Dot Physics

Tides. Why are they so hard?

Posted by Rhett Allain on August 3, 2009
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Every introductory astronomy text and most intro physics texts talk about tides. The usual explanation is something along the lines of:

  • The moon exerts a gravitational force on the Earth and all the stuff on the Earth.
  • This force decreases with distance (1/r2).
  • Thus the moon pulls greater on one side of the Earth than the other
  • This doesn’t matter except for oceans which can move.
  • BOOM. Two tides a day due to a bulge on the side close to the moon and the opposite side.
  • Oh, the Earth is slowing down.

Really, that is what almost all intro texts say. Go check for yourselves.

Yes, the tides are related to the moon and gravity (and the Sun also). However, this is not a really full explanation. I know what the textbooks would say:

“Well, of course the tides are more complicated than that. But this is an introductory level course and the full explanation is WAY too complicated.”

My answer is: then what is the point of putting that topic in intro texts? This is the same problem I have with wrong stuff in shows like Fetch with Ruff Ruffman. Why put wrong stuff in there? I just don’t get it. There are many other things that you could talk about that the audience could understand.

So what is the deal with the tides? I think I can explain the tides at a simple level. The tides are caused by fake forces. Yes, fake forces. You see, the Earth is moving even though we pretend like it isn’t. Not only is the Earth moving, it is accelerating.

  • The surface of the Earth is accelerating as it moves in a circular due to its daily rotation.
  • The Earth is accelerating as it moves in a circle (mostly) around the Sun
  • And finally, the Earth is accelerating as it orbits the moon.

I know you thought that the moon orbited the Earth, not the other way around. Well, actually they orbit each other. Both have mass so both exert a gravitational force on the other and both change momentum. The center of mass of the Earth-moon system is inside the radius of the Earth, but it is not in the center of the Earth. Here is a shot from a vpython program of the Earth-moon system (for the rest of this post, I am going to pretend like it is just the Earth and moon and the Earth):

i-26842aa2948fa2281a5710b06653ed87-vpython.jpg

The blue circle represents the Earth. The red dot is the center of mass of the Earth-moon system (the point about which the Earth orbits) and the blue line is the path of the Earth’s center as it orbits the moon. You can’t see the moon because I zoomed in. In the correct scale of the Earth-moon system, you can’t see much. Here is the whole thing.

i-deca1709b10d59ebd187af4db0f65f1b-vpython_1.jpg

Ok, let me finish my point and then I will show some more simulations. Suppose that all points of the Earth had the same external gravitational field and that all points had the same acceleration. In this case, no one on the Earth would really feel anything regarding the moon. It would be just like astronauts in orbit. If all parts are pulled the same, you don’t notice.

But…all points are not pulled the same and all points on the Earth do not accelerate the same. The Earth is somewhat rigid, so if I want to pretend like it is a point particle, then the center would be it. All other points (like the surface) would have to have some fake force to make up for the difference in accelerations. This fake force can be calculated as the difference in the gravitational force from the moon at the center and at the surface. Here is a plot of that:

i-19fb7f53c4cfa71914e309769f1d926c-vpython_3.jpg

Note: I scaled these forces so they would be “planet-sized”. So, you can see there would be two “bulges” of water due to this (the water can move). See – that explanation isn’t too difficult, right? Really, this is an effect anytime you have a rigid object accelerating where the acceleration should vary with position. What about the space station? It is rigid-ish and the gravitational force varies with position (so the acceleration would change). How about another simple calculation. What would the “tidal force” around the inside of the space station look like?

i-9935370c153775d37a75055af7bd97a6-untitled_1.jpg

I lied. These arrows are the “fake gravitational field”, not the fake gravitational force. Also, that was true for the Earth ones above. Anyway, these things are not very large. The have a magnitude on the order of 10-6N/kg (or m/s2 if you want to think about the acceleration). But you notice it is similar to the tides.

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Comments

  1. #1 Anonymous Coward
    August 3, 2009

    This is the same problem I have with wrong stuff in shows like Fetch with Ruff Ruffman. Why put wrong stuff in there? I just don’t get it.

    Sometimes things are oversimplified to the point of inaccuracy. Sometimes an accurate answer requires stuff that isn’t covered in the textbook (see below). In these cases, I agree that the subject should probably be omitted from the book.

    But sometimes people make honest mistakes; physics is tricky, after all. For example, I would point out that the tides on earth do not appear at all as drawn above in the blue-circle-with-yellow-lines figure. In fact, I’d venture that the OPPOSITE of that drawing is closer to the truth.

    If you hang out for a day at the beach and watch the moon and the tides (on the day of a new moon, so you don’t have to worry about solar vs lunar tides) you will quickly find that high tide occurs pretty close in time to moonrise and moonset, and low tide occurs pretty close in time to the moon being overhead.

    (If you’re a landlubber, you can check out the tides online at, say, http://www.protides.com/hawaii/1332/
    To simplify things, make sure to concentrate on a days with a new moon to avoid the complications of the solar and lunar tides fighting each other.)

    So why did your model lead to an answer that’s kinda the opposite of experimental observations on earth?

    Big hint: I think your picture would be pretty much correct if the earth wasn’t rotating.

  2. #2 Anonymous Coward
    August 3, 2009

    P.S. Great problem, by the way. A few years ago, I would have drawn the same picture you did, for the same reasons, and I’d wager that most physicists would do the same.

  3. #3 Peter J
    August 4, 2009

    As my fields include sea defence work, marine engineering, physics and energy, and I’ve been involved with the sea since my youth and am an international yachtsman on tidal waters, I suppose I may be a relative expert.

    Unfortunately, if you go into the dynamics in detail you’ll find them almost entirely consistent with classical theory. Saying a book should leave out a subject if it can only cover basics is a bit like saying nobody should mention cars unless they fully understand every engineering detail!

    Your combined centre of mass point ignores the sun, and motion. There are 5 such points in the solar system, the Lagrangian points, with zero gravity. We’ve just sent spacecraft to check out the main two, which lead and trail the earth’s orbit.

    The main point here is, just because you don’t understand how something works and can be consistant doesn’t mean to say there aren’t lots of people who do, and lots of other reference sources with the information in. Tides are basically simple but, as with quantum physics, very complex in the minor detail. The earth itself is also, by the way, affected by tidal forces. It’s just not so noticable.

    There are a few tiny issues, but they only relate to Relativity not being able to deal with angular momentum, and time dilation issues raised by jets in Galaxy M87. But when the ruling paradigms are corrected I promise you it won’t change the tides!

    Peter J

  4. #4 Uncle Al
    August 4, 2009

    The Equivalence Principle claims locality to banish tidal effects. Differentially loaded eight-cm diameter Eotvos balance rotors null to at least 1 part in 20 trillion difference/average. General Relativity flubbing angular momentum has three possibly observable anomalies:

    1) All common matter is fermionic. Isolated electrons and nucleons certainly, many atomic nuclei, cosmic atomic ortho-hydrogen, magnets. GR has no observed failures.

    2) Relativistic angular momentum. A millisecond pulsar (neutron star) equatorially spins around 11% lightspeed and is a supercon gigagauss magnet, too. PSR J1903+0327 and PSR J0737-3039A/B validate GR. Summed observation times are insufficient to date to decide relativistic spin-orbit coupling. GR has no observed failures.

    3) Geometric parity. Quantitative geometric parity divergence of a chiral atomic mass distribution is intimately coupled to its moments of inertia (J. Math. Phys. 40(9) 4587 (1999)). Nobody knows if local left and right shoes obey the Equivalence Principle. Crystallographic enantiomorphic space groups P3(1)21 and P3(2)21 are intensely opposite “shoes” – quartz. No resolved chiral astronomic bodies exist. Meat (chiral L-protein amino acids) and wood (chiral D-sugars) cancel an Earth-moon parity Nordtvedt effect. GR can detectably fail with a parity EP anomaly and not falsify any prior observation.

    Somebody should conduct a parity Eotvos experiment,
    http://www.mazepath.com/uncleal/erotor1.jpg
    This experiment has never been performed. More’s the pity.