5
$\begingroup$

The tides raised by the moon on the Earth's oceans are adequately and universally explained by:

  1. The shape of the equipotential surface whose peaks are at the sublunar point and its antipode.

  2. The sloshing effect of a real ocean, with viscosity, inertia, coastlines and an irregular bottom, as the points mentioned in 1 move across the surface of the Earth and the water tries to keep up.

…although Galileo, of course, described the whole idea of the moon influencing tides as lunacy.

My question is purely about the shape of the equipotential surface in 1 – or equivalently, and more graphically, about the difference in height between high tide and low tide if the Earth were entirely covered by an inertialess ocean with zero viscosity.

Wikipedia says "The theoretical amplitude of oceanic tides caused by the moon is about $54~\mathrm{cm}$" but gives no reference for this figure. Other sources (including this Stack Exchange site) talk about "$60~\mathrm{cm}$" or "just over half a metre" or even "a few feet", also without giving a reference - and without even clearly specifying whether "$60~\mathrm{cm}$" means "from $-60\,\mathrm{cm}$ to $+60\,\mathrm{cm}$" or "from $-30\,\mathrm{cm}$ to $+30\,\mathrm{cm}$".

What I am looking for, therefore, is a source which can be considered authoritative and is not just a referenceless repeating of "what everyone knows".

Improve this question
$\endgroup$
4
  • $\begingroup$ (The edit to "house style" has resulted in a misquotation from Wikipedia, which uses the word "centimetres" and not the abbreviation "cm": not important to the substance of the question, but important if someone tries to search the Wikipedia article for "54 cm" and finds nothing). $\endgroup$
    – Martin Kochanski
    Jul 20, 2016 at 11:17
  • 1
    $\begingroup$ There's no "house style", at least not with regard to this sort of thing; it's just one editor's judgment. You could revert that edit, or that part of the edit, but I wouldn't worry about it too much. I would expect that someone checking that quotation would have the sense to try searching for "theoretical amplitude" or "caused by the moon" or some such thing. $\endgroup$
    – David Z
    Jul 20, 2016 at 11:24
  • $\begingroup$ re Galileo : since lunacy originally meant intermittent insanity due to the changes of the moon (=luna), he wasn't far wrong. $\endgroup$
    – sammy gerbil
    Jul 20, 2016 at 21:59
  • $\begingroup$ Anyone have a reference for the quote from Mr. Galilei? $\endgroup$
    – cms
    Apr 22, 2018 at 4:43

4 Answers 4

Reset to default
3
$\begingroup$

The value of 0.54 m is derived in the supplement resources "Tidal Distortions" to Hale Bradt, Astrophysics Processes — The Physics of Astronomical Phenomena, Cambridge University Press, 2008. The computation of 0.54 m is done just after equation (20) on p. 13.

Improve this answer
$\endgroup$
2
+50
$\begingroup$

My question is purely about the shape of the equipotential surface in 1 – or equivalently, and more graphically, about the difference in height between high tide and low tide if the Earth were entirely covered by an inertialess ocean with zero viscosity.

Wikipedia says "The theoretical amplitude of oceanic tides caused by the moon is about 54 cm" but gives no reference for this figure. ...

What I am looking for, therefore, is a source which can be considered authoritative and is not just a referenceless repeating of "what everyone knows".

1. The University of Hawaii has a website: "Exploring Our Fluid Earth" which they describe as:

"Exploring Our Fluid Earth is based on the nationally recognized Fluid Earth/Living Ocean (FELO) aquatic science curriculum (Klemm et al., 1990; Klemm et al., 1995). The Exploring Our Fluid Earth curriculum is grounded in the inquiry approach to learning and examines marine and freshwater systems of the earth by studying the influence of water on the planet.".

They explain in simple terms that the exact answer depends upon:

  • The location of the Moon, it's elliptical orbit and the lunar declination

  • The location of the Earth with respect to the Sun, it's elliptical orbit and the local geographic features:

"Factors that influence tidal range occur not only on a solar system scale, but also on local scales. Tidal ranges vary considerably at different points of a coastline due to seafloor features. When oceanic tidal bulges hit wide, shallow continental shelves the height of the tide is usually magnified. Conversely, mid-oceanic islands that rise steeply from the seafloor and do not have continental shelves have smaller tidal ranges. Mid-oceanic islands often have very small tidal ranges of 1 meter or less.

In narrow mouthed basins that are connected to the ocean, the tides often rise higher than in wide bays and harbors. This is analogous to pouring an entire can of soda into a short wide glass and a tall narrow glass. The soda will rise higher in the narrow glass, because there is not as much area to spread out as in the wide glass. The same amount of seawater will rise higher in a narrow basin than in a wide-mouthed harbor.".

2. The Public Encyclopedia Services Home Page has a webpage titled: "Ocean Tides - The Physics and Logic" which explains in detail (while certainly leaving some things out) the math behind the calculations.

3. The National Oceanic and Atmospheric Administration, U.S. Department of Commerce, has over a dozen webpages: starting with the Welcome page: "Tides and Water Levels", and while seemingly devoid of any math, the eleventh page, titled: "Tides Roadmap to Resources", lists numerous references.

4. PhysicalGeography.Net has a far too short explanation with great graphics, that depicts a greater range:

Mixed Tides

5. The Wikipedia webpage you referenced "Tides" is indeed a bit sparse on math too, try this Wikipedia page: "Tidal Acceleration".

Improve this answer
$\endgroup$
1
  • $\begingroup$ The Public Encyclopedia Services page have a strange logic. The far side is less attracted than the near side and so it moves away from the moon. In other words, they are stating that: "Less attracted"="repulsion". The centrifugal force explanation in Hale Bradt 's book: Astrophysics Processes, makes more sense. $\endgroup$
    – Shaktyai
    Sep 18, 2022 at 11:08
-2
$\begingroup$

For a wave, amplitude is crest to mean or trough to mean and range is crest to trough. Amplitude = 1/2 range, so the answer to your question is +60 to -60 or +54 to -54. Not 30 to -30 or 27 to -27.

enter image description here Source.

I imagine you read the Wikipedia explanation on how much tides vary. That 54cm would require no sun, but still liquid oceans and the Moon to have a circular orbit around the Earth with uniform ocean depth and no land. It's a calculated number not a practical one, but it is 54cm above and below normal giving a 108 cm range.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I can't see an actual figure for tidal range in the source you quote. But it is a long and complex page and perhaps I have missed it. $\endgroup$
    – Martin Kochanski
    Jul 20, 2016 at 15:48
  • $\begingroup$ @MartinKochanski, sorry, I didn't feel confident running the math on a figure. I just answered the first half of the question on the amplitude being half the range. $\endgroup$
    – userLTK
    Jul 20, 2016 at 16:52
-5
$\begingroup$

Imagine a U tube full of water where the Moon's gravity pulls upward on one arm, but not the other. Water would rise in that arm.

magine

Water will not flow from one arm to the other if the pressure is the same at the bottom of each arm.

$\rho(g_{Earth} -g_{Moon})(h+\Delta h) = \rho g_{Earth} h$

From this you can show $\Delta h = h \frac{g_{Moon}}{g_{Earth} -g_{Moon}}$

You can look up $g_{Earth} = 9.81 m/s^2$

You can calculate $g_{Moon} = G\frac{m_{Moon}}{r^2}$ where r is the distance from the Moon to Earth.

h is the depth of the ocean.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Depth of the ocean does not matter. Nor of course $g$ at the surface of the moon. $\endgroup$
    – user137289
    Apr 15, 2018 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.