance (cm), gives a tidal locking time of 477 billion years for the Earth to the sun and 5 billion years for an exo-Earth around Epsilon Indi, again almost the same values, if you multiply the result by 100. I think a lot of this article is more or less photocopied from a book called "Habitability and cosmic catastrophes" By "Arnold Hanslmeier" "2009" so the formula probably must be accurate, in some way. — Preceding unsigned comment added by 86.185.215.187 (talk) 01:20, 29 November 2011 (UTC)

Your comment seems correct, the formulas are not contradictory. I derived the simplified formula from the first one. One needs to use $k_{2}\approx 3/2/(19\mu /(2\rho gR)),g=Gm_{s}/R^{2},\rho =m_{s}/(4/3\pi R^{3}),I=4m_{s}R^{2}/10$. Approximation for the Love number seems OK, since it is usually much smaller than 1. If I input μ for rocky planet, 3×10¹⁰ Nm⁻² and original ω = 1/3600/24 rad, Q = 100, I get 3.8 million years. The prefactor $6*10^{10}\;\mathrm {kg} ^{2}/\mathrm {m} ^{6}\mathrm {s} ^{2}\mathrm {year}$ should be probably different to give more realistic results, but from the direct calculation, I obtained $3*10^{10}\;\mathrm {kg} ^{2}/\mathrm {m} ^{6}\mathrm {s} ^{2}\mathrm {year}$, which is very similar. My personal guess is, that Q = 100 from the referenced article, is in CGS units and it is not valid in SI units. But the problem can be elsewhere. Irigi (talk) 09:09, 8 October 2014 (UTC)

Another issue with the Timescale portion as it stands, the tidal locking formula description shows a value for the mass of the satellite, but that value does not appear anywhere within the formula given. As the mass of the planet is given as a squared value, is this assuming the mass of the planet and satellite are the same? — Preceding unsigned comment added by 205.166.76.15 (talk) 18:25, 18 July 2014 (UTC)

Kant?

I'm dubious about [1]. "researched" appears a little over the top - speculated, or reasoned, might be closer. But the text too is uncertain: according to [2] Immanuel Kant, who took great interest in scientific issues, reasoned on the basis of pure theory that the action of oceanic tides must slow down the earth's rotation which isn't what the retarding forces of tides on satellite bodies says William M. Connolley (talk) 17:12, 29 September 2010 (UTC)

It isn't what Immanuel Kant says, either William M. Connolley (talk) 17:14, 29 September 2010 (UTC)

I agree with you and am removing the sentence. Not only is the way the claim is phrased dubious, the source does not show he was the first either. —Lowellian (reply) 03:27, 1 October 2010 (UTC)

Orbital changes

Can someone expand on this section, specifically what mechanism is responsible for an orbiting body moving farther from its parent as its rotation slows? The explanation that angular momentum is conserved just doesn't do it for me. An explanation of the forces that cause a body to speed up in its orbit like the preceding sections on Bulge dragging and Resulting torque would be nice. 128.32.99.173 (talk) 15:16, 5 November 2010 (UTC)

Sorry, but it is *precisely* the conservation of angular momentum that explains why this happens. We have various phenomena, and we have the principle that angular momentum is conserved. We see that the phenomena are aligned with the principle. It is common -- though incorrect perhaps -- to say that "aha, that is why X occurs, it is due to the conservation principle" but as far as a mechanism that aligns the phenomenon with the principle -- there isn't one that is known. In the same way, various phenomena that align with the principle that (global) entropy for a system tends to increase (but local entropy may decrease if it contributes to global entropy) lacks a mechanism. Also, things like Gauss's law -- that the distribution of charge in a large body tends to be such that the positive charge is on the surface (but the net charge for the body is still zero) lacks a mechanism of the detailed sort that you seek. Get used to it. Chesspride 172.164.20.73 (talk) 21:55, 9 February 2016 (UTC)

Time Scale Formula Useless

The timescale formula given is useless, because without units the value is meaningless. Iæfai (talk) 03:18, 20 April 2011 (UTC)

Pretty sure it is in seconds. — Preceding unsigned comment added by 149.169.221.113 (talk) 18:46, 25 September 2012 (UTC)

If the dimensionality of the formulae were correct (that is, if they were in units of time), you could use whatever units you wanted. However, the "complex" formula gives an answer in units of angle * time, while the "simple" formula gives an answer in units of distance^6 / mass^2 * time. As they are, both formulae are thus completely useless. 76.255.189.2 (talk) 20:51, 24 September 2013 (UTC)

Rotation of the locked object

It can be quite easily demonstrated that the apparent rotation of the moon about it's axis is actually a necessary result of the moons orbit and not an independent source of rotation, meaning that it is technically incorrect to say that the rotational period of a tidally locked object precisely matches it's orbital period, since it has no rotational period independent of it's orbital motion. Tidal locking actually slows the orbiting object's rotation until it completely stops rotating around it's central axis. DoC352 (talk) 07:24, 9 August 2011 (UTC)

Based on DoC352s' statement above, the following sentence from the first paragraph, "A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner.", is incorrect. As a result I am editing this sentence to say, "With Tidal Locking the smaller body does not rotate around its own axis as it revolves around the larger body. This is completely different than Tidal Resonance. With Tidal Resonance the smaller body does rotate around it's own axis as it revolves around the larger body.". There are some other statements about the Moon rotating further down in the article that I am also removing. If you disagree with these edits please let me understand your reasoning. Cvgittings (talk) 20:01, 9 March 2012 (UTC)

If the Moon or any other tidally locked body is going to be tidal locked, one of its hemispheres must always face its partner. To achieve this the tidally locked object must rotate about its axis. For example, point your finger at your screen and then rotate your hand about the monitor. If you don't rotate your hand about the axis through your palm there is no way to have your finger point at the monitor through out the entire revolution. The period of this rotation about your palm must match the period of the revolution of your hand around your screen or your finger will fall behind or get ahead of pointing straight at the screen. Q.E.D.

Phancy Physicist (talk) 18:56, 10 March 2012 (UTC)

Unfortunately this explanation is inadequate. Consider for instance a plane flying in a complete circle around earth at a specific altitude. As it flies "straight ahead" much like we might assume the momentum of the moon does, the gravity of earth causes the plane to fly around the curvature of the earth with the bottom of the plane always parallel to the surface of the earth. Does the plane therefore rotate about its own axis with the exact periodicity of its revolution around the earth? Such a conclusion seems to represent a fundamental misunderstanding of the difference between a geometric rotation and translation. When an object geometrically rotates about a distant central point, it maintains its same facing towards the central point without any separate rotation about it's own axis, much as we see with the moon relative to the gravitational pull from the Earth's center of mass. Alternatively, if it could be shown that gravity results in a translation rather than a rotation (i.e. if it can be shown that some force other than gravity acts on the plane to make the nose come "down" as it attempts to fly in a straight line out of the atmosphere, and that gravity from the earth alone would allow the plane to absurdly maintain the same absolute rotation relative to its starting point) as an object maintains forward momentum through the gravitational field, then Phancy Physicist's argument would be perfectly valid. DoC352 (talk) 04:13, 3 March 2013 (UTC)

First of all, the plane is not bound in it's path by gravitational forces. Or not only gravitational forces are keeping it in the sky. Second of all, rotation in mathematics has nothing to do with orbital mechanics. Rotation, as defined in the linked article, is used there as math jargon. Rotation as used here is used as astronomy jargon - i.e. there is a clear distinction between rotation and revolution. Furthermore, the geometrical rotation you're talking about requires a rigid connection between the rotating body and the rotation axis - i.e. all points of the rotating body must maintain the same distance to the point of rotation. With gravity, that's not the case. Finally, if you have a fixed (i.e. non rotating) reference frame, you can decompose the motion of the rotating body in two in two separate rotational motions - one about the central axis, and one about the body's central axis. Tidal locking means that these two motions are synchronized - that is they have a fixed ratio between the periods.95.76.220.229 (talk) 14:26, 7 January 2015 (UTC)Apass

Effect of composition and structure

In the article it says "μ can be roughly taken as 3×10¹⁰ Nm⁻² for rocky objects and 4×10⁹ Nm⁻² for icy ones.". What would the effect of a more or less substantial atmosphere, of (bodies of) surface or subsurface liquid(s), or of a body being mostly metallic be on this value? --JorisvS (talk) 12:39, 19 September 2012 (UTC)

Tidal locking in gas giants

Many of the extrasolar gas giants that were detected are very close to their primaries and are assumed to be tidally locked. But how this tidal locking is defined for planets that don't have fixed surface features (i.e. they don't really have a surface)? And what would be the effects on the planet? Visibly, I guess the planet would look pretty much as any other gas giant, with no visible cue that it's tidally locked. Apass 89.137.186.101 (talk) 20:25, 9 November 2012 (UTC)

Defining tidally locked rotation is, in principle, still straightforward for gas giants: Their rotation period must be the same as their orbital period. However, defining what is 'the' rotation period of a gaseous planet is far more tricky: The rotation period of Jupiter's cloud tops seems to be what is usually considered its 'rotation period', but this gives it an equatorial rotation period and a polar rotation period (Jupiter#Orbit_and_rotation). Tides will tend to synchronize a planet's rotational angular momentum to its orbital angular momentum, no matter its composition. Given that a gas giant's 'rotational period' is usually taken to be that of the visible clouds, other effects could affect a hot jupiter's apparent rotational period, possibly resulting in an apparently unlocked state. Strong winds resulting from high temperature differentials would likely be a factor here. --JorisvS (talk) 00:09, 13 November 2012 (UTC)

Oberon?

How come Oberon is both on "Locked by Uranus" and "Probably locked by Uranus" lists? — Preceding unsigned comment added by 90.151.131.81 (talk) 20:20, 28 March 2013 (UTC)

Language section

I've tried hard to change the Arabic corresponding language but received errors of usage by another item. The new way of modifying languages has made Wikipedia more complicated that before :(. Please the corresponding article in Arabic is ar:تقييد مدي--Almuhammedi (talk) 11:27, 22 July 2013 (UTC)

 Done, though I'm not sure this is the correct page to make such a request. — Reatlas (talk) 12:11, 22 July 2013 (UTC)

length of lunar month is getting shorter?

Under "Locking of the larger body", it says "Given enough time, this would create a mutual tidal locking between Earth and the Moon, where the length of a day has increased and the length of a lunar month has shortened until the two are the same.". Isn't the length of a lunar month actually getting longer? For example: RDV74 (talk) 04:26, 7 February 2016 (UTC)

Definition of tidal locking

According to:

Heller, R.; et al. (April 2011), "Tidal obliquity evolution of potentially habitable planets", Astronomy & Astrophysics, 528: 16, arXiv:1101.2156, Bibcode:2011A&A...528A..27H, doi:10.1051/0004-6361/201015809, A27.

"A widely spread misapprehension is that a tidally locked body permanently turns one side to its host." Further, "As long as 'tidal locking' denotes only the state of dω_p/dt [rotation rate change] = 0, the actual equilibrium rotation period ... may differ from the orbital period, namely when e [eccentricity] ≠ 0 and/or ψ_p [obliquity] ≠ 0." This statement differs from the definition in the lead of this article. Praemonitus (talk) 19:51, 7 April 2016 (UTC)

Regarding bodies A and B in the mechanism section

The mechanism section employs bodies A and B to explain the phenomenon and does a good job at it. The explanation can be improved if there was a figure depicting the two bodies A and B (or marked in the figure on the right - I hope that the person who updated this had such a figure in mind?). Also, what does red line depict in the figure? — Preceding unsigned comment added by Blackholebounce (talk • contribs) 19:06, 3 March 2017 (UTC)