Talk:Tidal acceleration

Evidence

This mechanism has been working for 4.5 billion years, since oceans first formed on the Earth. There is geological and paleontological evidence that the Earth rotated faster and that the Moon was closer to the Earth in the remote past.

What is this evidence that the earth rotated faster and the moon was closer? I'm not disputing it, but I would love to see this evidence. -- D. F. Schmidt (talk) 16:59, 23 August 2005 (UTC)

Good question. I think one piece is that the Earth's core is spinning slightly faster. This is consistent with the theory since the core is loosely coupled to the mantle and crust (since there is a fluid layer in between) and tidal forces are much stronger on the outer shell with its greater radius. Further, as I recall the deceleration of the Earth and the increasing seperation between Earth and Moon can and have both been measured to significant accuracy.

I have no idea what paleontological evidence could exist out there. Perhaps a fossil record of algae layers? Faster days would imply thinner layers though you'd need to adjust for what is thought to be lower solar output for the time. -- KarlHallowell 17:48, 23 August 2005 (UTC)

Incidentally, this claim should be backed up. Also remind the reader with a short phrase that the oceans dissipate energy. Maybe

This mechanism has been working for 4.5 billion years, since oceans first formed on the Earth and began dissipating energy through the tides.

-- KarlHallowell 18:09, 23 August 2005 (UTC)

But how like how did that happen 41.122.130.82 (talk) 17:58, 19 August 2024 (UTC)

Paleontological evidence is mentioned in [1]. Specifically, that 620 Mya the day was 21.9±0.4 hours, as deduced from rhythmites (alternating layers in sandstone). The average recession rate of the Moon was 2.17±0.31 cm/year from then until now, about half of the present recession rate. The citation is Williams, G. "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit". Reviews of Geophysics 38, 37 (2000). The Moon was formed 50 million years after the formation of the solar system by the collision of a Mars-sized planetisimal striking the Earth, causing the Earth to rapidly rotate with a roughly 6-hour day, as deduced from computer modelling. The oceans may have been formed by 150 million years (from the Origins miniseries on Nova on U.S. Public Broadcasting Stations [2]). The evidence is that the oldest zircon crystals (which is proof of continental crust) have a high amount of ¹⁸O, which means they were formed in water. A controversy exists between whether the oceans were formed by volcanic outgassing of steam or by comet strikes. But they could not have been formed solely by comets because comets measured so far have twice the amount of heavy water (deuterium oxide) than the present ocean has. I have only recently begun my search for these citations. Luckily, the Nova broadcast was last night, August 23. — Joe Kress 22:15, August 24, 2005 (UTC)

Excellent! I'm a bit too busy right now to help, but I'll put it on my watchlist FWIW. -- KarlHallowell 23:23, 24 August 2005 (UTC)

Moment of inertia

As a consequence, the polar diameter of the Earth increases, and since the mass and density remain the same, the volume remains the same; therefore the equatorial diameter is decreasing. As a consequence, mass moves closer to the rotation axis of the Earth. This means that its moment of inertia is decreasing.

I learned about this moment of inertia not long ago in college (this spring, actually), but I thought another thing: Since there are more people than there were before, two possibilities:

• Moment of inertia increases because mass increases around the fringe

or

• Rotation rate decreases because the rotational momentum remains the same

But maybe it's a blend of the two, and maybe they come to no net effect, in light of what is mentioned here. But still too much is being assumed. How can we know experimentally that the mass and volume and density remain the same? What if there are changes that are so minuscule so as to be undetectable. It has been known to happen. For instance, we can't detect microorganisms with our eyes, but we can with a microscope. In like manner, perhaps we cannot tell that these measurements are changing because our equipment simply doesn't allow us to? -- D. F. Schmidt (talk) 17:16, 23 August 2005 (UTC)

The key isn't whether the Earth is exactly the same mass, but whether mass changes are significant compared to the changing shape of Earth. It has been speculated that most of Earth's oceans were formed over the past few billion years by an accumulation of small comets. But even this accumulation of mass is dwarfed by the changing shape of Earth. -- KarlHallowell 18:01, 23 August 2005 (UTC)

Looking for algorithm

Hi,

I have been looking for accurate information on this subject to write a good Java class for a while. I am especially searching for day length on paleological scale (millions of years) and accurate data/equations. As it seems there is very few information on this subject on the web. Could anyone drop me an email at silvere@digitalbiosphere.com if he finds the correct equation or a pdf article to send me ?

Big thanks.

The info I cited above in Evidence is an average over 620 million years. Tidal acceleration is highly dependent on the particular continental configuration, but continents drift over millions of years, so tidal acceleration will constantly vary over time. I've seen some attempts to model the ocean basins for earlier continental configurations, but because most tidal friction takes place in shallow seas, those attempts would seem to be almost guesses with no assurance that the results are even in the right ballpark. — Joe Kress 04:04, 5 November 2005 (UTC)

Given that we know so little about the detailed bathymetry of the ocean and continental shelves say a billion years ago the best that you can do is use statistical arguments. The resonant structure of the ocean does depend on the topography etc but as the continents drift these are likely to have moved around in a random manner. The rate at which energy is dissipated is important but at present most of the energy is lost as soon as the tidal wave reaches a nearby resonant continental shelf (this usually takes less than 60 hours). Given there is nothing special about the present ocean, similar frictional damping is likely to have happened in the past. (There is also a small but significant amount of tidal friction in the solid earth but that would need another article).

What really changed in the past was the strength of the tidal forcing (the Moon was nearer) and the rotational speed of the Earth which was larger (a day originally lasted for about eight, maybe ten, hours). Increased strength by itself produces increased tidal friction and tidal acceleration. However because of the increased rotation rate, tidal wavelengths would have been a lot shorter and the efficiency with which they were excited by the long wavelength (half the Earth's circumference) tidal forcing would have been a lot lot smaller (it depends on the square). The net effect is that tidal accelerations may have been less over the last two billion years but increased before that when the extra tidal forcing term wins. David Webb 22:07, 18 August 2006 (UTC)

An influential variable that has fluctuated over millennia is the quantity and location of ice. Its effect on the tidal drag is clear: since it reduces the amount of liquid water in the oceans and itself responds to the tidal gradient much less, it reduces the drag. Its effect on Earth's moment of inertia (MI) is less clear cut: if ice accumulates at the poles MI decreases, but an accumulation of ice on a mountain range near the equator could be formed of water which had a smaller MI in its original position.

Above, Joe Kress has referred to the extent of shallow seas and continental layout whic I'd guess is more important than icecap size. Below is mention of global warming and sea rise. With regard to the length of the day, the Astronomical Almanac would be a starting point, but that's not over geological time. There are just a few data points to go by. NickyMcLean (talk) 10:51, 10 December 2016 (UTC)

A global freezing event has been proposed by several scientists, described in detail by Gabrielle Walker in her book, "Snowball Earth" (ISBNs 0 7475 6051 X Hardback and 07475 6433 7 Export paperback) and in summary at https://en.wikipedia.org/wiki/Snowball_Earth. One or more of such events are supported by evidence dating them to the period from 750 to 590 Mha. Depending on their extent, these could have reduced tidal drag by an order of magnitude. Kenastro (talk) 11:54, 9 December 2016 (UTC)

p.s. while I am on-line - what on earth is the "Explanation" section doing at the top of the main page? Is it someone's pet beef that the Moon is really another planet (like Pluto's companion)? David Webb 22:07, 18 August 2006 (UTC)

Good point. I've made some changes there, but it could still use more work. Deuar 16:04, 23 August 2006 (UTC)

Angular momentum

Quote:"however, most of the energy lost by the Earth is converted to heat, and only about one 30th is transferred to the Moon" Does anyone have a citation or justification for this statement? — Preceding unsigned comment added by 134.76.40.56 (talk) 10:11, 9 December 2011 (UTC)

I am confused about whether the earth's angular momentum remains constant or gradually decreases. The qualitative description in this article suggests that it is gradually decreasing but the quantitative description says it remains constant. Are the two sections discussing the same angular momentum? If so, which is it - constant or decreasing? 155.91.45.231 22:34, 10 November 2005

The last paragraph of the quantitative description is not discussing the same angular momentum discussed elsewhere. It only refers to the change in Earth's moment of inertia due to a change in its shape, like an ice skater bringing her arms inward. This portion of Earth's angular momentum is constant, but the rest of Earth's angular momentum is decreasing due to tidal friction. How to clarify this is problematic. To compound the confusion, the angular momentum of the Earth-Moon system is constant, so the decrease is Earth's rotational angular momentum is transferred to the Moon's orbital angular momentum as an increase. That last paragraph also places too much emphasis on relative changes in earth's polar and equatorial radii. Earth's north pole radius does not change, but radii at other high northern latitudes do (Canada and Scandinavia). Whether the south pole radius (Antarctica) is now changing or changed in the past has not been established. — Joe Kress 05:50, 11 November 2005 (UTC)

Proposed merge with Tidal Friction

Seems a good idea given the sparseness of the tidal friction article. However, it should be noted that the current tidal acceleration article talks almost exclusively about the Earth-Moon system. -- KarlHallowell 14:48, 31 December 2005 (UTC)

Perhaps the Tidal friction article should be revamped into "Tidal heating" to discuss things like Io's and Enceladus's heat source. Tidal friction as such is already being discussed in some detail here at tidal acceleration. Deuar 10:28, 29 May 2006 (UTC)

Tidal acceleration involves a transfer from one bodies rotational angular momentum to another bodies orbital angular momentum and, given Newtons's laws, this is only possible if tidal energy is lost to friction somewhere in the system. Thus tidal acceleration and tidal friction are just two sides of the same coin - and in my view should be combined. David Webb 21:18, 18 August 2006 (UTC)

Energy balance

The angular momentum and energy subsection was difficult to follow, and incorrect, so I have implemented a fix. Admittedly this is a rather confusing issue. However, there is a quantitative issue which I am not familiar with, and could use a definitive explanation:

How much energy is lost due to friction of the tides on Earth, and how important is it?

No, it was correct but you mis-understood Tom Peters 09:47, 28 May 2006 (UTC)

Well, it's better now, anyway. My "approximately conserved" statement is a bit embarassing. Late night shorthand that didn't work out :-(

There is still a problem with the way it is now. It is stated:

The excess rotational energy dissipates through friction of the tidal waters along shallow coasts, and is lost as heat.

What excess? Gravity is a conservative force, so the gravitational torques should neatly transfer the Earth's rotational energy to the Moon's orbital energy without the need for any friction. This friction appears to be an un-related extra. For exmple at PhysicsWorld, there is a quantitative calculation without any mention of friction [3]. I would suggest "Additionally, some rotational energy ... " Deuar 14:03, 28 May 2006 (UTC)

A recent overview of the subject by a world-renowned authority, Walter Munk ("Once again: once again—tidal friction", Progress in Oceanography 40 (1997), 7-35), leaves no doubt that the Moon's torque on the ocean's tides is dissipated as heat. Indeed, if it is not dissipated as heat, then it does not affect the Moon's motion. "In a steady state situation, the net work done by the Moon on the water is balanced entirely by net dissipation of oceanic tides." Here, 'work' has its normal physics definition of force times distance or torque and 'dissipation' means the generation of heat. Earth's rotation would not be able to drag tides ahead of the Moon without friction and hence heat. Without heat, the Moon's torque on the tides would pull the Earth's tides back into synchronization with the Moon. Heat exactly balances and maintains torque. — Joe Kress 22:05, 28 May 2006 (UTC)

The friction is not related to the Moon's migration, but may be relevant for the Earth's slowdown. Since it is known that the Moon is moving out, then it must be obtaining energy from the Earth (total energy increases with orbit size), and it can't do that from friction. But the friction must be there. Does anyone know how big is it's effect on the Earth's rotation compared to the loss of angular momentum + energy to the moon? Deuar 20:51, 26 May 2006 (UTC)

If you are competent to write on these matters, you should be able to compute that yourself. Tom Peters 09:47, 28 May 2006 (UTC)

Well I always wonder - suppose I see an inconsistency. Is it better to fix it according to what appears right, or leave it wrong? Deuar 14:03, 28 May 2006 (UTC)

Munk provides a formula that directly relates tidal dissipation (heat) to the Moon's acceleration:

D = - (dn/dt)(Ω - n)[r²MM_E]/[3(M+M_E)],

where n, M, and r are the Moon's orbital velocity, mass, and distance, and Ω and M_E are Earth's angular velocity and mass. He assumes that the lunar acceleration of lunar longitude dn/dt = -25"/cy², which is somewhat smaller than the lunar laser ranging value of -25.858"/cy². — Joe Kress 22:05, 28 May 2006 (UTC)

Thanks for the recent added explanations on receding Moon and energy dissipation. This is the first time I really have the impression I understand the phenomenon. And looks so simple now. Good work! −Woodstone 18:14, 29 May 2006 (UTC)

edits by Joe Kress 30 May 2006

Joe, I find your latest edits not at all an improvement:

• "net tide ... (integrated over all oceans)" does not clarify anything. I doubt that one can integrate a tide. IMNSHO just "tide" would be as accurate, and much more to the point.
• "tidal friction (dissipated as heat)": friction (a force) can not be dissipated as heat. Friction working on the rotating Earth dissipates energy in the form of heat. Also the extra qualifier "tidal" does not help at all here: in which respect is "tidal" friction different from just "friction"? The original text was more accurate and understandable.
• "(within 48 hours)": where does this come from? Are you sure it is not 47 or 49 hours? I don't think specifying an exact value contributes much here.
• Under historical evidence, you removed the line of reasoning that leads to the conclusion that the day was shorter. Now there is a big leap of thought between finding rhythmites, and the result of the day length. IMNSHO the explicative reasoning is necessary to bridge this gap.

I intend to revert these edits, unless you can clarify these points. Tom Peters (Wikipedia apparently already logged me out again) 82.170.18.66 22:43, 30 May 2006 (UTC)

Tom (after moving your edit to the end of the page):

• The problem with "tide" is that there is no such thing as an equilibium tide (an ocean in the form of a prolate spheroid) on Earth for the Moon to pull on. Earth's continents break up any equilibrium tide as they pass under the Moon. Instead tides of much greater amplitude rotate around each ocean basin. The Moon pulls on large tides that are both ahead of and behind it. An integration over the world's oceans is indeed preformed by computer, effectively subtracting these large tides from each other in order to find the net difference, only 3.23 cm. I intend to add a paragraph about this.
• I am using friction in the sense of one body rubbing against another. In normal usage, when you rub your hands together, for example, you get heat. The precise physical sequence would only complicate matters. But it is critical that heat dissipation be mentioned, and your wording was ackward. Friction normally exists between two solid bodies, but here the friction is between a liquid and a solid, so I think "tidal" is an appropriate modifier.
• "Within 48 hours" comes from Munk as cited. Specifically, he stated that "In a steady state situation, the net work done by the Moon on the water is balanced entirely by net dissipation of oceanic tides. With a global tide energy of 4 x 10⁵ TJ, [a dissipation of 2.4 TW] implies that once every 48 hours all of the tide energy is renewed!" However, the calculation actually yields 46.3 hours, so Munk was a bit loose. A less precise "within two days" would be more appropriate.
• I only removed a reference to numerical integration because the solar system becomes chaotic before some tens of millions of years, so it cannot be used as proof of a constant year over billions of years. I have now consulted the citation and can add their argument, for whatever it's worth (that the constant of gravitation has not changed).

— Joe Kress 06:33, 31 May 2006 (UTC)

Presumably the most likely scenario for year length changes is from perturbations by other planets changing a, not from a changing constant of gravitation, and this is what numerical integration was trying to check (I haven't read the article, my uni doesn't seem to subscribe to that journal). Even though semi-major axis is what is most resistant to orbital resonances. If significant changes had been seen then, yes their details could not be trusted due to the solar system's chaotic behaviour, but the general presence or absence of changes is fairly trustworthy. It's a matter of finding out whether the Earth is in an integrable or non-integrable part of phase space. In any case, I don't think it's worth belaboring this point in a tidal article, since tidal forces would not be expected to have much effect (the Earth's rotational angular momentum is peanuts compared to its orbital angular momentum, and you wouldn't expect tides on the Sun raised by the Earth to be significant). Deuar 10:21, 31 May 2006 (UTC)

What about the Sun?

Well, there's quite a difference between shift-tilda and control-tilda, so I'll try again.

In discussing the tides due to the Moon, those due to the Sun ought to be mentioned as the effect is about half that of the Moon.

Kenastro (talk) 16:47, 8 December 2016 (UTC) Indeed they should, the main reason, IMO, being that the article implies that the Earth - Moon system is closed for angular momentum, and that its total is therefore conserved. My understanding is that the Sun's tidal effect reduces the total angular momentum of Earth and Moon by contributing to the slowing of the Earth with small (negligible?) effect on the Moon's distance. I shan't give any references for this, because most people who know anything about the tides know that spring and neap tides occur at times when the Earth-Moon and Earth-Sun directions are, approximately and respectively, in-line and at right angles (I learnt this at school), and can deduce that the Sun has a tidal effect on the Earth's rotation rate. ::

I see that Herbmuell discusses the Sun's tidal effect below, in another context.  ::

Likewise, it should be mentioned that the tides as observed are the consequence of resonances in the ocean basins as built up by the history of the tidal forces, which (as is mentioned) would produce only a small rise alone.

In thinking about rotational and orbital resonances, I first thought that there might be "wells" that perhaps could not be escaped from, thus Mercury in a 3:2 resonance with only a small perturbation (there being no oceans of mercury slopping about, Hg or H2O) might be trapped in that state, whereas the wells associated with 354, 364, 363,... revolutions/orbit would be shallow. And what of the bizarre resonance of Venus, that turns the same (shrouded) face to Earth on closest approach? Then I convinced myself that the energy dissipation by friction into heat being irreversible, the resonance would be broken eventually. After pausing to confuse myself by wondering what would happen if the orbit and rotation were in the opposite senses, it seems clear to me that the slowing of rotation would not be constant but would wobble as resonances were entered and exited. That is, to start with the Earth's revolution would be too rapid so that it would shove the tidal bulge forwards and via frictional losses (with the bulge's position being held back by the Moon's gravity) would be slowed. Later, after slowing past the exact resonance the spin would be behind the Moon's schedule and so the slowing would abate.

This raises another point. Even with a 1:1 resonance of rotation and orbit, the Moon would not hover above a constant point of the Earth's surface but wander a little. Unless the orbit was exactly circular and there were no other masses in the Universe to peturb matters...

In general, I think explanations involving phrases such as "due to the conservation of ..." should be avoided because they are appeals to magical reasoning, as if there is some Universal Accountant observing all and enforcing its rules. A rocket advances, not because of conservation of momentum (even though it is upheld) but because of the pressures in the reaction chamber being unbalanced by the vent hole. Understanding this chain of reasoning is one thing, reciting "conservation of momentum" another.

Regards, NickyMcLean 23:45, 8 June 2006 (UTC)

I completely agree with you regarding not invoking conservation laws as an explanation. Hear, hear!

Regarding resonances, there is no reason for the resonance to be ever broken by small perturbations. An external perturbation transfers energy into the body's wobble, which is later lost through tidal friction and the body becomes nicely locked again. Provided the external perturbations give the body much less wobble energy per year than it can lose in a year it remains at the bottom of the well, only ever slightly wobbling.

Also, in almost all cases, the rotation slows asymptotically down to the locked rate without oscillating around it, because the system is usually overdamped. Deuar 17:00, 9 June 2006 (UTC)

I thnk you are misguided. Conservation laws determine the quantities of energy, angular momentum, etc that can be exchanged. The gravitational force is there, and there are boundaries on its work determined by the conservation laws. From conservation you can draw conclusions on what happens. Everything happens at once but there are different approaches to explain all that, which all lead to the same results. There is nothing "magical" about this reasoning: it is a quantitative sum. You cannot make conserved quantaties disappear from the system. So for instance you are forcced to conclude that the Moon moves away from the Earth. Regarding tidal locking: eventually the Moon will not be wobbling around some intermediate position, but will stay put over the surface of the Earth: the orbit will indeed be circular. Tom Peters 00:09, 10 June 2006 (UTC)

The "wobbles" I was referring to are due to perturbations by other bodies than Earth,Sun and Moon. Sure, they're tiny and possibly even un-observable from Earth with current instruments, but it's just a matter of principle. They have no qualitative effect. Deuar 12:36, 10 June 2006 (UTC)

Gravitational perturbations by the other planets and asteroids on the Earth and Moon are both detectable and have been calculated very precisely. See Jihad Touma, Jack Wisdom, "Evolution of the Earth-Moon system" (pdf, 2.33 MB), _Astronomical Journal_ 108 (1994) 1943-61, which obviously includes tidal friction. — Joe Kress 06:00, 11 June 2006 (UTC)

Regarding conservation laws, of course they're a correct and useful way to do the calculation and lead to the same results. duh. I've just found that when discussing things with people who are not familiar with physics, invoking conservation laws is usually not particularly insightful for them. A rather blank expression appears on their face, and i've lost them. On the other hand talking about forces and torques, action-reaction, etc. seems to be much more intuitive and work better. imho ;-) Deuar 12:36, 10 June 2006 (UTC)

Amplitude

This article states:

• Earth's net equilibrium tide has an amplitude of only 3.23 cm.

In contrast, the article tide states:

• The theoretical amplitude of oceanic tides due to the Moon is about 54 cm at the highest point

This is rather a big difference. The latter result comes form the referenced article: Myths about Gravity and Tides. The calculation is shown extensively and is verifiable. Where does the result in this article come from? −Woodstone 13:37, 9 June 2006 (UTC)

3.23 cm comes from Munk as cited. It is the net equlibrium tide measured from satellite data. The Myths figure is calculated for a hypothetical ocean-covered Earth. One obvious reason for the difference is the different Earth models. This indicates how far removed the simplistic ocean-covered Earth is from reality. Coincidentally, that same satellite data shows that the maximum tide in the open ocean (not that at shore) is around 80 cm (at only a few locations). According to Munk, tidal dissipation due to the Sun is 1 TW whereas that due to the Moon is 3 TW, hence the Sun's affect is 1/3 that of the Moon, not 1/2. One reason for the difference is that 1/3 includes all harmonics, not just the prolate spheroid M₂ 'harmonic'. Another difference is that the dominant solar tide is S₁, which has only one bulge per day, not two. — Joe Kress 06:20, 10 June 2006 (UTC)

I have confirmed that 3.23 cm is correct because earlier figures of 3.21 cm and 3.64 cm appear in "The Earth's Variable Rotation: Geophysical Causes and Consequences" by Kurt Lambeck (1980). However, S₁ appears to be a typo — it should be S₂, which would be a semi-diurnal tide like M₂. — Joe Kress 06:00, 11 June 2006 (UTC)

Thanks for the explanations. Perhaps some of it should be placed in the tide article (and in this one). It still leaves the question of the tidal acceleration in the sun-earth system and how it combines with the moon-earth effect. The earth's rotation cannot be in step with both. What would the theoretical final rotation be? −Woodstone 09:34, 11 June 2006 (UTC)

Currently, the Moon's tidal effect is the larger, however with time the Earth's revolution slows and the Moon drifts out so that its effect weakens. But, once the Moon's orbit 'around' the Earth (it is actually always concave towards the Sun) attains a simple ratio with the Earth&Moon around the Sun, its orbit will be disrupted. Currently the Moon's orbital period is between 13 and 12 cycles/year and presumably (disregarding Velikovsky) it had survived the 13/year resonance on the way to the current state. Smaller ratios are stronger resonances, and, with the Moon more distant from the Earth, its "binding energy" would be less. As a further point, I have read somewhere that one effect of the Sun on the Earth is to increase the rate of the Earth's spin (yes yes, slightly) via some atmospheric effect. Evidently, the the atmosphere is assymetrical in that the dawn side is cooler than the sunset side, but offhand, I can't think of mechanism that might have a spin effect. Weather patterns do affect the spin, but they are temporary (yes, I can wave the "conservation of momentum" flag too) - likewise movements of magma, earthquakes, continental rise/fall/drift, etc. NickyMcLean 05:08, 14 June 2006 (UTC)

The above does not answer my question. Let me rephrase. After earth's rotation has slowed down so it is locked with the moon's orbit, there is still a monthly tide on earth, caused by the sun. This would continue to slow down the earth's rotation. As a result, it then turns the other way relative to the moon. So the tide from the moon would start to run in opposite direction as the one from the sun. At a certain point the net effect of the two decelerations would cancel and the rotation would just wobble slightly around that value. Is this a correct analysis? What would the final rotation time be? −Woodstone 07:50, 14 June 2006 (UTC)

The referenced article Myths About Gravity and Tides puts the tidal lock for the Earth-Moon system at a day of fifty hours. But the solar tide would continue to slow the Earth whereupon the Moon would be wound back towards Earth finally disintegrating at the Roche limit. There would be no stable state along the way in that argument. NickyMcLean 20:44, 14 June 2006 (UTC)

"cy" is not a standard unit

I know that each time "cy" is used, it is explained ("cy is centuries"), but in fact according to metric standards it should be centiyears, rather than hectoyears. It's not ambiguous, it's simply non-standard.

Could we use "hy" (perhaps with explanation, "hy = hectoyears = centuries") instead?
Or get a conversion into (some power of 10 times) seconds?

-Patrick N.R. Julius 23:00, 17 January 2007 (UTC)

The article must use the unit used in the references, which is usually "cy". I have never seen either "hy" or powers of seconds used by any cited author, so they cannot be used in the article. Technically, the Julian century (100 Julian years) is used in the citations, each century having exactly 36525 days, each day containing exactly 86400 SI seconds, which could be added to the article as a footnote. Although their product is 3155760000 s, I have never seen that product mentioned by any cited author — indeed, many authors fail to specify the specific kind of century because other kinds of centuries, like a century of mean tropical years, 36524.22 days, would produce the same results given the limited number of significant digits to which those results are known. — Joe Kress 05:36, 18 January 2007 (UTC)

"hy" might not be used by any author, but it is at least correct according to SI prefixes, and given precedents for this kind of use of SI prefix I'd argue that we should use it. (After all, both "hecto-" and "year" are used very widely; the combination should be obvious.) Double sharp (talk) 14:08, 1 February 2015 (UTC)

Changed to "hy" throughout: to me "cy"'s connotation of "centiyears" is too dangerous for the reader. Double sharp (talk) 14:11, 1 February 2015 (UTC)

Acceleration vs. Deceleration

As far as I understand, when an object in orbit accelerates, it increases it's escape velocity, yet, I also understand that as per Kepler's 3rd law, an object orbiting at higher needs less speed to remain in orbit.

In one section in the article it states that the torque between the Earth and the Moon "accelerates the Moon in its orbit", yet in anouther section it states that as per Kepler's 3rd law the moon's "velocity actually decreases, so the tidal acceleration of the Moon causes an apparent deceleration of its motion across the celestial sphere". I note it says "apparent deceleration" but im a bit confused as to if it's actaully accelerating or decelerating since the 3rd law shows that it should infact be orbiting slower since its in a higher orbit.

Thanks, Zhatt 23:16, 19 January 2007 (UTC)

You are essentially right. The "apparent" is there because the Moon moves into a higher orbit, and even if it would keep the same linear speed, its angular velocity will decrease, i.e. deceleration but due to a perspective effect. Anyway, the main point in all this is that a deceleration is also an acceleration but by a negative amount; while in this case the Moon is really accelerated, i.e. a force is working on it that adds energy to its motion: but that goes into gravitational potential energy and not into kinetic energy, i.e. not into higher speed. Feel free to reformulate for improved clarity. Tom Peters 13:33, 20 January 2007 (UTC)

So, if I'm getting this right, the moon is nither really accelerating nor decelerating and is keeping the same linear speed; it is simply moving away from earth, but it's because of the "acceleration" that it's moving into a higher orbit? Now because it's in a higer orbit it is, lets say, "mathematically" moving slower since it's angular velocity has decreased? It's a little hard to wrap one's head around. I think I'm going to go read up some more on angular velocity.

Thanks,

Zhatt 21:58, 20 January 2007 (UTC)

No, I wrote "would". Kepler's third law: n^2 * a^3 = C , i.e. angular velocity n = sqrt(C/a^3); velocity v = sqrt( C*(2/r - 1/a) ). As a (and r) increase, the orbital velocity as well as the angular velocity decrease. But we can still call it an acceleration because the total energy (potential + kinetic) increases as a result of the torque force. Tom Peters 23:05, 21 January 2007 (UTC)

Ah! Ok, I get it now. The orbital velocity of the moon is in fact decreasing, yet it's gaining momentum and, therefore, technically accelerating. That basically what I assumed at first, but it seemed backwards so I wanted to make sure. Thank you very much. Zhatt 06:55, 23 January 2007 (UTC)

I question whether "tidal acceleration" is a good term for this phenomenon. "Acceleration" means gaining speed, not energy. I have tried to clarify the article a bit, but it would be clearer if it weren't called "tidal acceleration" at all! Eric Kvaalen (talk) 20:39, 21 October 2008 (UTC)

For the Earth-Moon system in isolation (neglecting the Sun etc.), the virial theorem, that the average potential energy is twice the average kinetic energy of translational motion (treating Earth as a mere particle), applies here. (Since there is no significant oscillation of either there is nothing to average.) For every two joules of potential energy gained by the Earth-Moon system as they move apart, one joule of translational kinetic energy is lost, most of it to the Moon. By conservation of energy the remaining lost joule must be divided between the Earth's decreasing rotational energy and frictional losses in the oceans and interior, in a proportion determined by conservation of angular momentum. If the Earth really were a particle tidal acceleration would be zero because the second lost joule would have nowhere to go. Vaughan Pratt (talk) 23:05, 18 November 2013 (UTC)

Effects of sea level rise due to global warming?

Since most of the dissipation takes place in the shallow seas, would a minor sea level rise have noticeable effects on the rate the Moon moves away from the Earth? Count Iblis 20:15, 24 September 2007 (UTC)

Probably not, because for any low lying land that is inundated, a corresponding amount of shallow sea would be deeper, thus their affects on tidal acceleration should cancel each other. Unfortunately, present knowledge does not allow an accurate prediction of the heat generated by tidal friction and hence its affect on the Moon's rate of retreat. LLR could detect it given enough time, but many other changes on Earth affect the Moon's distance, such as Earth's oblateness. — Joe Kress 19:18, 25 September 2007 (UTC)

I can think of two effects but I don't know their size. Firstly, polar ice sticks up higher than present sea level. On melting, its mass will become lower, closer to Earth's centre, so helping Earth spin faster (like that skater pulling in arms). Secondly, water moves more easily than ice so the newly-melted water, moving tidally, will add to the contribution of friction, slowing Earth's rotation over millions of years by a slightly greater amount than if the ice had stayed solid.Greenstone lizard (talk) 13:46, 15 February 2013 (UTC)

Polar ice doesn't have much angular momentum because it is near the poles. Changing the local altitude of a mass near the pole has even less effect because most of a polar-area altitude change (which is directed to the local vertical) is parallel to the spin axis, the radial change is sine(latitude) which is nearly zero at eighty degrees. Changing the height of the earth's oceans by say a metre is not a big change except for shore dwellers. However, what counts is that water melted near the poles changes the level of the whole ocean so that there is a nett motion of water from the poles to the equator - that is, a polar iceblock, on melting, is evenly spread over the whole ocean so it is not just a shift of mass from onshore glacier to nearby polar water. Thus, a mass spinning close to the poles becomes a shell over the whole earth (most of which is well away from the poles), and so the earth's spin slows. Calculation of this is straightforward, given formulae such as the moment of inertia of a sphere and thus of spherical shells. A complication is the rebound of the earth's crust when the weight of ice is removed (still ongoing since the last ice age), but this is slow whereas the water redistribution might as well be thought of as instant. NickyMcLean (talk) 20:25, 17 February 2013 (UTC)

Advance of tidal bulge

An anonymous editor quantified the advance of the tidal bulge as 3° which I removed. I agree that that is about the advance of the bulge of the solid earth. But the energy dissipated by tidal friction in the solid earth is a tiny fraction of the total tidal friction, 0.1 TW vs 4.0 TW. The energy dissipated in the oceans is dominant. If Earth were assumed to have a world-wide ocean, then the advance of the tidal bulge would depend on the depth assumed for that ocean. I'll have to check my sources to see if they mention any angle alongside the 3.23 cm for the height of the net equilibrium tide. — Joe Kress 06:41, 5 October 2007 (UTC)

I think this paragraph is more about the force from the displaced bulges, whether solid earth or water, than the friction, but anyway, because of the oceans it's not easy to give a clear statement about the size of the angular displacement. Deuar 08:44, 5 October 2007 (UTC)

An angle of 3° is simply wrong. The angle between the Earth-Moon line and the peak of Earth's solid tide or body tide (almost an equilibrium tide) is 0.16±0.09° according to Ray, Eanes, and Chao, "Detection of tidal dissipation in the solid Earth by satellite tracking and altimetry", Nature 381 (1996) 595–7. The angle between the Earth-Moon line and the peak of the degree 2 order 2 spherical harmonic component of the dominant M2 ocean tide (not an equilibrium tide) is 65° according to Munk, "Once again: once again—tidal friction", Progress in Oceanography 40 (1997) 7–35, p.8. More research is needed before I can discuss the amplitude of their bulges. — Joe Kress 05:52, 20 October 2007 (UTC)

Well found! Deuar 07:54, 23 October 2007 (UTC)

Boiling AWAY?

Fair enough that oceans will boil - but "away" ? Away where?? I think the following change is needed:-

... "the Earth's oceans to boil away to gas, removing reducing the bulk of the tidal friction"... 

203.206.137.129 (talk) 18:21, 14 March 2009 (UTC)

I've changed "boil away" to "vaporize". — Joe Kress (talk) 05:09, 15 March 2009 (UTC)

First

Kant seems to have been the first to mention tidal retardation. This was in 1754. —Preceding unsigned comment added by 87.194.34.71 (talk) 12:04, 22 April 2009 (UTC)

Tidal braking

Google shows that the more commonly used term is tidal braking as opposed to tidal deceleration, which this article uses. In fact some of the articles that link here use that term, and yet the term isn't used anywhere in this article.

For a few more reasons to use the term from a Google Books search for tidal braking:

• Close Binary Stars: Proceeding of the I.A.U. Symposium, No. 88, Toronto, Page 27 by International Astronomical Union Staff - Science - 1980 - 618 pages TIDAL BRAKING When a body spins with a period T in the presence of a ... The amplitude dependence of the tidal braking is in general very large (see ...
• Mechanics problems in geodynamics‎ - Page 460 by Ren Wang, Keiiti Aki - Science - 1996 - 396 pages Finally, the bounceback to its less oblate interglacial shape makes the Earth spin faster, overcoming a third of the tidal braking by the Moon and Sun. ...
• Critical observations versus physical models for close binary systems: a ...‎ - Page 7 by Kam-Ching Leung, National Science Foundation (U.S.), Zhongguo ke xue yuan - Science - 1988 - 472 Tidal braking of the Earth by lunar torques ...

Food for thought... -Miskaton (talk) 20:38, 17 September 2009 (UTC)

Also note that I've modified the redirect for tidal braking which used to link to tidal locking, which also doesn't mention the term anywhere. It now redirects here. -Miskaton (talk) 20:44, 17 September 2009 (UTC)

The title of this article is "tidal acceleration" with many more Google hits than either tidal braking or tidal deceleration. — Joe Kress (talk) 22:13, 19 September 2009 (UTC)

Yes, that's correct, but not relevant. In physics, there's really no such thing as "deceleration." Acceleration (in this case, angular acceleration) is a vector, and that vector can have a positive or negative sign depending on the context, but how you set that sign is arbitrary. OK, so physics 101 aside, this article is equally about tidal deceleration/braking and acceleration because they're the same thing. Because the literature uses the term "tidal braking," this article should as well. It's not Wikipedia's or Google's call, but the field of orbital mechanics. -Miskaton (talk) 20:31, 22 September 2009 (UTC)

kinetic & potential energy: dubious tag discussed

Quote: "Although its kinetic energy decreases by 130 GW, its potential energy increases by 260 GW." "GW" appears to be a measure of power, not energy, so the statement appears to employ units that are incommensurable with the quantity that it attempts to describe, in which case it can't be correct. If the editor who posted it would give the supposed reference (missing so far) then probably the matter could be sorted out. Terry0051 (talk) 21:34, 2 January 2010 (UTC)

The decrease is not a one-time event, so an amount of Joules would not make sense, but I agree the new "at the rate of" is an improvement.--Patrick (talk) 23:30, 2 January 2010 (UTC)

Specific_orbital_energy#Rate_of_change gives the formula for the rate of change of the specific orbital energy with respect to a change in the semi-major axis, to be multiplied by the mass of the Moon and the recession rate: (((((399e12 * 7.35e22) / 2) / (384e6^2)) * 3.84e-2) / 365.25) / 86400 = 1.21002696e11. Whatever the merits of [4], this little computation is confirmed under "Energy Loss Per Year" (loss of kinetic energy is gain of total energy).--Patrick (talk) 08:41, 3 January 2010 (UTC)

[From Terry0051] (1) Thanks for the clarification. I don't want to seem needlessly ungrateful, but the clarification reveals further cause for doubts. Both the dimensions and the numbers appear to be identifiably wrong, and the whole calculation also seems to be original research.

(2) Dimensions: The cited article-section Specific_orbital_energy#Rate_of_change does not give a rate of change of energy with time, it gives a rate of change of energy for an increment of orbital size. The dimensions are equivalent to energy/distance, not energy/time. So 'power', as energy/time, still appears to be incommensurable with what is being described.

As I mentioned and applied, the quantity with dimension energy/distance is multiplied by the recession rate, which is 3.84e-2 m/year or 3.84e-2 / 365.25 / 86400 m/s, the value you mention yourself in point (5).--Patrick (talk) 23:46, 3 January 2010 (UTC)

(3) Numbers: The cited formula can only be valid for "the rate of change of specific orbital energy with respect to a change in the semi-major axis", if what is considered is an ideal and unperturbed two-body system, not a perturbed multi-body system. The real, and strongly-perturbed, situation of the Moon is very far from an ideal hypothetical 2-body situation, and the unperturbed formula cannot be relied on, or at the least, it needs proof that it offers anything like a reasonable estimate for a very small incremental long-term property of the motion.

(4) (When such idealized formulae for hypothetical and unreal situations are applied, as here, to very rough estimates of what is in this case a single astronomical parameter, especially in the case of a very small incremental long-term property of the motion, then an abstraction is being made from a real situation that depends on many parameters, and sometimes on small differences between large terms. There is a big risk of making what can be called partial or double accounting for something in the picture somewhere, and only a good reliable source can really dispel doubts that this has happened.)

(5) A perhaps more realistic estimate of the change with respect to time might be obtainable directly from figures for lunar distance and velocity and their rates of change without assuming that the unperturbed energy integral applies. Of the the existing cited reliable sources (Chapront, 2002; Dickey, 1994), one is said to give the mean rate of recession of the Moon as +3.84 ± 0.07 m/cy (as cited in the article, I didn't check back to Dickey, 1994), and the other gives the mean longitudinal angular motion (in a near-inertial reference frame that leaves out general precession) (checked from Chapront 2002): 218◦d18'′59".8782 + 1732559343.3328”*tc −6.8700”*tc^2 +0.006604*tc^3 −0.00003169”*tc^4 (where tc is interval from 2000 Jan 1.5 (on the TDB time scale) in Julian centuries of 36525 days).

(6) The mean distance of the moon for J2000 is not included in either of these sources, but from the series of Paris-observatory lunar ephemerides (ELP) see ELPMPP02 overview one of the most recent estimates gives 385000.52719 km ('r0').

(7) (Digging into sources shows that this distance figure is appropriately larger than the nearest mean elliptical semi-axis 'a', by a factor of about (1+e^2/2) where e is the eccentricity, see Equation of the center#Analytical expansions. Mean values (over time) of distance and of reciprocal distance are not reciprocals of each other, in summary because of the unequal angular velocity, which reduces the times spent at the shorter distances.) Correspondingly, the mean orbital speed in an elliptic orbit is less than in a circular orbit of radius 'a' (see Orbital speed#Mean orbital speed, under 'elliptic motion' -- an item for which a source is given).

(8) The mean angular velocity n0 (in "/cy) is (1st deriv. of expression for mean longitude, truncated at 2nd term) 1732559343.3328" −13.74"/cy. The energy balance (or, possibly, imbalance) can then be estimated from (a) the kinetic component based on mean orbital speed. The speed in a circular orbit of radius r is (n.r), the instantaneous kinetic energy (1/2)Mm(n.r)^2, and its mean in the elliptical case is estimated by: (1/2)Mm(n0.r0)^2/(1+e^2/2). (b) The potential energy component given by the mean of -GMm/r, estimated by GMm(1+e^2/2)/r0, where r0 is the mean of r. Thus total energy: (1/2)Mm(n0.r0)^2/(1+e^2/2) -GMm(1+e^2/2)/r0.

(9) On this basis, an estimate in the rate of change (with respect to time) of the mean total energy, expressed in terms of the sourced data, could be obtained from the first derivative with respect to time of the energy integral (where n0', r0' are the time rates of change of n0, r0): Mm( (n0'.n0.r0^2 + n0^2.r0'.r0)/(1+e^2/2) + G(1+e^2/2).r0'/r0^2), ensuring compatibility of units between G, n0 and r0.

(10) I don't suggest putting forward calculations based on this formula for inclusion in the article, because there may still be an element of OR in it, even though it seems closer to relevant sourced data than the previous suggestion, and more likely to be realistic. I offer it only as a part of the reason for urging caution when making unsourced extrapolations from formulae that quite likely embody inappropriate assumptions. This is why I do suggest that the originally-added 'facts' about energy change should be either reliably verified or removed. Terry0051 (talk) 17:57, 3 January 2010 (UTC)

For a result with only two significant digits (120 GW) it seems we can simply assume an orbit that is circular, except that it is slowly spiraling out, with the understanding that, like the recession rate, it is a mean value over a revolution. For extra caution we could write "roughly 120 GW".--Patrick (talk) 23:55, 3 January 2010 (UTC)

But where is the source of any assurance that even the sign and magnitude are actually correct? The offered basis of calculation does not even show that it yields a rate of change with respect to time. Why does it seem that such a calculation makes a reasonable assumption when it also uses an 'unperturbed' assumption for a strongly perturbed orbit? How exactly is this different from just clothing a naked guess with a respectable garment of number? Terry0051 (talk) 00:06, 4 January 2010 (UTC)

Re: "does not even show that it yields a rate of change with respect to time": energy/distance multiplied by distance/time gives energy/time, how is this not clear?--Patrick (talk) 00:25, 4 January 2010 (UTC)

It's not apparent what makes it clear that the basis of calculation is valid as applied to the problem proposed, for reasons already offered. In the absence of a reliable source for calculating this way, the suggested fact appears to be unsupported or at best WP:OR, at least a synthesis of the kind deprecated in WP:NOR. Terry0051 (talk) 00:45, 4 January 2010 (UTC)

Ok, if +3.84 ± 0.07 m/cy is the rate of change of the mean value of the distance as a function of time, the recession rate in terms of the semi-major axis could be different, due to the factor of about (1+e^2/2), if the eccentricity changes systematically. That would affect the energy rate, so I will remove it.--Patrick (talk) 12:15, 4 January 2010 (UTC)

I was going to say much the same thing. There is not too much that goes into the calculation (Kepler's Laws, mass of the Moon, and the recession rate), so I wouldn't necessarily worry about calling it OR, although it should be made clear, perhaps in a footnote, how the calculation was done. The catch is that the energy depends precisely on the semi-major axis, whereas it is not clear what is meant in the sources by "recession rate". This could be the rate of change of the semi-major axis, but it could also be any number of other average values. That info is certainly available, but I couldn't find it in a brief search, and until we do, we should be cautious about the statements we make. --Art Carlson (talk) 13:05, 4 January 2010 (UTC)

Wait a sec. Maybe we can do this after all. In the second paragraph on p. 4 of Lunar Geophysics, Geodesy, and Dynamics, the value "37.9 mm/yr for total da/dt" is explicitly given for the rate of change of "the lunar semimajor axis a". This is even in the context of "[t]idal dissipation in the Earth and Moon", so it would be pretty safe to use the above calculation for the power. (I'm not saying we need to include this in the article, although I think it is a cool number.) --Art Carlson (talk) 13:37, 4 January 2010 (UTC)

Thanks. I added the ref. If someone can clarify which type of average the +3.84 ± 0.07 m/cy refers to we can keep that too, otherwise we can drop it, now that we have a value with a clear meaning.--Patrick (talk) 23:36, 4 January 2010 (UTC)

I'm updating to a much more recent reference, DE421 Lunar orbit, physical libations, and surface coordinates (2008), which also explicitly states it is the change in the semimajor axis. I'm dropping the values given by the 1994 and 2002 refs because they are too old. The value originally in the article, 3.84 cm/yr, has a typo since the value actually given by Dickey (1994) was 3.82 cm/yr. — Joe Kress (talk) 02:48, 11 January 2010 (UTC)

The same thing bugs me. GW is a unit of power, not energy. It is simply not possible for 'kinetic energy' to be reduced by 130GW. --Alastairdent (talk) 07:49, 23 September 2015 (UTC)

Acceleration of Earth due to tides on Sun

On 3 June 2008, 70.110.33.142 (talk · contribs) added the phrase "Tidal acceleration is also moving the Earth outward from the Sun" with this edit. I'm removing it because it seems to be wrong. According to the latest edition of Orbital Ephemerides of the Sun, Moon, and Planets, the Earth's orbit is changing due to the other planets' gravity much faster than the tidal acceleration could explain, and on the longest timescale quoted, it's actually moving towards the Sun. Melchoir (talk) 11:02, 10 December 2011 (UTC)

I have a related question. The change of earths distance from the sun due to tidal acceleration is negligible, says Melchoir, okay, but what about the change of the length of the day due to the sun? The moon creates a tide on earth, earth drags it forward, the moon pulls it back, earth slows down - says the article. Couldn't the same be said for the sun instead of the moon? When calculating the change in the length of day, shouldn't both the moons pull on "his" tide and the suns pull on "her" tide (and finally the change in earths moment of inertia) be accounted for? --Herbmuell (talk) 12:29, 22 November 2016 (UTC)

Deceleration

Could anyone give an estimation of how long it would take for retrograde objects and prograde objects that orbit more rapidly than their primary is rotating to crash into the primary? --JorisvS (talk) 12:21, 20 September 2012 (UTC)

Offhand, Phobos and Diemos orbiting Mars should be worth pursuing. No ocean slosh either. NickyMcLean (talk) 21:38, 20 September 2012 (UTC)

Centrifugal force explaining tidal bulge

The article included: "There is a water bulge on the other side of the Earth that is caused by the centrifugal force of the Earth rotating about the Earth–Moon barycenter". I believe this statement was wrong. If you work in a system of coordinate that rotates with the Earth-moon orbit, the centrafugal force on the Earth in that system of coordinates would explain very little of the bulge, the main reason of which is simply that on the other side of the earth the moon's attraction is weaker, as it goes like 1/R^2 with R being distance to the center of the moon. Both the centrafugal force and the gradient of the moon's gravity imply both bulges (the one pointing to the moon and the one pointing away from it), although intuitively it seems like attraction to the moon explains one bulge and centrafugal forces explains the other. It's important to keep in mind that to create a bulge what you need to consider is the difference in the force acting on different parts of the earth rather than to consider the direction of the force itself.

I edited the article to reflect what I said, then reverted the change as I realized I was wrong:

1) Plenty of explanatory sources over the internet explain it as "1 bulge due to centrafugal forces, 1 bulge due to attraction", trying to argue against it is sort of like original research. 2) My claim that centrafugal forces were not important is wrong. The centrafugal acceleration is w^2*R_earth w=2pi/T T=1month, giving 4.6*10^-5 m/sec^2 while the grvitational spaghettification is 2*M_moon*G/Distance-to-moon^3*R_earth ~ 1*10^-6m/sec^2, so the centrafugal forces seem to be dominant.

I still think "both effects cause both bulges" as what matters is the difference between forces on the center of the earth and forces on its periphery, and in light of my simple calculation I would guess the centrafugal forces dominate, the gravitational gradient negligible, but I guess I could still be wrong.

37.46.38.174 (talk) 11:09, 4 November 2013 (UTC) — Preceding unsigned comment added by 37.46.38.174 (talk) 11:01, 4 November 2013 (UTC)

One minor problem with your calculation I think is that it is using the radius of the Earth when you should be using that plus the distance from Earth's center to the barycenter, which adds another 4700 or so km and raises your 4.6E-5 number to 8E-5.

The second thing I'd worry about is the influence of Earth's spin on your calculation, which you've neglected but which develops a centrifugal force nearly a thousand times as much. The latter force is uniform around the Earth and so induces no lateral force, but for your calculation to be sound it would have to be the case that one can naively add centrifugal forces when they are compounded on different centers of rotation.

My third question is, does the centrifugal force simply raise the water? For completely incompressible water this could not happen unless the force exceeded that of gravity. However because of the geometry of the forces in the neighborhood, water would tend to flow from all directions towards the point on the Earth's surface furthest from the moon, thereby raising the amount of water there. (I imagine this is how Laplace initially saw the matter.) If the radially-inward forces driving this water were stronger for the gravitational gradient effect than for rotation about the barycenter, due to the very different geometry of the two forces, then you might have been right originally.

Just asking. I don't know the answers but they seem nontrivial to work out. Maybe it's in Munk or somewhere. However the absence of the word "barycenter" from the article Tides is troublesome, particular given the accuracy of the theory's experimental confirmation. Moreover continents get in the way of such inwardly-directed flows, making the problem even harder! Vaughan Pratt (talk) 01:25, 19 November 2013 (UTC)

The Tide article clearly explains that both bulges are caused by a differential gravitational field, and includes a link to "Tides and centrifugal force: Why the centrifugal force does not explain the tide's opposite lobe (with nice animations)".192.249.47.204 (talk) 16:48, 30 June 2015 (UTC)

 Unless I am missing something, the sizes of the effects seem relatively straightforward if we expand algebraically; normalising both radius and force, we have:  
 net force towards larger object = 1/(1+deltaR)^2 - (1+deltaR)  = -3deltaR + 3.deltaR^2/2 + H.O.T.  
 In so far as the idea of separating the two effects is meaningful, we can see that, at least along the axis, each effect indvidually causes almost exactly the same amount of bulge on each side, and the variation due to (inverse square) gravity is twice the variation due to the centripetal effect.PhysicistQuery (talk) 23:43, 30 November 2016 (UTC)

Tidal Acceleration as a cause for Leap Seconds

The article states: "As a result of this process, the mean solar day, which is nominally 86400 seconds long, is actually getting longer when measured in SI seconds with stable atomic clocks. (The SI second, when adopted, was already a little shorter than the current value of the second of mean solar time.[9]) The small difference accumulates every day, which leads to an increasing difference between our clock time (Universal Time) on the one hand, and Atomic Time and Ephemeris Time on the other hand: see ΔT. This makes it necessary to insert a leap second at occasional, irregular intervals." Later, the article state that the change in the length of the day due to this process is approximately 2 ms per 100 years. This would require one leap second approximately every 50000 years. So, while the bold statement is not actually false, it's quite misleading. The Leap Second article explains that leap seconds are necessary due to "irregularities in the Earth's rate of rotation" and reiterates the minor contribution due to tidal acceleration on a scale of about one every 50000 years. In fact, of the 26 or so leap seconds implemented so far, none have been due to tidal acceleration, and won't be for many centuries.192.249.47.204 (talk) 16:21, 30 June 2015 (UTC)

The calculation of 192.249.47.204 is not correct. Supposing that the mean solar day was actually 86400 SI seconds long in 1885 (read from graph in McCarthy & Seidelmann, TIME: From Earth Rotation to Atomic Physics, Wiley-VCH, 2009, p. 54) and the day gets shorter by 1.7 ms per 100 years (same page) then in 2015 the day would be shorter by 1.7 × 1.3 centuries = 2.21 ms, giving a day length of 86399.9979 seconds. Multiplying by 365 days per year gives 31,535,999.19 SI seconds in a 365 day calendar year. This compares to 365 × 86400 = 31,536,000 mean solar seconds in a 365 day calendar year, if no leap seconds were inserted. The year length measured in SI seconds is about 0.8 seconds less than measured in mean solar seconds. The leap second to be inserted today will delay the beginning of July by one second, thus keeping the dates determined by SI seconds in rough synchronization with the dates determined by mean solar seconds. Jc3s5h (talk) 18:03, 30 June 2015 (UTC)

Thank you. I was missing that the 2 ms is of course applied to every day; 2 ms * 365 is approximately 3/4 of a second over the course of one year.192.249.47.204 (talk) 12:24, 1 July 2015 (UTC)

The statement "This lead to the introduction of the leap second in 1972." is not misleading it's just plain wrong. In fact the citation (http://www.timeanddate.com/time/leapseconds.html) is a furphy as it does not even mention tidal acceleration or the moon. The statement should be removed or replaced with something like:

Although leap seconds are used to adjust for changes in the Earth's rotation one would only be required every about every 50,000 years due to tidal acceleration. All leap seconds introduced so far have been due to other effects on the Earth's rotation. — Preceding unsigned comment added by 110.174.138.88 (talk) 04:42, 9 July 2022 (UTC)

in the section: Quantitative description of the Earth–Moon case

I would like to see references for the values given herein.

In particular, are the units correct on:

ΔT = +31 s/century^2  ??

It appears that this should be on the order of ms/century^2 ??? — Preceding unsigned comment added by DRG9143 (talk • contribs) 17:15, 2 November 2016 (UTC)

Figure of 1.7 ms per century - should it be 1.8 ms per century?

It's a broken link and this is an article from 2016 says 1.8 ms per century: [5] Robert Walker (talk) 07:57, 1 December 2017 (UTC)

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Retardation and length of day

Under “Retardation of the planet’s rotation”:

“For the earth-moon system, taking sin(2α){\displaystyle \sin(2\alpha )} of 1/13 and k=0.2, we get 4.5x10−22sec−2. For a 24-hour day, this is equivalent to 17 seconds in 1 million years, or 1 hour (i.e. lengthening the day by 1 hour) in 210 million years.”

The connection between the change in momentum and the relationship to the change in the length of the day isn’t very clear. — Preceding unsigned comment added by 49.195.75.72 (talk) 01:32, 12 December 2019 (UTC)

Errors in the article

In the section "Effect of the Sun", both formulae are in error. In the first eqn, the numerical factor must be not 9/2 but 3, see eqn (148) in In the second eqn, the coefficient must be not 27/4, but 15/4. See the first term in the third line of eqn (116) in or of eqn (120) in, and also keep in mind that the moment of inertia of a homogeneous sphere is (2/5)MR^2. — Preceding unsigned comment added by 71.163.138.195 (talk) 20:11, 26 June 2022 (UTC)

Thank you. The section in question was part of a "Theory" section, removed as original research. ★NealMcB★ (talk) 03:24, 19 September 2022 (UTC)

References