Talk:Tidal acceleration

message: :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)
replies: text: 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)
thread_title:  Moment of inertia 

message: 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)
replies: text: 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)
thread_title:  Angular momentum 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Proposed merge with Tidal Friction 

message: 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)
replies: text: 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)

text: 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)
replies: text: 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)

text: 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)
replies: text: 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)
thread_title:  Amplitude 

message: 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)
replies: text: 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)
replies: text: "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)
replies: text: 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)
thread_title:  "cy" is not a standard unit 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Effects of sea level rise due to global warming? 

message: 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)
replies: text: 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)

text: Well found! Deuar 07:54, 23 October 2007 (UTC)
thread_title:  Advance of tidal bulge 

message: 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)
replies: text: I've changed "boil away" to "vaporize". — Joe Kress (talk) 05:09, 15 March 2009 (UTC)
thread_title:  Boiling AWAY? 

message: 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)
thread_title:  First 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Tidal braking 

message: 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)
replies: text: 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)

text: 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 , 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)

text: 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)

text: 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)

text: 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)

text: 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)
replies: text:  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)

text:  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)
replies: text: 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)
replies: text: 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)
thread_title:  kinetic & potential energy: dubious tag discussed  

message: On 3 June 2008,  added the phrase "Tidal acceleration is also moving the Earth outward from the Sun" with . 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)
replies: text: 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)
thread_title:  Acceleration of Earth due to tides on Sun 

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

message: 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)
thread_title:  in the section: Quantitative description of the Earth–Moon case 

message: It's a broken link and this is an article from 2016 says 1.8 ms per century:  Robert Walker (talk) 07:57, 1 December 2017 (UTC)
thread_title:  Figure of 1.7 ms per century - should it be 1.8 ms per century? 

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thread_title:  External links modified (January 2018) 

message: 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)
thread_title:  Retardation and length of day 

message: In the section "Effect of the Sun",https://en.wikipedia.org/wiki/Tidal_acceleration#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 https://ui.adsabs.harvard.edu/abs/2019CeMDA.131...30B/abstract 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 https://link.springer.com/article/10.1007/s10569-011-9397-4 or of eqn (120) in,https://arxiv.org/abs/1105.6086 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)
replies: text: Thank you. The section in question was part of a "Theory" section, removed as original research. ★NealMcB★ (talk) 03:24, 19 September 2022 (UTC)
thread_title:  Errors in the article