Talk:Tidal force

Weyl Tensor
Why is the Weyl Tensor (or its simplification of differentiating Newton's force law equation) necessary? —Preceding unsigned comment added by 208.242.126.115 (talk • contribs) 19:29, 23 December 2002

Similar point: Some physicists, e.g., Richard Feynman in Lectures on Physics, vol 1, attribute the ocean bulge on the 'side' of the earth opposite the moon, to the centrifugal effect due to the rotation of the earth around the earth-moon center of mass. This seems to be a more cogent explanation than some mysterious 'cinching at the waist causes the the other parts to bulge' notion, especially where we are dealing with fluids rather than solids. Alternatively, a 'gravity warps the very fabric of space' argument may make some sense here. —Preceding unsigned comment added by 66.151.236.129 (talk • contribs) 16:17, 22 March 2004

Sun, Earth, and Moon
The accuracy of this section has been disputed, but no reason has been given. Could you or anyone else provide more details about the problem? Comfr (talk) 05:39, 23 September 2018 (UTC)
 * I made a statement that violated Newton's third law. I feel embarrassed.  Unless someone else notices another mistake, I will remove the tag in a few days.

Free fall
Anon added a subsection Free fall, which was first unsourced. After my revert anon restored the content  with some source, but most of the added content is not in that source, and it doesn't look very reliable to me, as no author is mentioned. Only the formula is there. I removed per wp:unsourced and wp:no original research. See also wp:CALC. - DVdm (talk) 07:03, 3 July 2019 (UTC)


 * That was me. What seems to be unsourced is just a practical example of application of the sourced formula


 * $$ F = G M \frac{ a^2 - b^2 }{ a^2 b^2 } $$


 * If we replace all the variables with the correct numbers, where


 * we obtain the following formula:


 * $$ F = 6.67408 \times 10^{-11} \times 1.5 \times 1.98847 \times 10^{30} \times \frac{ (7411270 + 5)^2 - (7411270 - 5)^2 }{ (7411270 + 5)^2 \times (7411270 - 5)^2 } $$


 * which returns the force of


 * $$ F = 9.780336560799131 $$


 * which is very close to 1 g.


 * Of course this is not a coincidence, since I had chosen the distance and the size of the spaceship exactly in order to have the value of 1 g.


 * I am pasting here the paragraph removed, which I would like to see back in the article. At least the formula deserves to be there, it's not possible that an article about tidal force does not have a formula that describes how two bodies in free fall are pushed away from each other. --82.6.195.35 (talk) 02:37, 5 July 2019 (UTC)

In free fall
Two negligible masses in free fall towards a massive third body experience a force pulling them apart from each other equal to:


 * $$ F = G M \frac{ a^2 - b^2 }{ a^2 b^2 } $$


 * where $$G$$ is the gravitational constant, $$M$$ is the mass of the massive body both objects are falling towards, and $$a$$ and $$b$$ represent the distances of the two bodies from the third massive body.

For instance, imagine an ordinary 1,5 solar mass neutron star, and imagine a spherical spaceship, 10 m in diameter, in stable orbit around it at a distance of 7 411 270 m. In such conditions two astronauts, one lying on the near side of the spaceship and the other one lying on its far side, will experience a g-force towards the walls equal to 1 g, despite the spaceship remains in stable orbit, uninfluenced by the possible difference of weigh between the two astronauts.

At very close distances to a massive body the tidal force can become so strong to stretch atoms apart (see spaghettification).

Discussion

 * Regarding "... just a practical example of application of the sourced formula", see wp:CALC. As for the formula itself, a proper textbook source would be needed, see wp:verifiability. Shouldn't be hard to find in the literature... see wp:BURDEN - DVdm (talk) 09:55, 5 July 2019 (UTC)
 * As for "see wp:CALC", the page clearly says "Routine calculations do not count as original research. [...] Basic arithmetic, such as adding numbers, converting units, or calculating a person's age are some examples of routine calculations." – so it seems we have the answer. Regarding "As for the formula itself, a proper textbook source would be needed", what is wrong with the University Physics textbook cited? --82.6.195.35 (talk) 10:39, 5 July 2019 (UTC)
 * Your addition is much more than just routine calculations: the variables in the equation have precise physical meanings and the equation can be used in certain circumstances. It could be that you are applying a physical formula where it cannot be applied. This is way beyond "routine calculations". For instance, free fall, spaceships and astronauts are nowhere mentioned in chapter 13-6 of the source. Your text and your footnote (!) would be a schoolbook example of wp:original research.
 * As for the source from openstax.org, I think it is reliable for the equation itself. At least the authors are listed there, whereas I couldn't find any in the original source at courses.lumenlearning.com. So I think we can take the formula on board, but not in a context of free fall. Try to stick as close as possible to what is actually in the source. - DVdm (talk) 11:09, 5 July 2019 (UTC)
 * Note; regarding openstax.org, see also Reliable sources/Noticeboard. - DVdm (talk) 11:45, 5 July 2019 (UTC)
 * Initially I didn't take that formula from a book, I calculated it myself. And since the tidal force is only the "difference in strength in gravitational field" (Wikipedia says, using words, not numbers), it means that
 * $$ F = \frac{ G M }{ r_a^2 } -  \frac{ G M  }{ r_b^2 } $$,
 * which gives the formula above:
 * $$ F = G M \frac{ r_a^2 - r_b^2 }{ r_a^2 r_b^2 } $$
 * Actually before calculating it I had come to Wikipedia looking for it, but since Wikipedia had only a narration of it and I needed an actual equation, I had to calculate it myself.
 * Then you asked for a source, I googled "Tidal force formula", and I got exactly the same equation I had calculated (I wasn't so wrong after all). The only difference is that I removed the falling masses from the equation, since I was looking for an acceleration, not a force – yes, we will have to rename the "F" into an "A" before putting it back (my bad). So I put the source among the references. But now it seems that the reference is not good enough…
 * What happened to that proverb that said that mathematics is not an opinion? Does even a subtraction need to be referenced? --82.6.195.35 (talk) 12:58, 5 July 2019 (UTC)
 * Don't get me wrong. I think that the reference is good enough, and the formula appears in the source—changing variables is no problem. I asked for a second opinion about the source at wp:RSN anyway. Can't do much harm, right? - DVdm (talk) 13:15, 5 July 2019 (UTC)
 * @DVdm Thanks, although I really think it is not needed for such a simple formula (it is obvious that the equation is correct – after renaming "F" to "A"). Btw, here is a more classical example of application of that formula. Instead of a spaceship the entire Earth is used, and instead of a neutron star there is the Moon (very typical situation indeed). The free fall situation (orbit) remains unchanged (Earth and Moon are in free fall). And that doesn't change the equation: on the near and far sides of a rigid body orbiting another body you experience a tidal force according to the equation above. The only big change is the fact that differently from the spaceship, Earth has its own gravity, many orders of magnitude stronger than the tidal force caused by the Moon. --82.6.195.35 (talk) 16:18, 5 July 2019 (UTC)
 * Blogs don't count as wp:RS . - DVdm (talk) 16:23, 5 July 2019 (UTC)
 * Can we try to put a bit of order to the discussion? What exactly do you think it would be needed to be referenced? 82.6.195.35 (talk) 21:27, 5 July 2019 (UTC)

Just chiming in here per the notcieboard. I agree with User:DVdm. In order to add a section about free fall, there should be a better source used than the online "OpenStax University Physics" source. Actually the section cited "13.6 Tidal Forces" does not have the formula arranged in the format User:82.6.195.35 put in the article. Obviously it requires algebraic derivation and some assumptions to get that format, but that type of mathematical work is clearly not WP:CALC. The policy on WP:CALC applies to simple straight forward things like basic arithmetic or converting numbers into different units, not doing mathematical derivations. Many readers will not know if the derivation is correct. A textbook or academic article showing the exact formula is needed.

The example of the astronauts looks like WP:OR and certainly is not needed if someone finds a reliable source with the formula derived already. Wikipedia is a public source with anonymous editors whose expertise in physics is unknown. However, we certainly can trust in the expertise of textbooks or handbooks because they usually the authors of such texts have verifiable expertise in physics and certainly know what they are doing.Ramos1990 (talk) 21:35, 5 July 2019 (UTC)

2nd bulge
Thanks for the great page.

It would be nice to explain in more detail why there is the second bulge on the opposite side of the Earth. OK, the equipotential mentioned under the picture is probably enough. But I like something like here:

https://oceanservice.noaa.gov/education/tutorial_tides/tides03_gravity.html

explaining inertia effect. — Preceding unsigned comment added by Sykkor (talk • contribs) 08:47, 1 September 2019 (UTC)


 * Yes I agree with Sykkor; as with many articles explaining tides, this Wikipedia article covers the changes in the moon's gravitational field across the earth in some detail, but fails to adequately explain the opposing bulge. This cannot be explained by differences in gravitational forces by themselves, since they are acting in the same direction. This would be like expecting water to flow uphill!


 * The opposing CENTRIFUGAL force to the moon's gravity, caused by the earth's orbit around the gravitational barycentre of the Earth-Moon system explains the opposing bulge. Without this centrifugal force the tides would only bulge on the moon side of our planet. Therefore, this effect is almost as important as gravity.


 * The article in the following link is the clearest explanation which I've come across which addresses all the key issues, so I think we should base at least part the article on this.


 * https://noc.ac.uk/files/documents/business/Double-Bulge-Explanation.pdf


 * A crucial detail explained here is that the Earth-moon system doesn't rotate around the barycentre like spinning a giant dumbbell, because in that scenario each side of a dumbbell would be rigidly fixed and themselves rotate. This would mean the centrifugal forces would differ throughout the earth and set up far stronger tides than could ever be caused by differences in the moon's gravity. To represent the correct situation, the Earth must be rotated around the barycentre WITHOUT rotating the Earth around it's own axis. This causes the centrifugal forces to remain constant throughout the earth. Since the centrifugal force only precisely opposes the moon's gravity at the centre of the earth, smaller residual forces remain at the edges, particularly where the moon is nearest or furthest than the centre. So all this causes the familiar bulge, where there is a net force pulling the water towards the moon at the side of the Earth facing the moon, and an identical net force pulling the water at the side of the earth facing away from the moon.


 * Because the centrifugal forces are constant, and don't set of a force differential by themselves like gravity, they are often ignored in tidal explanations, but they are essential in explaining the opposing bulge and magnitude of the tides.


 * Obviously, this concept applies to all two body systems in orbit around a common barycentre, including the earth-Sun system, although the barycentre is inside the Sun in this case.
 * --Andromedean (talk) 10:59, 15 July 2020 (UTC)


 * I've compiled a Google spreadsheet here based on mass, distance and radial velocity to illustrate the balance of gravitational and centrifugal forces near the centre of the earth and the imbalance further away. This only refers to the centreline of the Earth moon and Earth Sun system. See the two graphs at the bottom. Notice the gravitational and centrifugal forces are very much larger than the difference between them. Feel free to play with this, I have a copy.
 * https://docs.google.com/spreadsheets/d/1jIktTsgEa9ANfUJDWqkVK6G3bsc9vmR43rJu7HlKkMo/edit#gid=1205982223
 * Andromedean (talk) 10:33, 18 July 2020 (UTC)

Hello,

Considering I have been very disappointed with explanations of tidal forces that I have read elsewhere, I would like to make the Wikipedia article about the subject into the most understandable one. It is considered of great importance, but is still class C. I will not try to edit it myself, just list the questions I have, hoping to be positive and askingn for help. I'm not pushing the work onto others, I wish to avoid making errors at this stage. When I read the article I'm behaving as a "candide": although I do understand most of it, I'm trying to imagine what a novice would ask. I'll quote:

"Thus, the tidal force is also known as the differential force, as well as a secondary effect of the gravitational field." This sentence can probably be left out entirely, especially the terms "differential" and "sceondary" are confusing at this point in the text.

The next paragraph, starting "In celestial mechanics… and ending "exerted by the third body on the first." really threw me off.  It may be relevant, but not at this point.  I called up this article in order to understand the tides as caused on the Earth by the Moon.  Of course the Sun is there too,

The explanation section could do with a drawing of the two bodies (I volunteer to make one), whereas figure 3 is probably superfluous. (also: why is the numbering as it is, starting with figure 4? Again, I'm willing to edit, but at this stage I would like to avoid making errors)

The section "Formulation" gives me the most trouble: the whole phrase in parentheses, starting "(In other words, the comparison… " and ending with "… a geocentric reference frame.)"  is extremely confusing.

The next sentence "Tidal acceleration does not… " is perfect and very important.

But then we get "Consider now the acceleration… ". A drawing is required (I volunteer to make one).

The phrase "For simplicity… " clashes with the complexity of "The Maclaurin series… " a little later on.

The caption of figure 7 says "…once the field at the centre of the sphere is subtracted…" but there is no reason given for subtracting a uniform field. The reason is given in the later sentence "This term does not affect the observed acceleration of particles on the surface of m because with respect to M, m (and everything on its surface) is in free fall." but one needs to add that m should be thought of as a rigid body, all its particles move as one, under the acceleration experienced as in its centre. The particles at the surface experience the same acceleration only because they are pushed by their surrounding particles, and thus act as if they were in a uniform field. The forces between the particles of the rigid body m that maintain this uniformity need not be accounted for, but they do exist and deserve mention. It would also be less confusing to use M for the body considered and m for the body causing the tidal forces on M. Again, a drawing would greatly help and I volunteer to make one.

So, thanks to this article I have finally understood tidal forces, including the "second bulge", but we need to work on it. I have also made a spreadsheet which in simple ways calculates the vectors shown in figure 4 and makes a graph. If this can be useful, I'll donate it.

(as a bonus: the Moon's tides cause sufficient deformation of the Earth's crust that this was measurable in the performance of CERN's LEP accelerator (https://accelconf.web.cern.ch/p93/PDF/PAC1993_0044.PDF), as well as the LHC which is housed in the same underground tunnel; also the tides were measured inside the ATLAS experiment when a water level was used to level the 50m long machinery) — Preceding unsigned comment added by RobertCailliau (talk • contribs) 19:51, 3 January 2021 (UTC)

" Tidal action on bath tubs, swimming pools, lakes, and other small bodies of water is negligible"
This isn't true of the Great Lakes. Are there others it isn't true of? We should probably find a source saying which lakes it is and isn't true of. פֿינצטערניש (Fintsternish), she/her (talk) 23:01, 16 June 2023 (UTC)