10.1: Tide-Generating Forces

Tides continuously raise and lower the sea surface. The vertical movement of the water surface is as much as several meters, twice a day in some areas. Clearly large amounts of energy are necessary to move the water involved in this process. The energy is supplied by the gravitational attraction between the water and the Earth, moon, and sun—but how? Tides are cyclical, but they do not come and go at the same times each day, and the height to which the water rises or recedes varies from day to day. Why does this happen? To answer these questions, we must first learn some basic information about gravity and the motions of the moon around the Earth and the Earth around the sun.

Gravity

Newton’s law of gravitation states that every particle of mass in the universe attracts every other particle of mass with a force that is proportional to the product of the two masses and inversely proportional to the square of the distance between the two masses. The gravitation force (F) between two masses can be expressed as

F = G × (M₁ × M₂)/D₂

where G is the gravitational constant (a constant defined by Newton’s law and equal to 6.673 ×10^-11 m³•kg^-1•s^-2), M₁ and M₂ are the two masses, and D is the distance between them.

For approximately spherical objects, such as planets, the entire mass can be considered to be at the object’s center of gravity, and the distance D is measured between the centers of the two masses. The gravitational force increases as either mass increases or the distance between the two objects decreases.

We are familiar with gravity because it is the force that attracts all objects on the Earth’s surface, including ourselves, to the much greater mass of the Earth itself. In addition, the Earth and every object on it are subject to gravitational attractions toward the sun, the moon, other planets, and every other celestial body. In the following discussion in this chapter, we refer to Earth’s gravitational force that we experience on its surface as gravity, while we refer to the gravitational force between celestial objects such as the Earth and moon as gravitational attraction. This is done simply to minimize possible confusion as the phenomena are the same and the laws of gravity apply to both.

Gravitational forces are extremely small in relation to the gravity we experience on Earth’s surface. Although the sun, stars, and some planets each have a mass much greater than the Earth’s mass, they are much farther away from us than we are from the Earth’s center. For example, gravitational attractions that the sun and moon exert on any object at the Earth’s surface are approximately 0.06% and 0.0003% of the Earth’s gravity, respectively. Gravitational attractions of other celestial bodies are much smaller.

Although they are almost negligible in comparison with the Earth’s gravity, gravitational forces exerted by the moon and sun on the Earth are the cause of the tides. To understand how these forces cause the tides, we must first consider the characteristics of orbits in which celestial bodies move.

Orbital Motions and Centripetal Force

Any two bodies that orbit together in space, orbit around their common center of mass. This works like a seesaw. If the two riders are of equal weight, the seesaw balances when the riders are at equal distances from the center. Similarly, two planetary bodies of equal mass would orbit around a point exactly midway between them. However, if one rider is heavier than the other, the heavier rider must sit closer to the center of the seesaw to make it balanced. Similarly, two planetary bodies of unequal mass orbit around a point closer to the more massive body. The Earth’s mass is 81 times that of the moon. Therefore, the common point of rotation is about 4680 km from the Earth’s center, or approximately 1700 km below the Earth’s surface (Fig. 10-2). Similarly, because the sun has far greater mass than the Earth, the common point of rotation of the Earth and the sun is deep within the sun.

Figure 10-2. Any two bodies in space orbit around a common point of rotation. Just as, on a seesaw, the heavier person must sit nearer the pivot point to provide balance, the common point of rotation for two bodies orbiting in space must be closer to the more massive body. (a) Thus, the Earth’s center must be much closer to the common center of rotation than the moon’s center is. (b) Because the Earth is much more massive than the moon, their common point of rotation is approximately 1700 km below the Earth’s surface.

As two planetary bodies orbit each other, they must be constrained in their orbit by a centripetal force that prevents each from flying off in a straight line (CC12). The centripetal force is supplied by the gravitational attraction between them. The centripetal force required is the same for all parts of each of the two bodies. Because centripetal force varies with distance from the center of rotation (CC12), this may be difficult to understand. However, within either of the orbiting bodies, all particles of mass move at the same speed in circular paths of the same diameter. We can understand this motion if we examine the motions of the two objects as they orbit each other without considering other motions, such as the Earth’s spin on its axis or orbits that involve a third body.

Figure 10-3 shows a non-spinning planet (a hypothetical “Earth”) orbiting with another planetary object (a hypothetical “moon”). In this diagram, the common center of rotation, which is also the center of mass of the two bodies, is located beneath the surface of the larger planetary body (the hypothetical Earth). Careful examination of Figure 10-3 reveals that, as the two bodies rotate around each other, each particle of mass on the planet moves in a circular orbit, and all of the orbits have the same diameter. However, as Figure 10-3 shows, the centers of the circular orbits are all displaced from the common center of rotation, except the orbit of the planet’s center of mass. Because the particles of mass in the planet move in circles of the same diameter and complete the circles in the same amount of time, all particles require the same centripetal force to maintain their orbits. This is true no matter where the common center of rotation is. It is true for the Earth-moon system, where the center of rotation is inside the Earth, and for the Earth-sun system, where the center of rotation is inside the sun.

Figure 10-3. This figure is a representation of the location of the Earth in relation to the moon as they orbit around their common center of mass, which is inside the Earth. In this representation the Earth is assumed to be nonrotating (not spinning on its axis). As the two bodies rotate around their common center of rotation, points A and B, which are at the Earth’s surface, and point C, which is the center of the Earth, would move through the color-coded locations shown in the figure. Notice that all points on or within the Earth move in circular paths that have the same diameter. This is also true for all points on or within the moon, or on or within any other orbiting bodies. Because the orbital diameter and rotation rate are always the same, the centripetal force needed to keep each body in orbit is equal at all points within each of the two bodies. Note that a centripetal force is also needed to maintain parts of the Earth in the circular motion of the Earth’s spin, but this is a separate motion with a different distribution of its centripetal force and does not affect the centripetal force between the Earth and moon.

The Balance between Centripetal Force and Gravitational Force

The total gravitational force between the moon and the Earth (or the sun and the Earth) must be equal to the total centripetal force needed to maintain these bodies in their orbits, or else the orbits would change. Although all particles of mass within the Earth are subject to the same centripetal force, the gravitational force on each particle of mass within the Earth varies with its distance from the moon or sun. For example, the distance from the moon’s center to the Earth’s center is approximately 384,800 km, and the Earth’s diameter is approximately 12,680 km. Therefore, a point on the Earth’s surface nearest to the moon is only 378,460 km from the moon’s center, and a point farthest from the moon is 391,140 km from the moon’s center. Because the gravitational force is inversely proportional to the square of the distance, the change in the moon’s gravitational force between the point on the Earth nearest to the moon and the Earth’s center is 378,4602 divided by 384,8002, or about 1 to 1.034. Hence, the moon’s gravitational attraction is about 3% greater at the Earth’s surface nearest the moon than it is at the Earth’s center (the average gravitational attraction between moon and Earth). Similarly, the moon’s gravitational attraction at the Earth’s surface farthest from the moon is about 3% less than the average gravitational attraction.

The average centripetal force for each orbiting body must equal the average gravitational force between the two if the orbit is to remain stable. Because the centripetal force is the same at all points on the Earth and gravitational force is not, there is a net excess gravitational force on the side of the Earth nearest the moon, and a net deficit of gravitational force on the side farthest from the moon (Fig. 10-4). These imbalances are tiny in relation to the Earth’s gravity and are easily compensated in the solid Earth by small changes in pressure gradient (the gradient of pressure within the atmosphere, ocean, or solid Earth that increases toward the Earth’s center). Pressure within the Earth (and other objects) represents the force needed to resist the gravity tending to pull all material toward the Earth’s center). For people, the compensation in the pressure gradient causes a change in weight. Each of us weighs about 0.000,000,1 kg (0.1 mg) less when the moon is over the opposite side of the Earth than when it is directly overhead —much too small a difference to notice.

Figure 10-4. The total gravitational force between the Earth and moon (or sun) must equal the total centripetal force needed to maintain the two bodies in their common orbit. However, because the gravitational force varies slightly at different points on the Earth, whereas the centripetal force is the same at all points, there is a slight imbalance between them everywhere except at the exact center of mass of the Earth. On the side of the Earth nearest the moon, the gravitational force due to the moon is slightly higher than it is at the Earth’s center. The slight excess of gravitational force over centripetal force at this point is easily compensated by the pressure gradient. Similarly, there is a slight excess of centripetal force over gravitational force on the Earth at the point directly away from the moon. Thus, objects at points directly under the moon weigh very slightly less than they do at points where the moon is directly overhead.

Unlike solid objects, ocean water can flow in response to the imbalances between gravitational force and centripetal force. These flows are the tides. For convenience, we refer to the net excess of gravitational force over centripetal force as the “tidal pull” or “tide-generating force.”

Distribution of Tide-Generating Forces

At the point on the Earth’s surface nearest the moon, there is a slight tidal pull. Water is pulled upward toward the moon at this point. However, the Earth’s gravity, which is comparatively very strong, acts directly opposite the tidal pull at this location. Therefore, the tidal force at this point (and at the opposite side of the Earth, which is farthest from the moon) has little effect, just as the tidal force has only an extremely small effect on our weight.

At other points on the Earth’s surface, the tidal pull is exerted in a direction different from that of the Earth’s gravity because the direction between that point on the Earth’s surface and the moon, and the direction between that point and the Earth’s center of mass, are aligned at an angle. Thus, at every point except directly under the moon and exactly on the other side of the Earth, the tidal pull has a component that acts parallel to the Earth’s surface (Fig. 10-5). This component of tidal pull, which is referred to as the “tidal force” cannot be compensated by the Earth’s gravity and therefore causes water to flow in the direction of the force.

Figure 10-5. Horizontal tidal forces. (a) Components of gravitational attraction, centripetal force, and the Earth’s pressure gradient. The vertical component of the tidal force is easily compensated by a minute change in the pressure gradient. Magnitudes of the tidal force and tidal force components are highly exaggerated in the figure. (b) The horizontal tidal force is zero at a point directly under the moon (or sun) and a point located exactly on the opposite side of the moon. It is also zero in a ring (annulus) around the Earth that is equidistant from these two points. The horizontal tidal force increases to a maximum and then decreases between the point directly under the moon and the annulus where the force is zero. The same is true between the point directly away from the moon and the annulus, but the tidal force is in the opposite direction.

FIGURE

The net result of the tidal forces acting on the Earth’s oceans is to move water toward the points nearest to the moon and farthest from the moon (Fig. 10-5b). This movement creates bulges of elevated water surface at these points, and a depression of the water surface in a ring around the Earth approximately halfway between these points. The diagrams in this book necessarily greatly exaggerate the bulges. Even on an Earth totally covered by oceans, the bulges would be less than a meter high.

The tide bulges are oriented directly toward and away from the moon. Because the Earth is spinning, the point on the Earth facing the moon is continuously changing. Therefore, as the Earth spins on its axis the bulge remains aligned with the moon, but the locations on the Earth that are under the tide bulge change.

Relative Magnitude of the Lunar and Solar Tide-Generating Forces

Although all planets and stars exert tidal forces on the Earth, only tidal forces of the moon and the sun are significant. Tidal forces are caused by the small difference in gravitational force from one side of the Earth to the other. This difference depends on the distance between the Earth’s center and the other body’s center, and on the other body’s mass. Although we need not perform the calculation here, we can show mathematically that the magnitude of the tidal force is proportional to the mass of the attracting body (sun or moon) and inversely proportional to the cube of the distance between the centers of the two bodies (r³). This means that the tidal force is much more dependent on the distance between bodies than it is on mass.

For the Earth–moon system, the tide-generating force (F_M) is given by:

F_M  =  K × (Mass of Moon) / (Earth-to-moon distance)³

and for the Earth–sun system, the tide-generating force (F_S) is given by:

F_S =  K × (Mass of Sun) / (Earth-to-sun distance)³

where K is a constant that is always the same for tidal forces between the Earth and any other planetary body.

The sun’s mass is about 27 million times greater than the moon’s. However, the sun is about 390 times farther from the Earth. The ratio of solar to lunar tidal force can be calculated with the following equations:

F_S / F_M = (Mass of Sun)/(Mass of Moon) × (Earth-to-moon distance)³/(Earth-to-sun distance)³

= 27,000,000/1 × (1 × 1 × 1)/(390 × 390 × 390)

=0.46

Despite the sun’s much greater mass, its tide-generating force is only about 46% of the moon’s tide-generating force because the sun is much farther away from the Earth.