Some books and websites describe the tides as being a result of
"centrifugal force" due to the Earth and Moon moving. But
that's not actually correct, because the tides would exist even if the Earth
and Moon weren't moving! "Velocities" are not really involved.
So those books are actually wrong, even textbooks! If the Earth
and Moon were somehow standing still, there would still be tidal bulges, at
least for a while until they later crashed together due to gravity!
Even the tide that exists on the backside of the Earth would exist!
Newton's equation for Gravitation is all you need. It is in all books that describe gravitation. F = G * M_{1} * M_{2} / R^{2}. Gravitation therefore depends on the "mass" (most people incorrectly think of it as weight) of each of the objects involved. It is an "inverse square" relationship, where the attractive force depends on the distance between the two objects. There is also a constant 'G' that just makes the numbers come out right for the system of measurements (feet, meters, seconds, etc) that we use.


One way of looking at it is that the water from other Oceans would try to flow to that spot, because of that "excess acceleration" is essentially trying to "pile up water" there. So, even if the Earth and Moon did not move, or rotate, it would form a "hill of water" or "tidal bulge" at that location, directly under the location of the Moon, due to the upward "excess acceleration" due to the Moon's gravitation and that lesser distance. It turns out that that "tidal bulge" would be a little over two feet high, not noticeable in the enormity of the Earth!
But the Earth and Moon both move! Specifically, the Earth rotates once every day. This makes the Earth actually rotate UNDERNEATH that (unavoidable) tidal bulge that the Moon's gravitation constantly causes. It seems to us that there are tides that move across the oceans, but the actual tidal bulges do not really move very much, and it is actually that the Earth (and us) are rotating past that bulge that makes it seem to be moving to us.
By the way, each time the Moon is overhead, YOU are also being attracted UPWARD by the Moon, as well as downward by the Earth. It actually changes your weight! But by only by a really, really tiny amount! (around one part in nine million, one onehundredth of a gram!)
Since R is small compared to D (about 1/60 of it) this is nearly equal to G * M_{moon} * (2 * D * R) / D^{4}, or (a constant) / D^{3}. Tidal acceleration is therefore approximately proportional to the CUBE of the distance of the attracting body. If the Moon were only half as high above the Earth, the tidal accelerations would be EIGHT TIMES as great as they are now!
As it happens, both the Moon and the Sun cause such tides in our oceans. The Moon is about 400 times closer than the Sun, so it causes a tidal acceleration equal to 400^{3} or 64 million times that of an identical mass. But the Moon's mass is only about 1/27,000,000 that of the Sun. The result is that the Moon causes a tidal acceleration that is about 64,000,000/27,000,000 or 64/27, or a little more than double of that of the Sun. That's why the tides due to the Moon are larger than those due to the Sun.
At different times of a month, the Sun and Moon can cause tides that add to each other (called Spring Tides) (at a Full Moon or a New Moon, when their effects are lined up.) However, at First Quarter or Third Quarter Moon, the tides cause by the two sort of cancel each other out (with the Moon always winning!) and so there are lower tides then, called Neap Tides.
It's possible to even estimate the size of the tides, using that same result above, G * M_{moon} * (2 * D * R) / D^{4}, or G * M_{moon} * 2 * R) / D^{3}. We know the values of all those quantities! In the metric system, G = 6.672 * 10^{11}, M_{moon} = 7.3483 * 10^{22} kg, R = 6.37814 * 10^{6} m, and D = 3.844 * 10^{8} m. This gives a result for the relative acceleration as being 1.129 * 10^{6} m/s^{2}, about one ninemillionth of the acceleration of the Earth's gravity on us.
In case you are curious, we know that the acceleration due to gravity of the Earth is around 9.8 m/s^{2}. So the actual downward acceleration on us and water is reduced by about 1/8,700,000, when the Moon is at its average distance, and the Moon is overhead. If you happen to weigh 200 lbs (90 kg), your scale would show you lighter by about 1/2500 ounce (or 0.01 gram), when the Moon was overhead as compared to six hours earlier or later.
That 1.129 * 10^{6} m/s^{2} is the distorting ACCELERATION, and that number is precise. Of course, it depends on the current distance of the Moon which is in an elliptic orbit, and also the celestial latitude of the Moon for the specific location. A truly precise value is difficult since it changes every minute!
We can now compare two locations on Earth regarding this matter, one directly above the other. We know that we have F = G * M_{1} * M_{2} / R^{2} for both locations. By eliminating the mass of our object, we have a = G * M_{1} / R^{2}. So we could make a proportion for the two locations,
a_{lower} G * M_{earth} / R_{lower}^{2} _____ = _________________ a_{higher} G * M_{earth} / R_{higher}^{2}
The mass of the Earth and G cancel out and we have the proportion of the
accelerations equal to R_{higher}^{2} /
R_{lower}^{2}.
a_{lower} R_{higher}^{2} _____ = _________________ a_{higher} R_{lower}^{2}
In our case, we know the proportion of the accelerations, different by 1.129 * 10^{6} m/s^{2} /9.8 m/s^{2} so the left fraction is 1.0000001152. The radius R therefore has to be in the proportion of 1.0000000576 so that its square would be the acceleration proportion. If we multiply this proportion (difference) by the radius of the Earth, 6.378 * 10^{6} meters, we get 0.367 meter, which is the required difference is Earth radius to account for the difference in the acceleration.
If the Moon were NOT there, the equilibrium height of the ocean would have been the radius of the Earth. We have just shown that due to the differential acceleration of the Moon on the water, we have a new equilibrium radius of the Earth's oceans there which is around a third of a meter greater.
This suggests that the oceans' tidal bulge directly under where the Moon is should be around 0.367 meter or 14.5 inches high out in the middle of the ocean. The Sun's tidal bulge is similarly calculated to around 0.155 meter or 6.1 inches. When both tides match up, at Spring Tide at Full Moon or New Moon, that open ocean tide should be around 0.522 meter or 21 inches high. When they compete, at Neap Tide at First or Third Quarter Moon, they should be around 0.212 meter or 8.5 inches high in the open ocean. These numbers are difficult to measure experimentally, but all experiments have given results that are close to these theoretical values.
However, the Earth itself also "bends" due to the tidal effects and so the Earth itself (the ocean bottom) has an "Earth tide" that occurs. It is not well known as to size, and it is certainly small, but estimates are that it is probably just a few inches in rise/fall. The actual distortion that occurs in the Earth and in the waters of the oceans are extremely complex, so we are just using an estimate here. (We are preparing another webpage that discusses the actual flow patterns of the tides in the oceans of the Earth.) In any case, this then results in open ocean tides as being around a foot or foot and a half high, which seems to agree with general data, such as at remote islands in the middle of oceans.
For the water, we have been assuming that water can flow fast enough to balance out, therefore the level of water would be at an isogravitational constant value. This is not really the case, and water takes a while to build up that tidal bulge. If the Earth rotated REALLY slowly, it would be more true. But at the Equator, the rotation of the Earth is over 1000 miles per hour, and the depth of the oceans causes a limit on the speed of deep ocean waves to be around 750 mph.
This results in the tidal bulge always being dragged or carried along by the rapidly rotating Earth, to a location which is actually not underneath where the Moon is! The tidal bulge is NEVER directly under the Moon as it is always shown in school books! Instead, the reality is that the tidal bulges are generally several thousand miles away from that location. We see reason to cut some slack in school books regarding this, as it would just add an additional complication into the basic concept of the tides.
Notice that no centrifugal force was ever mentioned in this calculation, and it is a direct application of Newton's equation for gravitation. The books that, in attempting to describe why tides exist, DO choose to use arguments that involve centrifugal and centripetal forces, can therefore be rather misleading. There IS a way to do that and get an approximately correct result, but I think that it really overlooks the fact that it all is purely a simple gravitational result, and it gives the impression of tides only occurring due to a "whirling" effect of the Earth and Moon, which is NOT true. The only real value I see in those approaches is that the tide on the backside of the Earth might SEEM to be a little more logical. The mathematical description given above accurately describes that opposite tide, too, and even shows that it is always slightly less high, which might not seem as obvious when trying to use the "whirling" idea.
The same is still true when considering the "solid earth" and a particle of water in an ocean opposite where the Moon happens to be. It might seem amazing, but the Moon actually gives the entire Earth a greater acceleration, because of being closer, than it does for that puny bit of water in that backside ocean! And so all the equations above can be used again, with the one difference that our new distance from the Moon to the water is D + R instead of D  R.
When you plug in the numbers, you get a slightly smaller differential acceleration, 1.074 * 10^{6} m/s^{2} (instead of the slightly larger 1.129 * 10^{6} m/s^{2} that we got for the front side water.) Using the same calculation as above, we get a Mooncaused tidal change in radius of 0.349 meter (instead of 0.367 meter as before) or 13.8 inches. That is only 0.018 meter or around 3/4 of an inch difference in the heights of the front and back (Mooncaused) tides. The rear side tide IS therefore slightly smaller, but not by very much! The same reasoning and calculation applies to the Suncaused tidal bulges, so the total Spring Tide difference is actually around one inch.
(I have never seen any other presentation explain why the rearside tide is slightly smaller than the Moonside tide is, or the calculation of exactly how much that difference is!)
Again, please notice that we have not only proven that the rear side tide exists but even calculated its size, without having to use any centrifugal force! It is frightening that even a lot of school textbooks present a REALLY wrong explanation!
Well, water has a lot of friction, both with itself (viscosity) and with the seafloor (drag), so the tidal bulges travel (relatively) at slower speeds across the oceans! Since the Earth rotates so fast, this results in the tidal bulge(s) lagging and being "dragged forward" of that location. Also, instead of the actual water shooting all over the oceans at extremely high speed, the actual water tends to move relatively little, and the wavefront of the tide can pass at such high speed (often around 700 mph in deep open ocean) without much noticeable effect of actual movement of water. (No sailor ever senses any water shooting past his ship at 700 mph! Instead, he basically senses nothing, as his ship is gradually raised up about a foot over a period of several hours.) The 700 mph speed cited here is dependent on the DEPTH of the water, by a wellknown and rather simple formula giving the maximum speed of wave velocity due to the local water depth. For the common depths of the central oceans, that speed is around 700 mph.
The fact that friction between the water and the seafloor causes the tidal bulges to be shifted away from the line between the Earth and Moon has some longterm effects. That friction is actually converting a tiny amount of the Earth's rotational energy into frictional heat energy, which is gradually slowing down the Earth's rotation! Our days are actually getting longer, just because of the Moon! But really slowly. A thousand years from now, the day will still be within one second of the length it is now. The longterm effects of this are discussed below.
By the way, a "tiny amount" of the Earth's rotational energy is still quite impressive! The amount of frictional energy lost from the Earth's rotation is actually around 1.3 * 10^{23} Joules/century (Handbook of Chemistry and Physics). That is 1.3 * 10^{21} wattseconds per year, or 3.6 * 10^{14} kilowatthours each year. For comparison, figures provided in the World Almanac indicate that the entire electric consumption of all of the USA, including all residential, industrial, commercial, municipal and governmental usages, and waste, totaled around one onehundredth of that amount (3.857 * 10^{12} kWh) in 2003! (This fact, that natural oceanseafloor frictional losses due to tidal motions are taking a hundred times the energy from the Earth's rotation as all the electricity we use, and still barely having any effect on the length of our day, inspired me to seriously research the concept of trying to capture some of the Earth's rotational energy to convert it into electricity. It would be an ideal source of electricity, with no global warming, no pollution, and no rapid consumption of coal, oil or nuclear just to produce electricity! A webpage on that concept is at Earth Spinning Energy  Perfect Energy Source)
There are actually two situations regarding tides approaching land, which are actually closely related. Say that our one+ foot high tidal bulge of extra water is traveling at VERY high speed across the open ocean, and then it arrives at a Veeshaped Bay, like the Bay of Fundy in Canada. Say that the widest part of such a bay is 100 miles wide, so we start out with a wave that is one foot high and 100 miles wide. Remember that water is incompressible! Now, follow this wave as it enters such a bay. As it gets around halfway up the bay, the bay is only half as wide, 50 miles. But there is still just as much water in the oncoming wave. In order for all that water to squeeze into a 50 mile width, do you see that it must now be TWO feet high? Continuing this logic, if the bay narrows to two miles wide at the end, all that water would have to still be there, and it would now have to be a fiftyfoottall tide.
There are additional compounding effects in that waves travel at speeds that depend on the depth of the water. As the waves are moving up such a bay, the water keeps getting shallower, and the wave velocity (called celerity) greatly slows.
The very inner end of the Bay of Fundy actually has such tides! This "funneling effect" of the shape of that bay causes it. All during that process, though, there is a lot of friction, with the bottom of the bay, and among the waters and surf. These energy losses actually reduce the growth effect described above, and for bays that do not have such funnel shapes, tides in those bays are relatively moderate in size. But in those uniquely shaped bays, the tides are very impressive. At the very inner end, the tide comes in so quickly and so strongly that it even has a special name, a "tidal bore". A fairly large river at the end of the Bay of Fundy winds up flowing BACKWARDS briefly about twice each day due to the intensity of the tidal bore there!
The other situation is when a tide from the open ocean runs into a continent straight on. Due to natural erosion and many other effects, most shorelines gradually slope outward, getting deeper and deeper as you go farther from the shore. When an approaching tidal bulge gets to this vertically tapering area, a situation a lot like the funnelshapedbay width effect occurs. The wave essentially gets lifted upward as it moves up the "ramp". The actual shape of the contours of the slopes near continents greatly affects this process. If the slope is too shallow or too steep, or if it has irregular slope, much more energy is lost in friction and minimal tides are seen at the shoreline. But for some locations, it explains why the measured tides are much higher than the open ocean tides are. Most of the East Coast of the US has tides that are several feet high.
This discussion should help to explain the incredible complexity of the actual tides seen. Even worse, erosion and deposition are continuously modifying the contours of the ocean bottom, so these effects change over time. All these effects contribute to making the precise prediction of the size and arrival time of tides a very complex and imperfect science.
There is a way to calculate that eventual length of day/month. It relies on the fact that angular momentum must remain constant in a system that has no external torques applied to it. That means we must first calculate the total angular momentum of the EarthMoon system. There are four separate components of it, the rotation and the revolution of each of the Earth and Moon.
Angular momentum is the product of the rotational inertia (I) and the angular velocity (ω). For two of these situations, the revolution components, the rotational inertia is I = M * r^{2}. The r dimension is the distance between the barycenter of the system and the center of the individual body. (The Moon does NOT actually revolve around the Earth, but they BOTH revolve around each other, or actually a location that is called the barycenter of the system. In the case of the EarthMoon system, the barycenter happens to always be inside the Earth, roughly 1/4 of the way down toward the Center of the Earth, on a line between the centers of the Earth and Moon.)
For the Earth, this gives I = 2.435 * 10^{38} kgm^{2}.
For the Moon, this gives I = I = 1.049 * 10^{40} kgm^{2}.
For these two components, the angular velocity is one revolution per month, or 6.28 radians in 27.78 days, or 2.616 * 10^{6} radians/second.
That makes these two (revolution) angular momentum components:
Earth = 6.371 * 10^{32} kgm^{2}/sec
Moon = 2.745 * 10^{34} kgm^{2}/sec
The rotational inertia of the Earth and Moon are somewhat more
complicated to calculate, primarily because the density gets
greater toward the center of each. You can find the derivation in
some advanced geophysics texts. For the Earth, the currently
accepted value is:
I = 8.07 * 10^{37} kgm^{2}
Since the Earth rotates once a day, the angular velocity of it is
7.27 * 10^{5} radians/second.
That makes the Earthrotational angular momentum:
Earth = 5.861 * 10^{33} kgm^{2}/sec
It turns out that the component due to the Moon's rotation is extremely small. The actual rotational inertia of the rotation of the Moon is actually not known, but it is around 6 * 10^{34} kgm^{2}.
That makes the Moonrotational angular momentum:
Moon = 1.6 * 10^{29} kgm^{2}/sec
Totaling up these four components, we have 3.391 * 10^{34} kgm^{2}/sec as the TOTAL angular momentum of the system. The laws of Physics say that this angular momentum must be conserved, must always exist with the same total.
The four terms are now:
Earth_{rev} = m * R^{2} * (ω) or
6 * 10^{24} kg * (D/60.37)^{2} * 6.28 / (Daylength)
or 1.03 * 10^{22} * D^{2} / (Daylength)
Moon_{rev} = m * R^{2} * (ω) or 7.34 * 10^{22} kg * ((D*59.37)/60.37)^{2} * 6.28 / (Daylength) or 4.458 * 10^{23} * D^{2} / (Daylength)
Earth_{rot} = I * (ω) or 8.07 * 10^{37} * 6.28 / (Daylength) or 5.03 * 10^{38} / (Daylength)
Moon_{rot} = I * (ω) or 6 * 10^{34} * 6.28 / (Daylength) or 3.77 * 10^{35} / (Daylength)
Totaling all four components, the two pairs can be combined:
4.561 * 10^{23} * D^{2} / (Daylength) + 5.03 * 10^{38} /
(Daylength). This total must equal the previously calculated total
angular momentum of 3.391 * 10^{34} kgm^{2}/sec.
It turns out that Kepler discovered a relationship between the time interval of revolution and the distance between them. The square of the time interval is proportional to the cube of the distance between them. In our problem, this results in the D^{2} term being able to be replaced by (DayLength)^{1.333} * 4.589 * 10^{8}.
This results in our equation becoming:
2.093 * 10^{32} * (Daylength)^{0.333} = 3.391 * 10^{34}
which solves to a Daylength of 4.208 * 10^{6} seconds, which is equal to about 48.7 of our current days.
This arrangement would have a spacing of 5.58 * 10^{8} meters or around 347,000 miles, as compared to the current 238,000 miles.
This then indicates that the Moon will continue to very slowly move outward, apparently for hundreds of millions of years, until it eventually gets to that distance. At the same time, the length of our day will continue to very slowly get longer until it becomes about 49 times as long as now! All these results are directly calculated from the basic laws of Gravitation!
I have seen previous estimates where the final distance would be about 400,000 miles and the period would be 55 days. Those numbers were apparently calculated over a hundred years ago, by the mathematician Sir George Darwin. It is not obvious that anyone else has done these calculations, a central reason for my composing this presentation. The possibility exists that the figure for the rotational inertia of the Earth was not known as well by Darwin as it is now. I believe that the above calculations reflect the current knowledge of the values.
There is also a factor regarding the Earth losing kinetic energy of rotation due to frictional heating of the ocean tides against the ocean bottoms and the continents. This is considered to be about 1.3 * 10^{21} Joules (wattseconds) per year. (Handbook of Chemistry and Physics). (For comparison, this energy consumption is equal to approximately 100 times all the electricity used in the USA!)
Researchers have rather accurately determine that each year is around 16 microseconds longer than the year before. Interestingly, the friction of the tides against the seafloor and the continents should cause a slowing of around 22 microseconds each year but that there are some effects that actually cause a secular increase in the rate of spin of the Earth (which we shall not discuss here!) We therefore have a day that (should) increase in length due to this effect by 22 microseconds each year.
The amount of friction between the water and the moving Earth under and in front of it depends on the speed of that differential motion. It is a reasonable assumption that it depends directly (proportionally) on the speed of that motion, that is the rotation rate of the Earth. This being the case, we can apply some simple Calculus to do a little differentiating and integrating to establish that there is a simple exponential relationship. Specifically, we find that ln(86400  0.000022)  ln(86400) = k * t, where t is the number of seconds in a year, that is 3.1557 * 10^{7}. This lets us calculate the value of k to be 8.0806 * 10^{18}. This value uses base e, and we can convert to a formula which uses base 2 by simply dividing this value by the value of ln(2), which then gives 1.16578 * 10^{17}.
We then have that the ratio of the rate of the Earth's spin is equal to 2^{t * k}. To find how long it will be until the Earth would be rotating at half the current rate (twice as long a day) (due ONLY to the tidal friction effect!), just set this equal to 1/2. For this, clearly t must equal the inverse of k, so that we would have 2^{1} which is 1/2. Therefore, that would be 1 / (1.16578 * 10^{17}) seconds from now, which is 8.59138 * 10^{16} seconds or 2.722 billion years! Quite a while! If you're still around 2.7 billion years from now, each day figures to be twice as long as it is now!
Going back the other way, 2.7 billion years ago, the Earth was spinning much faster. Again, if we consider just the effects of tidal friction, and if the oceans and continents were similar to as they are now, the Earth would have been spinning twice as fast, with 730 days each year! Probably even faster, because the Moon was then closer, and therefore the tidal effect would have been still greater. If the oceans formed around 4 billion years ago, then this equation gives a t * k product of around 1.5, and our spin ratio would have been 2.83, so the Earth must have then spun in around 8.5 of our modern hours, nearly three times as fast as now. And probably even faster than that, because the Moon was certainly much closer then and also those giant tidal bulges must have had tremendous friction when running into a continent at 3,000 miles per hour! I betcha that beach erosion must have been something really impressive then! Every four hours, a giant tidal wave (ACTUALLY a tidal wave and not the misnamed tsunami!) would crash into each continent at the 3,000 mph rotation speed of the Earth. Wow!
In any case, as a bonus, we have shown that the early Earth must have rotated probably more than three times as rapidly than today, at least until the oceans formed and the Moon started having the tidal effects that has slowed down our rotation to what it is today.
And also that the slowing effect is pretty slow, and the exponential equation given above indicates that around 9 billion years from now, the Earth will have slowed to around 1/10 the current rate (36 REALLY long days every year!) and that would still be far from the eventual day length when the Earth and Moon will have finally gone into being lockedup facing each other forever. Since we believe the Sun will have used up its Hydrogen fuel far before that, it would occur in total darkness!
If you are an inquisitive sort, you could use the angular momentum conservation analysis given way above to figure out just how close the Moon must have been at that time when the oceans were forming! I WILL give you a clue that the month was then around 5 modern earth days long, and the Moon was HUGE in the sky, ballpark around 1/3 as far away as now! Do it! Show me that you can!
Keep in mind that the precise values were probably not actually these, as we have only considered the Moon's tidal effects and have ignored various other effects that affect the rotation rate of the Earth. In fact, we made a basic assumption regarding the effect BEING PROPORTIONAL TO THE RELATIVE SPEED. That is not necessarily true, and for example, it might instead be proportional to the SQUARE of the relative speed, which would materially change these calculations. The general time scales would still be extremely long though.
With really accurate modern equipment, it has been found that even the seasonal difference of the weight of snow accumulating near the pole in winter has a measurable effect on changing our total rotational inertia (as water, much of that mass of water would have been nearer the equator, slightly changing I) so we speed up and slow down for a lot of such reasons, but all of those effects are really pretty small!
In addition to all this, we know that continents wander around due to Plate Tectonics over these same long time scales. It seems clear that there must have been times when the arrangement of continents were better or worse than they are now regarding interfering with the tidal flows around the Earth, which is another factor which cannot be decently estimated. So numerical results such as these are necessarily very approximate, which is part of the reason we chose to use the simplest of all assumptions regarding the relative speed factor.
The reasoning presented above, based on that assumption, would indicate that the length of the day on Earth would have been 16 (modern) hours around 1.6 billion years ago. I have recently been told that someone has recently apparently done a similar analysis where their results imply that the day was 16 hours long around 0.9 billion years ago. I am not familiar with the specific analysis used to get such a number, so it is not possible to comment on its likely accuracy. (The analysis given here implies that the day was around 19 [modern] hours long at that time.)
I am not really concerned regarding the precise accuracy of either result, but rather have interest in the PROCESS of the analysis, so that any reader of this presentation should be capable of doing the complete analysis for him/herself.
Link to a thorough mathematical discussion of assorted Moon issues, is on The Origin of the Moon  A New Theory
Link to a slightly unrelated subject, that of trying to capture some rotational energy of the Earth for generating electric power Earth Spinning Energy  Perfect Energy Source
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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago