Copyright © 2004, Richard J. Hanak
An Ocean Tide Model
Revision 2
By Richard J. Hanak
October 5, 2004
Even the ancients knew that the moon is the primary cause of tides. On the side of the earth facing the moon there is always a high tide. Gallileo correctly deduced that the movement of the tides is caused by the rotation of the earth. But if a high tide at a location is caused by facing the moon, what causes a second high tide about 12 hours later when that location has rotated to the side of the earth facing away from the moon?
I recently read a contemporary explanation of tides accounting for two daily high tides with negative gravity, gravity exerting a repelling force, and the direction of a gravitational force depending on a reference other than its source. Those notions are as unappealing to me as the very old idea of the moon pulling the opposite side of the earth away from under the ocean to produce a high tide on that side. Many people still believe that old explanation, so let's examine it more closely.
The high tide on the side of the earth facing the moon is explained as caused by the gravitational attraction of the moon. In that explanation the ocean water moves closer to the moon while the earth beneath says put. On that side of the earth the ocean has become deeper. The water for the increase on that side came from the regions of low tide. That explanation seems reasonable enough.
If we are at the seashore at 9 p.m. and the moon is overhead there is a high tide. About 12 hours later, at 9 a.m., we see a second high tide though the earth has rotated so that the moon is now over the opposite side of the earth. On many seacoasts there are two high tides a day.
One of the explanations given for the high tide on the moonless side of the earth is as follows. Yes, the moon's gravitational attraction pulls the ocean toward the moon and closer to the earth on that side. However, because the gravitational pull of the moon is stronger on the ocean bed than on the ocean surface—the ocean bed being a little closer to the moon than the ocean—the earth is pulled out from under the ocean, so making the ocean higher there.
But if the earth is pulled out from under the ocean, what has happened to the earth? Surely the center of the earth doesn't move closer to the moon. Has that opposite side of the earth under the ocean bed been compressed enough to account for the high tide? If the ocean bed were pulled toward the moon, wouldn't the ocean merely follow, rather than stay where it was?
The question of whether or not the high tide on the far side of the earth can be caused by lunar gravitation alone is easily resolved. Figure 1 depicts the earth, moon, and the lunar gravitational forces (F1, F2, F3) on a unit of mass at the far side, center, and near side of the earth. The lengths of the force lines indicate the relative strengths of the forces.
The ocean must be pulled in a direction away from the center of the earth in order to produce a high tide. The moon's gravitational force (F3) on the near side could pull the ocean away from the center of the earth. However, on the far side the moon's gravitational force (F1) would pull the ocean toward the center of the earth. Force F3 would be a tensile force, while force F1 would be a compressive force.
Newton wrote "Law III. To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts." In explanation he wrote "If you press a stone with your finger, the finger is also pressed by the stone." Were the force of lunar gravitation (F1) alone to pull the far surface of the earth toward the center, the earth beneath the far surface would press back against that pull, so placing the far surface under compression. The claim that lunar gravitation alone could raise a tide on the far side of the earth, that it could cause an outward tension in the ocean on the far side, cannot be correct because it would contradict Newton's third law in regard to force F1. Furthermore, no point in an object can simulatneously be in compression and tension on the same line.
Lunar gravitation alone would cause any unit mass between the far side and center of the earth to be in compression. It would also cause any unit mass between the center and near side to be in tension. At any given point of an an object, tension and compression are mutually exclusive states. If lunar gravitation alone were to cause tension on one side of the earth and compression on the opposite side, it would produce neither tension nor compression at the center of the earth. Therefore, lunar gravitational force (F2) at the center of the earth would have no effect on tides.
Thus far we have only considered forces F1, F2, and F3. According to Newton's law of gravitation, every particle of the earth and every particle of the moon are mutually attracted to each other. If gravitational force were the only force, the moon and earth would have been drawn together long ago. Clearly, there must be a force that opposes gravitational attraction. The anti-gravity force we seek is an inertial force, the centrifugal force arising from orbital motion. The centrifugal forces that affect tides are described below.
The unacceptable idea that gravtitational forces alone caused tides stimulated me to see if I could successfully model tides using ordinary physics. As a starting point I chose gravitational force and centrifugal force—the existence of the latter now denied by many—as the two effective forces for interaction between objects as massive as the sun, earth, and moon. I am indebted to a correspondant on the Bad Astronomy bulletin board for correcting my earlier misconceptions regarding the distribution of orbital forces on the earth.
The following analysis of tides uses a simplified model. Neglecting solar system dynamics and ocean dynamics, it gives a snapshot in time. Other real features it initially neglects include the effects of the elliptical orbits of the moon and earth, the inclination of the earth's axis to its solar orbit, the inclination of the moon's orbit of the earth to the earth's orbit of the sun, the oblate shape of the earth, and the influence of other solar system planets. This model assumes a spherical earth for which the earth's gravity, being the same at all points of the surface, would have no tidal effect.
The forces presented here are: 1. expressed as newtons per kilogram of ocean water, 2. taken as positive if they raise a part of the ocean and negative if they lower it, and 3. resolved as their earth-radius vector components (in the direction of the raising and lowering). We will only analyze the forces on the equator of the earth in this revised model.
The situation we will examine is depicted in Figure 2, which is not to scale. Note that the sun and moon are on the same side of the earth and all lie on a straight line. Different position relationships of the earth, moon, and sun do not alter the principles involved and do not change the order of magnitude of the tides. In calculating the forces, the distances between the centers of the sun and moon from the center of the earth are taken as their average distances.
Any explanation of tides must account for the fact that many coastal areas on earth experience two high tides a lunar day. That fact really embraces two facts: the first that there are two high tides; the second that they occur in a lunar day. A lunar day — the time between upper transits of the moon, or between its risings, or between its settings — averages 24 hours and 50 minutes and varies as much as 15 minutes either way. The primary objects affecting ocean tides are the sun, earth, and moon.
As the earth circles the sun once a year and the moon circles the earth every 27.322 days, the earth spins on its axis once a day. The forces of the moon and sun at any location in the earth's space change only a little in the course of one day. The earth's daily spin causes different parts of the earth to pass through the same portions of earth's space in that day. It follows that the daily spinning of the earth causes the movement of the tides. The north and south poles, however, do not move in earth's space during the course of a day. They should not experience daily movement of tides.
Although we often consider the moon to be circling a 'stationary' earth, in
reality they both circle their common center of gravity or barycenter. The
barycenter, located about 4670 kilometers from the center of the earth, is about
1708 kilometers beneath the earth's surface. We usually think of the center of
the earth as orbiting the sun. Actually the barycenter, the center of gravity of
the earth-moon system, orbits the sun. Figure 3 depicts the earth and moon
orbiting the barycenter. The line connecting the centers of the earth and moon
always passes through the barycenter.
Were it not for the earth's daily rotation, the earth would always face the same direction. Revolution of the earth's center around the barycenter does not cause the earth to to have a monthly rotation. Thus, only the center of the earth orbits the barycenter. All other points of the earth move in unison in the same direction as the center of the earth. Therefore, at any instant in time each point on earth is in an orbit the same size as the barycentric orbit of the earth's center. At any instant the center of a point's orbit is unique to that point. I have referred to that center as the point orbit center. A particle's orbitting of its point orbit center gives rise to a centrifugal force acting on that particle. At the surface of the earth the point orbit centrifugal force and the lunar gravitational force combine to give a resultant tidal force.
Figure 4, not to scale, illustrates how the earth and moon move around the barycenter with the earth's daily rotation omitted. The black dot represents a fixed point on the earth. Though the center of the earth orbits the barycenter and the fixed point orbits its point orbit center, that fixed point always faces the same direction since the earth has no monthly rotation.
Many dictionaries reinforce the misunderstanding that tides are "caused by the attraction of the moon and sun." Gravitational attractive force is always directed toward its source or center. The gravitational forces of the moon and sun, then, can only raise the ocean from the earth on the side of the earth facing them. Their effect on the opposite side of the earth is to draw the ocean toward the earth, to depress it. Were it not for a second kind of force acting there, the water on the opposite would flow to the near side. Gravitational force can account for only one of the two 'daily' high tides. Although the moon is much closer to earth than the sun, we will find that the sun's enormous mass makes its gravitational effect on the earth almost 200 times that of the moon.
A non-gravitational force is also at work producing tides. Centrifugal forces, always directed away from their centers of revolution, give rise to the second daily high tide. The earth revolves in three different ways: its daily rotation on its axis, its 'monthly' (really 27.322 days) orbit of the earth-moon barycenter, and its annual orbit of the sun. Each of those revolutions generates centrifugal forces.
The daily spin of the earth on its axis generates enough centrifugal force to cause the earth to be an oblate spheroid with an equatorial radius about 21 kilometers greater than its polar radius. When we say that something is moving uphill we mean that it is moving farther from the center of the earth. Therefore, the mighty Mississippi River flows uphill while travelling from north to south, pushed by that centrifugal force. At the equator the earth's axial spin generates a lifting centrifugal force of 3.44x10-3 newtons on each kilogram weight of material, whether that be water or dry land. That is the force that can lift water 21 kilometers.
Figure 5 illustrates the trigonometric relationships involved in calculating the various forces acting at any position on the equator of the earth. The tide determining force components appear on the earth radius to that position. Angle A determines the longitude of a point on the equator. The force vectors and their components, as shown, illustrate their trigonometric relationships rather than their relative intensities.
The forces causing tides are much smaller than the gravitational and centrifugal forces that give rise to them. The tidal components of the forces resulting from earth-moon interaction are at most about one thirtieth of those forces. The tidal components of the forces resulting from earth-sun interaction are on the order of one ten-thousandth of those forces.
For example, for angle A = 0, the following forces are obtained.
| Lunar centrifugal | -3.373490x10-6 |
|---|---|
| Lunar gravitational | 3.488117x10-6 | Lunar total | 11.46x10-8 | Solar centrifugal | -6.049261x10-4 |
| Solar gravitational | 6.049781x10-4 | Solar total | 5.19x10-8 | Tide total | 16.65x10-8 |
Figure 6 shows the lunar forces and their total around the equator in 18-degree increments. The data for the equatorial model can be found on the data page accessible from the previous menu. We can quickly observe that the lunar (point orbit) centrifugal force and the lunar gravitational force almost cancel each other. The total force line shows the total of the lunar centrifugal and gravitational forces. Note that the lunar total force is always positive (directed away from the earth's center). Note also that the lunar centrifugal and gravitational forces each have only one maximum and one minimum.
In the present model a force of about 1.67x10-7 newtons acts to raise the level of the high tide above the level of the low tide. The centrifugal force of 3.44x10-3 newtons per kilogram of ocean water caused by the earth's daily
rotation raises the oceans 21 kilometers. Accordingly, the difference in high and low tide level caused by 1.7x10-7 newtons would then be about 1.02 meters. The difference in level between high tide and the ocean's average level would the be half that, or about 51 centimeters in the middle of the ocean. Coastal and river tides
are usually much greater than that because of dynamic effects. At a coast the
water in an advancing high tide cannot move into a preceding low tide region and
therefore accumulates to produce a higher local tide. At the mouth of a river
large amounts of water are forced into a narrowing channel, so amplifying the
tidal effect. At the Bay of Fundy, for example, those dynamic effects can
produce a tide as high as 15 meters.
The earth and moon orbit the barycenter once every 27.322 days and orbit the
sun every 365.25 days. The earth spins on its axis once a day, so causing the
tides to move in what is primarily a longitudinal direction. Latitude affects
the magnitudes of those longitudinally travelling waves.
As mentioned previously, the model we are considering is a simplified version
of reality, though not so condensed as to miss the primary effects. The
principal simplifications relate to the earth's orbit of the sun and the moon's
and earth's orbits of the barycenter, as well as the inclinations of the earth's axis, the earth's solar orbit, and the plane of the earth and moon orbits of the barycenter.
More changes could be added to this model. For example, the earth's axis is tilted 23 degrees from vertical to the the plane of its solar orbit, the moon's orbit of
the earth is inclined 5 degrees to the plane of the earth's orbit of the
sun...... and the sun is not always on the same line as the moon and earth......
the solar and lunar orbits are elliptical...... the distances change during the
course of a year......time could be included as a variable...... three
dimensional drawings......the other planets, ......
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