Tidal Misconceptionsby Donald E. Simanek
Note: Authorities disagree on whether and when the words "sun", "earth" and "moon" should be caitalized. I have chosen not to capitalize them when they are preceded with the word "the".
The word "tide" has two different meanings.
Confusions begin 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 : "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 effects.
Another thing that causes distortion of the shape of the earth is its axial rotation. 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 23 kilometers (0.4%) 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 shape is the reference baseline against which real tidal effects are measured.
Common misleading textbook treatments of tides.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.
The subject of tides is complex, perhaps too complex to treat fully and satisfactorily in a freshman-level textbook. For this reason, many such books wisely ignore the subject entirely. Even some advanced undergraduate level mechanics texts dismiss the subject with a few sentences and the disclaimer "Consideration of the details would lead us too far astray." That's prudent.
Terminology. Most places on earth experience two tides per day, called a semidurinal 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.
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 vectorscentrifugal forces. 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 done in a rotating coordinate system.
We will see later that even when a rotating coordinate system is assumed for the purpose of analysis, the size of the centrifugal force is the same size anywhere on or within the earth. 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.
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 from parallel.
Misconceptions lead to false conclusions
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 gravitational field of the moon, and the fact that this field has varying direction and strength over the volume of the earth.
These bulges distort the shape of the solid earth, and also distort the oceans. 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, but not because of inertial effects, as textbooks would have you think. Without continents, the water in the ocean would lag behind the rotation of the earth, due to frictional effects. But with continents the water is forced to move with them. However, the frictional drag is still important. Water in ocean basins is forced to "slosh around", reflecting from continental shelves, setting up ocean currents and standing waves that cause water level variations to be superimposed on the tidal bulges, and in many places, these are of greater amplitude than the tidal bulge variations.
What's missing?Too often textbooks try to toss off the tides question with a superficial analysis that ignores some things that are absolutely essential for a proper understanding. These include:
So why are there tidal bulges on opposite sides of earth?For a while we will set aside the complications of the actual earth, with continents, and look a the simpler case of an initially nearly spherical earth entirely covered with water. If this rotates on its axis there's equatorial bulge of both earth and water, but we will treat this as a "baseline" shape upon which tidal bulges are superimposed. It's shape is produced by its own gravitation and rotation on its axis. The distortions of this baseline shape are called tidal effects and are entirely due to the gravitational forces of the moon and sun acting upon the earth.
The distortion of water and earth that we call "tidal bulges" 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 tidal force.
Let's concentrate on the larger effect of the moon on the earth. To find how it distorts any volume of the earth's material body we must do the calculus operation of finding the gradient of the moon's gravitational potential (a differentiation with respect to length) upon that volume.
Since we have taken the near-spherical earth as a baseline, and the tidal effects are superimposed on that, we can ignore the earth's own gravitational forces on itself, leaving only the forces due to the moon. They are the forces causing tidal effects.
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:
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.
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 force is directed horizontally. At this point there's no component of tidal force to produce radial compression stress, and the radius of the earth there is nearly the same as the radius of the unstressed earth. The tangential components of tidal force push material toward the highest part of the tidal bulges.
The above description is appropriate for solid elastic materials. But for liquids the situation gets more interesting. Fluids move when forces are applied. They strongly resist compression or expansion. Water is very nearly incompressible and is clearly not rigid. So the tidal bulges arise because water has moved into 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 a liquid body, the tractive forces dominate, but the end result is still two tidal "bulges" when equilibrium is achieved.
How does this apply to the real earth?In the real earth, we have a solid crust with thin layers of ocean bounded by continents. The solid earth tides are dominated by the compressive-expansive radial components of the tidal forces. The large oceans are dominated by the tractive tangential components of the tidal forces. The interior of the earth behaves, in this context, like a solid elastic body, for mass movement of even the plastic materials cannot occur quickly enough. In either case, at equilibrium, the gravitational forces on each portion of matter are balanced by internal tensile forces.
The tidal bulges are very small, seemingly insignificantly small, compared to the radius of the earth. But over the huge area of one of the oceans, the tidal bulges alone still raise a huge amount of water. We have discussed these using the conceptual model of a stationary earth-moon system without continents, but 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 motions 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 water volume 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 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. But most important is the fact that the whole complicated system, including the coastal tides, are 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, recognizing the importance of "rotation", it is the rotation of continents (and their coastal geometry) with respect to water that gives rise to the complicated water level variations over the seas. 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 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 wanted was 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.
A picture of tidal forces.
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:
In any of these interpretations, similar force summation is happening throughout the volume of the earth. Tidal forces stress the materials of the earth (earth and water), distorting the earth slightly into an ellipsoid. These diagrams are necessarily exaggerated, for if drawn to scale, the earth, even with tides would be a more perfect sphere than a well-made bowling ball (before the holes are drilled). Quincey has a good discussion of this, with diagrams. We can see from the diagram above how these combined forces distort the earth into an ellipsoid. But we can see from this photograph of earth from space, that all of the distortions due to rotation, and due to 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.
The equilibrium theory of the tides.Our simple analysis above also showed the importance of the relaxation of earth materials to achieve an equilibrium between gravitational forces and cohesive forces of materials. In more detailed analysis, we find that the figure (shape) of our idealized earth model at equilibrium is a constant shape consisting of two bulges nearly oriented in alignment with the moon. Underneath this equilibrium profile, the earth turns on its axis once a day, so the bulges move with respect to geography.
The fuller treatment of all of this is called the equilibrium theory of the tides. It is usually carried out in a coordinate system, rotating about the barycenter of the earth-moon system. In this coordinate representation, the earth and the moon are considered stationary with respect to each other.
In this representation we can treat this system as if it were an inertial system, but only at the expense of introducing the concept of centrifugal force. It turns out that when this is done, the centrifugal force on a mass anywhere on or within the earth is of constant size and direction, and is therefore equal to the size of the gravitational force the moon exerts on the same amount of mass at the center of the earth. We'll look at this in more detail below.
A closer look at centrifugal forces.So what about those centrifugal forces so many books make such a fuss about? You'll notice we never mentioned them in our simple explanation. Should we have?
Many misleading accounts of the tides result from a common confusion about centrifugal effects due to rotation. Let's be very clear about this. The only real forces that act on the body of the earth are:
Polar coordinates are most convenient when doing problems such as this. The term "centripetal force" is nothing more than a name for the radial component of the net real force on a body. It is not some new kind of mysterious force.
Every part of a rotating body is accelerating, so it is not in equilibrium, and there's a net nonzero force on it. Each part of a rotating body is moving in a curved path, and therefore has a component of the net force acting on it. This component (called the centripetal force) is directed toward the instantaneous center of rotation. The sum of all real forces on each part of a rotating body is not zero.
When using the rotating non-inertial coordinate representation, it is customary to introduce "fictitious" forces called "centrifugal" forces. This dodge has the handy result that when you add the fictitious forces to the real forces Fnet = ma holds, where Fnet is the sum of real and fictitious forces. This turns a problem about a non-inertial system into one that can be analyzed as if it were an inertial system. 
Textbooks that introduce "centrifugal" language, but do not do any mathematical derivations, and do not explain or use rotating coordinate representations. are very likely to mislead the student into thinking that these "centrifugal" forces are actual real forces, arising from some mysterious causes. These books may even equate these forces to "inertia", which doesn't help anyone understand anything. The very worst offenders even describe centripetal forces and centrifugal forces as "reaction" force pairs, as in Newton's third law. This makes no sense at all, for they are acting on the same body. The action/reaction forces described by Newton's third law act on different bodies, by definition.
This figure, from French, explains the importance of centrifugal forces, which turns out to be of no importance at all unless you choose to do an analysis of the problem in a rotating coordinate system. As we said above, you don't have to.
We ignore the effects of the earth's rotation about its own axis, which of course underlies everything. 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. The motion of the earth about the earth-moon center of mass, causes every point on or within the earth to move in an arc of the same radius. This is a geometric result some books totally ignore, or fail to illustrate properly. Therefore every point on or within the earth experiences the same size centrifugal force. A force of constant size and direction throughout a volume cannot give rise to tidal forces (as we explained above). The size of the 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.]
So the bottom line is that centrifugal forces on the earth due to the presence of the moon 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.
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 each day. And all of these deformations sit "on top" of the equatorial bulge that goes all the way around the earth, due to the earth's daily rotation.
Tidal variationsThe oceans don't cover the entire earth, but "slosh around" daily within the confines of their shores. Timing of the ocean tidal bulges even at mid-ocean can depart considerably from the idealized tides we have described. Reflections from shores can set up interference patterns farther out in the ocean. Coastal tides have considrable local variations due to difference of shoreline slope, and ocean currents. But the driving force for all of these complications is still those two "daily" lunar tides (12 hours 25 minutes apart), which we have explained above, combined with the two much smaller daily tides (12 hours apart) due to the gravitational field of the sun.
More misleading textbook illustrations.An oceanography textbook has this diagram that at least shows centrifugal forces of equal size.
One is tempted to think "This book has it right!" But reading the text makes one suspicious. Then on the very next page we see this diagram in which the author identifies one tide as being from gravitation, the other from inertia.
Comparing the two pictures, one sees that they contradict. The one that shows forces clearly suggests that the moon's gravitational force is responsible for both tides.
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, for misunderstanding of centrifugal effects originates in some elementary-level physics textbooks. Nowhere does this book even suggest that rotating coordinate systems are being assumed.
Other lunar misconceptions.
This has another important effect. The moon exerts a retarding torque on those tidal bulges. This is in a direction to gradually slow the earth's rotation. And the bulges exert an opposite torque on the moon, increasing its distance from earth, and therefore reducing its velocity, as required by the law of conservation of angular momentum.
This exaggerated diagram, from a web site, shows an angle of about 45° between the bulges and the moon's position. But it does show correct rotation relationships. Such diagrams are often seen in textbooks. Of course, the relative sizes of earth and moon and their distance of separation are not to scale either. An accurate scale diagram wouldn't show any of these relatively small-scale phenomena. So exaggeration must be used to get the idea across. This wouldn't be so bad if the accompanying text clearly indicated that the distance and the angle have been exaggerated, but somehow that disclaimer is often forgotten. We have even seen some texts that don't anywhere tell us that the angle is only 3°.
Often textbooks say something like this:
The moon pulls the ocean on the near side of the earth more than it pulls on the center of the earth. The pull on the ocean 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, the earth and moon remain a nearly constant distance apart; and 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. Oh, the 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" refereed to in the example above is continual also. Now they bring in acceleration, and say that the lunar side of the earth accelerates most, the opposite side least. So, the student reasonably infers, 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 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.
Some dirty little secrets textbooks fail to tell you.The "tidal trivia" summary below puts things into perspective. The so-called equatorial bulge due to the earth's axial rotation lifts the equator about 23 kilometer. The moon's gravity gradient lifts water mid-ocean (where the ocean is deep) no more than 1 meter, that's only 1.6 × 10-2% of the earth radius. Why do we fuss about this? Because over an ocean of large area, that represents a very large volume of water. Also, it's the driving mechanism that controls the periods of the much larger tides at shorelines.
The moon's gravitational forces act 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. In a rotating coordinate system the centrifugal force of moon on earth is constant in size and direction over the volume of the earth, therefore it has no tide associated with it.
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 take underlying theory much into account. It is much like the pretty weather maps you see on TV, computer generated without a detailed use of all the physical details. The task is just too complicated for even our best 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 within 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.
SuperpositionAstrophysicists also need to understand the physics of tides. They must deal with tides in a more general way, such as tidal forces acting on binary stars, and on rings of Saturn. Consider this clear description from Josh Colwell's web site, intended for his students.
If the tidal force can tear apart a strengthless fluid object (as in the derivation of the classical Roche limit), then it still applies some stretching force to solid, intact bodies, such as moons. First consider the static situation where neither the moon nor the planet are moving or rotating: the tidal force stretches each object resulting in bulges along the line connecting the two objects. These are called tidal bulges.
This clear description uses a valuable conceptual approach to understanding tides—the principle of superposition. This principle is useful when several processes act together and the results of them are linearly additive. It allows us to consider each process separately and then combine the results. We first consider a separated earth and moon with no motion at all. They would have to be fastened in place somehow, but that's an unimportant detail. Such a static situation will cause tidal bulges on both bodies. This clearly tells us that such bulges are not due to motion, but are entirely due to gravitation.
The second paragraph looks at what happens to the moon when you add the rotation of the moon. But the same arguments can be extended, to see what happens to the earth as we add the rotation of the earth. The earth rotates much faster than the moon revolves, so the earth's tidal bulges "track" the moon, lying close to, but slightly ahead of, the line joining the earth and moon. They are "dragged" ahead by friction. The tidal bulges move across the earth's geography, and friction forces dissipate energy in the earth, slowing its rotation.
Then we can look at the gravitational torques acting on the tidal bulges. These act to slow the rotation of the earth also. But they also act to increase the moon's distance from earth and decrease its velocity. This is required by conservation of the total angular momentum of the earth-moon system. If the axial spin rotational momentum of the earth decreases (due to energy dissipation), then the orbital angular momentum of the moon must increase to compensate. Astronomy texts may be consulted for the very interesting long-term outcomes of these interactions.
Web sites with reliable information.
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.
Final Exam.1. These pictures are from various internet sources. Find the misleading features of each.
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 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. Textbooks sometimes say the tide on the side of earth opposite the moon is smaller than the tide on the side nearest the moon, because the moon's gravitational pull is weaker there, farthest from the moon. The moon is about 60 earth radii away from the earth. How much smaller is the weaker high tide compared to the stronger one?
11. 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.
12. This picture is commonly seen in elementary textbooks. It 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?
13. 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?
Exam answers.1. The first picture shows the actual tides being the sum of two tidal bulges, implying that those bulges have independent origins. We have shown this is not so. The second picture speaks of "rotational force", which may mean centrifugal force, but we can't be sure. We also have no clue whether "gravity" means the moon's gravitational attraction, the earth's gravity, or both together.
2. The tidal bulges in this static 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 lags the moon by only 3°, so if this were so at shorelines, the tides would arrive late 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.
5. Arons: "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. Materials differ in elastic modulus. Water levels are affected by tractive forces (the tangential component of the tidal force), which physically moves water into the tidal bulges.
7. Textbooks don't tell you this, do they? The high tide level in water is usually measured from low tide. Coastal tide levels are measured with respect to solid land (not shifting sand) on the shore.
8. There would be no tides in a uniform field. A field gradient is required for a tide.
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?
10. The ratio of the tidal forces is (59/61)3 0.90, so they differ by 10%. Compare the difference in gravitational force across the diameter of the earth: (59/61)2 = 0.94, or about a 6% difference.
11. 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.
12. 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 obviously, 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 tides.
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.
As you notice, these questions were designed deliberately to expose misconceptions arising from misleading textbook and website treatments.
Footnotes* The photo of the double lighthouse is a fake. There's only one lighthouse at Folly Beach, the Morris Island Lighthouse. However, in keeping with the spirit of this document, these lighthouses ought to be named "Centripetal" and "Centrifugal". [Photo © 2002 by Donald E. Simanek.]
 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.
Significant edits: 2005, 2006, 2008, 2009, 2011, Aug. 2015.
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