Tidal Misconceptionsby Donald E. Simanek
The study of tides is the tomb of human curiosity.
Note: Authorities disagree on whether or when the words "sun", "earth" and "moon" should be caitalized. I have chosen not to capitalize them when they are preceded by the word "the".
Introduction.Since the time of Galileo and Newton physicists have been fascinated with the cause and mechanism of ocean tides. It is a complex problem, but one that students and others find perplexing and seek simple answers. Textbook and website authors try to oblige, but too often they invent simplistic explanations that are simply wrong.This document will show examples of misleading explanations from textbooks and websites, and then show the correct reasons for ocean tides. Our goal will be to show a correct account that will promote genuine understanding.
The problem.The word "tide" has two different meanings.
Confusion begins when a textbook discussion of tides fails to define the word "tide", apparently assuming that everyone knows its meaning. One of the few books that clearly defines "tide" at the outset is The Planetary System by Morrison and Owen [1966]: "A tide is a distortion in the shape of one body induced by the gravitational pull of another nearby object." This is definition (2) above. It clearly says that tides are the result of gravitation, without any mention of rotation of the earth. And, perhaps most important in any discussion of tides, we must distinguish between land tides and ocean tides, for they have different mechanisms, though both are a result of the same gravitational forces. Since most flawed textbook treatments focus only on ocean tides, we will save discussion of land tides for later. Shoreline tides are a periodic rise and fall of water level, with a period equal to the period of the moon in the sky, Many locales also have tides with half this period. This is clear evidence that the moon's position in the sky is responsible for the tides. So what is the mechanism for this synchronicity? Moonbeams? Not likely. Gravity is the cause. We will see that gravity causes ocean waters to lift about 1 meter on the side of earth facing the moon, and also on the opposide side of the earth. We call these "tidal bulges". These are the driving force for shoreline tides. Inquiring minds want to know why there are two bulges on opposite sides of the earth. And lazy minds perpetrate wrong answers found on web sites and even textbooks. When these suggest that the bulge opposite the moon is due to inertia, rotation or centrifugal force, don't believe them. Rotation of the earth does distort its shape, but this is not a tide. Rotation changes the stress on water and land due to acceleration of these materials as they move in a circular path. This is responsible for the so-called "equatorial bulge" due to the earth's axial rotation. This raises the equator some 7 kilometers above where it would be if the earth didn't rotate. This is not a "tidal" effect, for it isn't due to gravitational fields of an external mass, and it has no significant periodic variations synchronized with an external gravitational force. This oblate spheroidal shape is the reference baseline against which real tidal effects are measured.
Getting it wrong.
First, let's look at those textbook and web site treatments that generate misconceptions. Some of them, we strongly suspect, are the result of their author's misconceptions. |
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Terminology. Most places on earth experience two tides per day, called a semidiurnal tide cycle. One tide occurs when the moon is overhead. Another occurs when the moon is on the opposite side of the earth, which means the tide is on the opposite side of the earth from the moon. This is called the antipodal tide. It is the occurance of the antipodal tide that puzzles many people, who want an explanation. We note that there are a few places on earth that experience only one tide per day (a diurnal tide cycle), due to complications of shoreline topography and other factors. The gulf coast of Mexico is one example.
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Any student looking at this textbook illustration would conclude that the tidal bulge nearest the moon is entirely due to gravitation, while the bulge opposite the moon is due to "inertial effects". Sounds neat, and the diagram looks impressive, but it just doesn't stand up to analysis.
The diagram below compounds this error by breaking the diagram into three diagrams, and adding even more mistakes. The top figure shows a supposed single tide due to the moon's gravitational attraction. The second figure (below) shows a single tide "due to rotation of the earth" about a "balance point" that is the center of mass of the earth-moon system (the barycenter). What are those arrows shown in the figures? Context suggests that they are force vectors—centrifugal forces. But centrifugal force is a concept that is only applicable to solution of problems in rotating (non-inertial) coordinate systems. The accompanying text does not say whether the earth is assumed to be rotating with respect to the moon. It doesn't say whether the analysis is being done in a rotating coordinate system. In fact, such books don't really do any mathematical analysis, they just engage in verbal hand-waving.
We will see later that even when a rotating coordinate system is assumed for the purpose of analysis, the centrifugal forces have the same size and direction anywhere on or within the earth. So they cannot raise tides. The figure shows the arrows as clearly of different sizes, larger at points farthest from the barycenter. So what can they possibly mean?
Now it could be that the arrows are only meant to suggest the displacements of water. If so, the caption should have said so. This diagram has many elements that can lead to misinterpretation, and strongly suggests the author or artist also had such misconceptions.
At this point, we strongly urge you to read, or at least review, a document explaining centripetal and centrifugal force.
Why can't they be consistent?
Many textbook pictures show the moon abnormally close to the earth. Therefore the arrows representing the moon's gravitational forces on the earth are clearly non-parallel. But in the actual situation, drawn to scale, the moon is so far away relative to the size of the earth that those arrows in the diagram would be indistinguishable (to the eye) from parallel.
Misconceptions lead to false conclusions
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These pictures, and their accompanying discussions, would lead a student to think that tides are somehow dependent on the rotation of the earth-moon system, and that this rotation is the "cause" of the tides. We shall argue that the "tidal bulges", which are the focus of attention in many textbooks, are in fact not due to rotation, but are simply due to the combined gravitational fields of the earth and moon, and the fact that the gravitational field due to the moon has varying direction and strength over the volume of the earth.
These bulges are due to distortion the shape of the solid earth crust, and also distortion of the oceans, but these two distortions have different reasons. If the oceans covered the entire earth uniformly, this would almost be the end of the story. But there are land masses, and ocean basins in which the water is mostly confined as the earth rotates. This is where rotation does come into play in shoreline ocean tides, but not because of inertial effects, as textbooks would have you think. Variations in ocean level reflect from continental shelves, setting up standing waves that cause more complicated water level variations superimposed on the tidal bulges, and in many places, these are of greater amplitude than the tidal bulge variations.
Tidal bulges move around the earth in synchronism with the moon and sun. But do not think of these as vast oceans of water moving with respect to continents. It is only the variations in water level—the surface profile of water—that follows the positions of the moon and sun in the sky.
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The distortion of water and earth that we call a "tidal bulge" is the result of deformation of earth and water materials at different places on earth in response to the combined gravitational effects of moon and sun. It is not simply the size of the force of attraction of these bodies at a certain point on earth that determines this. It is the variation of force over the volumes of materials (water and earth) of which the earth is composed. Some books call this variation the differential force or tide-generating force (TGF) or simply tidal force.
Let's concentrate on the larger effect of the moon on the earth. To find how it distorts shapes of material bodies on earth we must do the calculus operation of finding the gradient of the moon's gravitational potential (a differentiation with respect to length) upon each part of the earth.
If this procedure is carried out for all places around the earth, a diagram of tidal forces can be constructed, which would look something like this:
Tidal forces due to a satellite moon. [From the Wikipedia] The relative sizes of forces are exaggerated, but the directions are correct. |
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This diagram shows only the stress forces at the surface, but stress forces are distributed throughout the entire volume of the earth. One can now easily visualize how these shape-distorting stresses produce tidal bulges at opposite sides of the earth. The deformation of the earth's crust reaches equilibrium when the internal elastic forces in the solid crust become exactly equal to the tidal forces. The deformation of the water reaches equilibrium when it moves to minimize its potential energy. Remember that water is nearly incompressible. It does not compress or stretch.
Tidal forces have radial (along the direction of earth radii) components and tractive (tangent to the earth's surface) components. The radial components stretch or compress solid materials in the direction of the tidal force. The tractive components stress solid materials laterally. But it is the tractive components that physically move ocean water to form the tidal bulges.
At about 54.7° from the earth-moon line, the vector difference in the forces happens to be parallel to the surface of the earth. There the tidal forces are directed tangentially. At this point there's no component of tidal force to increase or decrease radial compression stress, and the radius of the earth there is nearly the same as the radius of the unstressed earth.
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Fluids can flow when forces are applied to them. They strongly resist compression or expansion. Water is very nearly incompressible and is clearly not rigid. So the tidal bulges in water arise because some water has moved toward the bulges from elsewhere, that is, from other regions of the ocean. This should not be surprising, for we know that water moves from higher to lower pressure regions in all situations, moving toward a condition of equilibrium at lowest possible potential energy. For ocean water, tractive forces are the cause of the tidal bulges.
The tangential components of tidal force push liquid material toward the highest part of the tidal bulges. This necessarily depresses the ocean surface elsewhere outside of those bulges. Tidal forces do not change the density or volume of water, they just move it around.
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The tidal bulges in the ocean should not be thought of as due to "lifting" of water, or due to compression and decompression of water. They are the result of water moving toward the regions of the tidal bulge. But do not think of "moving" as something like converging ocean currents rushing into the bulge. A tidal bulge is maintained by small displacements of huge amounts of water, over a huge area.
Also, the tidal bulges in the ocean raise water by very small, seemingly insignificantly amounts, compared to the radius of the earth. But over the huge area of one of the oceans, the tidal bulges contain a huge amount of water. We have discussed these using the conceptual model of a stationary earth-moon system without continents, with a uniform depth ocean covering its entire surface. We do this to emphasize that these tidal bulges are not due to rotation, but simply to the variation of the moon's gravitational field over the volume of the earth.
When we add continents to this model, the ocean bulges reflect from shorelines, setting up currents, resonant motion and standing waves. Standing waves of a liquid in a shallow basin have regions of high amplitude variation (antinodes) and regions of zero amplitude variation (nodes). So it's not surprising that in oceans we see some places where the tidal variations are nearly zero. All of this ebb and flow of the water surface affects ocean currents as well. Yet it is all driven by the tidal forces due to the moon's changing position with respect to earth.
Coastal topography (sea-floor slope and mouths of rivers and bays) can intensify coastal water height fluctuations (with respect to the solid land). In fact, these effects are usually of greater size than the tidal bulges would be in a stationary earth-moon system—sometimes ten times higher than the tidal bulge. But most important is the fact that this whole complicated system, including the coastal tides, is driven by the tidal bulges discussed above, caused by the moon and sun. It is a tribute to the insight of Isaac Newton, who first cut through the superficial appearances and complications of this messy physical system to see the underlying regularities that drive it.
Even when we look at this more realistic model, including the Earth's rotation, it is the rotation of continents (and their coastal geometry) with respect to the tidal bulges that gives rise to the complicated water level variations over the seas and shorelines. It is not some mysterious effect of "centrifugal force" or "inertial effects" as some textbooks would mislead you to think.
We have ignored the stress due to the gradient of the earth's own potential field, because it is nearly the same strength anywhere on the surface of the earth. We have also ignored the equatorial bulge of the earth, for we are treating that as the baseline against which the tidal effects are compared.
If all you want is the reason there are two tidal bulges, you needn't read further. I've sketched out an even shorter treatment as a model for textbooks that have no need to go into messy details.
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Remember, when you see this diagram of tidal forces, that it shows not the gravitational forces themselves, but the differential force, often called the tide-generating force. Similar pictures are found in other textbooks, but one must be careful not to mix the several different interpretations of the picture. These include:
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In any of these interpretations, similar force summation is happening throughout the volume of the earth. Tidal forces stress and push the materials of the earth (earth and water), distorting the earth's shape slightly—into an ellipsoid. These diagrams are necessarily exaggerated, for if drawn to scale, the earth, even with tidal bulges, would be smoother than a well-made bowling ball. Quincey has a good discussion of this, with diagrams. We can see from this photograph of earth from space, that all of the distortions due to rotation, mountains and ocean trenches, and tides, are really very tiny relative to the size of the earth. Keep this photo in mind as you look at the drawings, which are necessarily greatly exaggerated.
Exercise: How closely does the earth compare with a bowling ball. For the necessary data about bowling balls, see Bowling ball specifications. Accordidng to this source, the diameter of a bowling ball 13 lb. or greater is 8.55 inch with a tolerance of 0.01 inch. That's a 0.12 % tolerance. The difference between earth's polar and equatorial diameters is 23 km, or 0.4 %. By bowling ball standards, the earth doesn't quite meet the required roundness tolerance (due to its equatorial bulge and polar flattening). But this departure from sphericity is still too small to be noticed in photographs. Mountains and ocean trenches are much smaller, and tides far smaller still.Some photographs of the earth from space are computer synthesized composites of many photographs taken from orbiting earth satellites near the earth. The photos that are the best direct evidence of earth's roundness are unmanipulated single photos taken from a great distance, as from the moon, taken with a well-corrected camera lens. Even these show an earth indistinguishable from a sphere.
The tidal forces that are the sole cause the tidal bulges. These are the difference between gravitational forces and their average. Gravitational forces due to earth and moon distort ocean water. No other forces contribute to the tidal bulges. Don't be seduced by false explanations that use the words "inertia" and "centrifugal". |
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An alternative treatment of this is called the equilibrium theory of the tides. It is carried out in a coordinate system rotating about the barycenter of the earth-moon system. In this non-inertial coordinate representation the solid earth and the moon are considered stationary (in equilibrium) with respect to each other.
In this model we can treat the earth-moon system as if it were an inertial system, but only at the expense of introducing the concept of centrifugal force, technically called a "fictitious" force to distinguish it from "real" forces that are due to physical interactions between material bodies. This is handy when the measurements of a problem are with respect to a rotating frame of reference and the desired results are measurements with respect to that same frame of reference. The rotating earth is such a convenient frame of reference. Typically one chooses a polar coordinate system fixed on the earth.
It turns out that when this is done, the centrifugal force on a mass anywhere on or within the earth is, at every instant, of constant size and direction. So it cannot raise tides, nor can it deform the shape of material objects. Only real forces can do that.
As a result, we often hear students who have been so misled ask, "Why doesn't the motion of the earth around its barycenter give rise to centrifugal forces that might cause tidal bulges or contribute to them?"
We are interested in the earth-moon system, and for the time being we temporarily ignore the motion of this system around the sun.
In an inertial reference frame, the monthly motion of the earth is such that each piece of earth moves in a circle. At any instant all of these circles have the same radii and all radii are parallel. [These circles do not have a common center, however.] Therefore at any instant the centripetal forces are the same size and direction on every piece of earth. This force field of parallel and equal forces has no spatial gradient, and cannot raise a tide.
This figure, from French, shows the geometry. The dotted arcs A, B, and C have the same size and same radius. At any instant all of their radii are parallel.
In a non-inertial rotating reference frame, in which the earth and moon are both stationary, the same conclusion is reached even if fictitious forces could raise tides. A more detailed account follows.
We ignore the effects of the earth's rotation about its own axis. The equatorial bulge it produces is the baseline against which tidal variations are referenced. We are now focusing on the effects due only to the earth-moon system. We are also, still, assuming an idealized earth covered entirely with an ocean of constant depth. Therefore coastlines, ocean depth variations, and resonance phenomena are not issues.The motion of the earth about the earth-moon center of mass (the barycenter) causes every point on or within the earth to move in an arc of the same radius. This is a geometric result that some books totally ignore, or fail to illustrate properly. Every point on or within the earth experiences a centripetal force of the same size and direction at any given time. A force of constant size and direction throughout a volume cannot give rise to tidal forces (as we explained above). The size of the net centrifugal force is the same as the force the moon exerts at the earth-moon center of mass (the barycenter), where these two forces are in equilibrium. [This barycenter is 3000 miles from the earth center—within the earth's volume, about 3/4 of the earth's radius from its center.]
Centrifugal forces do not raise tides. So the bottom line is that centrifugal forces on the earth due to the moon's gravitational attraction are not tide-raising forces at all. They cannot be invoked as an "explanation" for any tide, on either side of the earth or anywhere else. So why do we find them used in "explanations" of tides in elementary-level books? Could it be because these text's authors are often misled by their own pretty diagrams? Once they launch into the rotating coordinate mode and start talking about centrifugal forces, they seem to forget that the earth's own gravitational field is still present and acting on every portion of matter on earth. They also forget that the non-uniformity of moon's gravitational field over the volume of the earth is alone sufficient to account for both tidal bulges, bulges that would be essentially the same if the earth-moon system were not moving, and the earth and moon were not moving relative to each other.
Physicists call centrifugal forces "fictitious" forces, because they are only conceptual/mathematical aids for the analysis of rotating systems that we choose to analyze in a non-inertial coordinate system. [We didn't have to do it that way.] In such a system fictitious interpretations may arise, such as the notion that the tidal bulge opposite the moon is due entirely to "inertial" (read "fictitious") forces, and the implication that gravitation has nothing to do with that bulge. [See the many "bad examples" earlier in this document.]
It must also be understood that these textbook pictures are static diagrams, "snapshots" of a dynamic system. The daily rotation of the earth underneath these "tidal bulges" causes the bulges to move around the earth's surface each day. All of these deformations sit "on top" of the equatorial bulge that extends all the way around the earth, and is due to the earth's axial rotation.
But then there's the curious caption that says that inertia is "sometimes" called centrifugal force.
Then on the very next page we see this diagram (below) in which the author identifies one tide as being from gravitation, the other from inertia. In what universe? Is this inertia centrifugal force? Two tidal bulges are shown extending all the way to the earth's poles. They should only extend 45.7 degrees from their maximum height. Earth's north pole is labeled, tempting one to think the tidal bulges are always centered on the equator.
Unfortunately, like so many other books, this book fails to tell the student the origin of these centrifugal forces, and fails to emphasize that they are not "real" forces, but only a useful device to do problems in rotating coordinate systems.
Here the chickens come home to roost. Misunderstanding of centrifugal effects originates in many elementary-level physics textbooks. Nowhere does this book even suggest that rotating coordinate systems are being assumed.
It is too easy to blame these errors on the artist. Don't authors proofread the books which will carary their name?
This may seem counter-intuitive. The earth's rotation and the moon's revolution are both counter-clockwise as seen from above the N pole. The earth rotates faster than the moon revolves around the earth, so the earth drags the high tide bulge "ahead" of the moon. Therefore, as we move with respect to both tidal bulge and moon (and faster than both), the moon crosses our meridian nearly 12 minutes after we experience the highest tide.
Textbooks and websites usually show misleading diagrams, like this one, with the symmetry axis of the tidal bulges making an angle of 30° or more with the moon's position. In fact, the angle is only 2.9°, so the tides are early by about 24(2.9/360)60 = 11.6 minutes. We doubt that even the most avid surfer would consider this of great significance. [However, resonance effects and effects due to shorelines and water depth can cause wide variations in the arrival time of high and low shoreline tides at various places on earth. Even in mid-ocean, there are variations due to resonance.]
This has another important result.The moon's gravitational attraction exerts a retarding torque on those tidal bulges. This is in a direction to reduce the earth's angular momentum and gradually slow the earth's rotation. The bulges also exert an equal size and oppositely directed torque on the moon, gradually increasing its angular momentum. The angular momentum of the earth-moon system is conserved.
Push-Pull language.
Often textbooks say something like this:
The moon's pull on objects on the near side of the earth is greater than on the center of the earth. Its pull on objects at the far side of the earth is smaller still. This causes the near ocean to accelerate toward the moon most, the center of the earth less, and the far ocean still less. The result is that the earth elongates slightly along the earth-moon line.This conjures images of motion of parts of the earth moving continually toward the moon. But in the actual situation, this distance doesn't change appreciably during a lunar cycle.
This misleading "explanation" is often found in lower-level physics texts that try to use "colloquial" language to describe things too complex for such imprecise language. Some of these books even say, as if it were a definition: "A force is a push or a pull". To the student mind this implies motion. These textbooks do consider forces acting on non-moving objects, but the harm has already been done by the earlier statement that the student memorizes for exams.
This "differential pulling" language exists in textbooks in several forms. Sometimes the phrase "is pulled more" or even "falls toward the moon faster" is used. All begin with the assumption that earth and moon are in a state of continually falling toward each other, and that's a correct statement, though not likely to be clearly understood by students. But if this "falling" is continual, then the "pulling" refered to in the example above is continual also. Then these texts bring in acceleration, and say that the lunar side of the earth accelerates most, the opposite side least. So, the student reasonably infers that the acceleration difference is continual.
Now if two bodies move in the same direction, the one with greater velocity will move more and more ahead of the other one. It's gain is even greater if the lead one has greater acceleration. If this "explanatory" language were to be believed as applying to the earth, the earth would continually stretch until it is torn apart.
This explanation goes astray because it doesn't acknowledge (1) the earth's own gravitational field acting to preserve the earth's approximately "round" profile and (2) tensile forces in the body of the earth. Also, it uses "force" language, without adhering to the fundamental principle of doing force problems: You must account for and include all forces acting on the body in question.
And, we suspect, the authors of these explanations may themselves have been misled by a misunderstanding of rotation and centripetal and centrifugal forces.
Though this stretching model is not valid for the earth-moon and earth-sun systems, it is valid for the interaction of a massive body and a smaller body that has weak internal cohesive forces. The encounter of a comet passing near a planet is an example. Here the comet may be torn apart by the non-uniform gravitational force due to the planet.
The moon's gravitational force acts in two ways on the earth:
Since the earth's axial rotation affects only the "baseline" level of land and water, against which tidal variations are referenced, a discussion of tides does not need to mention centrifugal forces. That only invites confusion and misconceptions. Centrifugal forces are not tidal (tide-raising) forces. Even when analyzed in a rotating coordinate system the fictitious centrifugal forces of moon on earth are of constant in size and direction over the volume of the earth at any time, therefore they can not raise tides.
The folks who do tidal measurements don't get into the physics theory much. Tide tables are constructed from past measurements and computer modeling that does not usually take underlying theory into account. It is much like the pretty weather maps you see on TV, computer generated without any detailed use of physical laws. The task is just too complicated for even our largest computers, and the data fed into them is far from the quality and completeness we'd need.
You might think that with global positioning satellites we'd know the measurements of water and land tides accurate to a fraction of a smidgen. You'd be wrong. If you check the research papers of the folks who do this, you see that they are still dissatisfied with the reliability of such data even over small geographic regions. We can map the surface of land to less than a meter this way, and get relative height measurements equally well, but absolute height measurements relative to the center of the earth are much poorer. Many of the numbers you see tossed about in elementary level books are copied from other elementary level books, without independent checking and without inquiring whether they were guestimates from theory or from actual measurement.
You may also think that modern computers have made tide prediction more accurate. In fact, the analog (mechanical) computers devised for this purpose in the 19th century did nearly as good a job, even if they have ended up in science museums.
2. If the Earth were not rotating, and the Moon stopped revolving around it, and they were "falling" toward each other, would the Earth have tidal bulges? If not, why? If so, would they be significantly different from those we have now? In what way?
3. Here's an example of how untrustworthy textbooks are. This is from a 1939 college level introductory college physics text.
From this explanation (previously given) it would seem that the tides should be highest at a given location when the moon is directly overhead (or somewhat more than 12 hours later). In fact, high tide always occurs when the moon is near the horizon. The reason is that the friction of the rotating earth tends to hold the tides back so that they always occur several hours later than we should expect.Find the serious error(s) in this short paragraph.
4. A web site has this gem of wisdom: "As the earth and moon whirl around this common center-of-mass [the earth/moon barycenter], the centrifugal force produced is always directed away from the center of revolution." Is there anything wrong with this statement?
5. [From Arons, 1979] If our moon were replaced by two moons half the mass of our moon, orbiting in the same orbit, but 180° apart, would the earth still have tides? If not, why not? If so, how would they compare with the tides we now have?
6. If the tides may be thought of as a "stretching" of the earth along the axis joining the earth and moon, then why are all materials not stretched equally, resulting in no ocean tides? If elastic strain is the reason for tides, then since the elastic modulus of water is so much smaller than rock, wouldn't you expect that rock would "stretch" more than water, causing water levels to drop when the moon is overhead? Explain.
7. When we say that the tide in deep mid-ocean is about half a meter, what is this measured with respect to? (a) a spherical earth, (b) an oblate earth with equatorial bulge, (c) the bottom of the ocean, (d) the ocean's shores (e) low tide.
8. If the earth were in a rotating, uniform (parallel field lines, constant strength) external gravitational field (don't ask how we might achieve this), would we have tides at the period of earth's rotation? Would we have tides at the half-period of earth's rotation?
9. If a huge steel tank were filled with water, and a sensitive pressure gauge were put inside, would the pressure gauge register tidal fluctuations with a period of about 12.5 hours?
10. The picture and text below are from the NOAA-NOS website. Your tax dollars at work to propagate misconceptions.
Gravity and inertia are opposing forces acting on the earth's oceans, creating tidal bulges on opposite sides of the planet. On the "near" side of the earth (the side facing the moon), the gravitational force of the moon pulls the ocean's waters toward it, creating one bulge. On the far side of the earth, inertial forces dominate, creating a second bulge.
Identify the specific misconceptions in the picture and the text.
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11. This picture, commonly seen in elementary textbooks, shows the lunar gravitational force large on the side of earth nearest the moon, smaller at the earth center, and even smaller on the side opposite the moon. What's misleading about this?
12. A textbook says "Tides are caused by the moon pulling on the ocean waters more strongly on the side nearest to the moon." If this were so, one would assume the catastrophe illustrated in the cartoon below. Why doesn't this happen?
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13. If the moon were covered with an ocean, would it have tidal bulges?
2. The tidal bulges in this situation would be essentially the same size as those we have now in mid-ocean. Of course, they wouldn't move across the earth's surface, so the complications due to oceans sloshing around within their shorelines would be absent.
3. A 90° lag would put the moon near the horizon at high tide. The tidal bulge leads the moon by only 3°, so if this were so at shorelines, the tides would arrive early by about 24(3/360)60 = 12 minutes. However, coastal and resonance effects modify this greatly, and there are places where the tides are highest when the moon is at the horizon, but this is not typical. Blackwood uses the word "always", which is clearly inappropriate.
4. "The center of revolution" is ambiguous. It is not one point. Each point on earth revolves around its own center of revolution. Only the center of the earth revolves around the barycenter. And if you made a map of the centripetal forces everywhere on earth, they would all be parallel to the earth-moon line.
5. Arons' answer: "The tide-generating effects now have the same magnitude and the same symmetry as in the existing situation." This is only approximately true, and ignores some small differences due to divergence of the fields. It's useful to think of this using the superposition principle. A moon of half size produces half as much tidal force. Two such moons 180° apart restore the original situation, approximately. Where the present tides on opposite sides of the earth are slightly unequal, the tides due to two opposing half-size moons would be of equal size on opposite sides of the earth.
6. Water has a high elastic modulus. It flows easily, but rock does not. Water levels are affected by tractive forces (the tangential component of the tidal force), which directed toward the tidal bulges.
But it has been observed that the level of water in stone wells drops at the time of high tide. The stone well is stretched but the water in it is not, so the water level, relative to the well, drops.
So why doesn't the same happen in oceans? Ocean water tidal bulges are due to water relaxing tangential to the earth, not to stretching or lifting.
7. Textbooks don't tell you this, do they? The high tide level in shoreline water is usually measured from low tide or from the mean water level there. Coastal tide levels are measured with respect to solid land (not shifting sand) on the shore.
8. There would be no tidal bulges in a uniform field. A field gradient is required for a tidal bulge.
9. Yes. The elastic modulus of steel and water are different, so this would alter the water pressure as water and steel respond differently to tidal forces. Follow-up question: Would the water pressure inside be higher at high tide, or lower?
Answer: At the time of high tide the steel tank is stretched and its volume increases. So the water pressure inside is reduced. [See question 6.]
In an open container of solid material containing water we see another interesting effect. At high tide, the water level in the tank decreases because of the solid tank's increased volume.
10. The picture suggests that the near bulge is only due to gravitation, the other one only due to "inertial forces". The text speaks of "inertial forces", without saying that such a term has no meaning except in a non-inertial coordinate system. The phrase "pulls the ocean waters toward it" implies "motion toward it". The moon exerts gravitational forces on the far side bulge not much smaller than on the near side, and if these forces are "pulling" toward the moon on the near side, they are also pulling toward the moon on the far side. No mention is made of that.
11. The three arrows show gravitational forces due to the moon. No other forces are shown. This leaves the impression that these are the only forces responsible for the tides. But, as we have shown, earth tides are due to the combination of gravitational force due to the moon, gravitational force due to the earth, and tensile forces in the material body of the earth.
a. If the forces shown in the diagram were the only forces acting, then the points A, B, and C would have different accelerations (by Newton's F = ma), and the earth would soon be torn apart.
b. Does the picture represent how things are in an inertial frame? If so, then in view of the above observation, these can't be the only forces acting on the earth. So where are the other forces in the diagram, and what is their source?
c. Does this represent how things are in a non-inertial frame, perhaps rotating about the earth/moon barycenter? If so, then the centrifugal and Coriolis forces should be explicitly shown, for they must be included when doing problems in such a frame of reference.
Gravitational forces due to the moon, gravitational forces due to the earth, and tensile forces of materials are the only real forces acting on the material of the earth. These alone account for the tidal bulges. Rotation plays no role.
So that raises the question in the student mind: What accounts for the motion of the earth around the earth/moon barycenter. The answer is simple: the net force due to the moon on the body of the earth is solely responsible for that. (We are here ignoring the sun.) It must be so, for (aside from the sun) the moon's gravitational force is the only external force acting on the earth. As students learn in freshman physics, internal forces cannot affect the motion of the body as a whole, for they add to zero in action/reaction pairs. Therefore they need not be included in the equation of motion of the body itself.
I think what irks me about textbook treatments of tides is that they undo the good work we try to accomplish in introductory physics courses. We emphasize correct applications of Newton's laws of motions. First we tell the students to identify the body in question, the body to which we will apply Newton's law. We stress that they must identify the forces on the body in question and only the real forces, due to bodies external to the body in question. We ask students to draw a "free-body" vector diagram showing all these forces that act on the body in question. One must not include forces acting on other bodies. Then sum these forces, to apply F = ma. If the net force on the body is non-zero, then it must accelerate. This analysis, done in an inertial system, is adequate to understand the tidal forces, in fact that's the way Newton did it when he discussed tides.
12. This is, of course, a joke. However, as with so many absurd notions, this isn't easy to explain.
13. Yes, there would be tides on a lunar ocean. In fact, there are land tide bulges on the moon. These were "frozen in place" when the moon solidified.
As you notice, these questions were designed deliberately to expose misconceptions arising from misleading textbook and website treatments.
I also find the textbook I used when I took freshman physics, "College Physics" by John Eldridge, (Wiley, 3rd edition, 1947). In a brief paragraph "The Fictitious Centrifugal Force" and a footnote, he defines the two meanings of centrifugal force: (1) The real inward axial force that counters the centripetal force, and (2) a fictitious outward force in a rotating reference frame. Then he cautions, "Because of this ambiguity in meaning the beginninng student is advised not to use the term." In his brief treatment of tides he does avoid the term, but instead uses the meaningless argument that the moon "pulls" the water and earth away from each other, a completely fraudulent argument found in many elementary textbooks.
Listing a link here does not imply total endoresement of everything found there, nor of anything by the same author on other subjects. But that should go without saying.
[1] Here the terms "fictitious force" and "real force" are being used in the technical sense. Real forces are those that satisfy Newton's law F = ma when the acceleration is measured in an inertial reference frame. Fictitious forces are those we introduce as a mathematical and conceptual convenience when doing problems in a non-inertial reference frame. The centrifugal and Coriolis forces are fictitious forces in this context. We do not wish to get into murky philosophical waters with the question "What is 'real' really?" Nor are we using the words "real" and "fictitious" in the colloquial sense. See any undergraduate text in Classical Mechanics that discusses non-inertial reference frames. Or see fictitious force in the Wikipedia.
Uncredited pictures and quotations are from internet and textbook sources. We assumed their authors would rather remain anonymous. However, if anyone wants credit for them, we'll be happy to oblige.
Text © 2003 by Donald E. Simanek. Input and suggestions are welcome at the address shown to the right. When commenting on a specific document, please reference it by name or content.
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