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## Tidal forces associated with the moon

In the following discussion the effect of the earth's rotation about its own axis will be ignored. The effect produced by this rotation has already been taken into account in the expression for gravity. Thus we can regard the earth merely as moving without rotation along a circular path relative to the earth/moon system. Every point of the irrotational body depicted in Fig. 2.04-2 is subect to the same centripetal

acceleration since the orbits of each point are of identical radii. This acceleration is given by Eq. (27b)

If we introduce the concept of centrifugal force per unit mass to replace this centripetal acceleration, then at the point A (see Fig. 2.04-3) on the earth's surface the force of

attraction on a unit mass due to the moon is partially offset by the centrifugal force on the unit mass at this point. The net force per unit mass at A is

and this is directed towards the moon. The centrifugal force per unit mass at point B of Fig. 2.04-3 is the same but the attractive force due to the moon is less. The net force per unit mass at B is

and this is directed away from the moon. The expressions (36) anbd (37) can be written as

with no sensible loss of accuracy. Now, if we use the approximation

then

Inserting the values of , , a and d from Table 2.04-II we obtain

Thus at the points A and B the net effect is to produce a reduction of gravity in the amount of 0.00011 cm/sec only.

The lunar tidal force at other locations can be analyzed by reference to Fig. 2.04-4.

The distance from point P on the earth to the moon's enter is given by

where is the angle formed by the lines joining the center of the earth with P and with the moon's center. The components of the attractive force of the moon in the directions parallel and normal to the line joining the centers of the earth and moon are, respectively,

where

Because of the very small value of the ratio a/d (about 1/60), the following approximations are in order

The net value of the two forces acting parallel to the line OL is

Making use of Eq. (42) we find

The net force components can be written approximately as

It is more convenient to deal with the radial and tangential components and of the net tidal force at the earth's surface. Referring to Fig. 2.05-5 we see that:

Introducing the approximation (39) and simplifying Eqs. (49) leads to

When , and reduces to the value given by Eq. (40). For :

indicating a slight increase in gravity (negative being towards the earth's center).

Actually, it is the tangential (horizontal) component which leads to the development of the tides since the slight effect on g by the vertical force has virtually no effect. The maximum tide-producing force associated with the moon occurs at a value of of and has the value

This force is equivalent to the component of gravity acting along a slope of 8 cm per 1000 km which is about twice in magnitude the mean slope of the sea surface along the equator in the Pacific, but is of the order of one fourth the slope of the surface across the major currents such as the Gulf Stream and Kuroshio Current. Consequently, the horizontal tide force is far from negligible compared with other horizontal forces in the sea.

Next: Tidal forces associated with Up: Tidal forces Previous: Equilibrium of the earth-moon

Steve Baum
Mon Dec 1 08:50:29 CST 1997