Tides
REFERENCE: Barger & Olsson
Consider the gravitational forces acting on a mass element m on the surface of the earth. The earth itself exerts a force to draw m into a perfectly spherical configuration. But concurrent with that spherically-symmetric force, the moon attracts m. Of course, even as the moon is attracting m, it also attracts the earth. As far as the earth-moon system is concerned, when m is on the far side of the earth (away from the moon), on average the moon attracts the earth with more force per unit mass than it attracts m. When m is on the near side, the opposite occurs. It is this difference in moon force that causes the tides.
To analyze the tidal force we pick an inertial reference frame (primed) somewhere in space. We let r1 and r2 represent the position vectors of m and ME respectively. Then r represents the position vector of m relative to an earth-bound coordinate system.
Note that the relationship between the position vectors is r1 = r2 + r, which can be rewritten r = r1 - r2. Its second derivative yields
r = r1 - r2.
From Newton's law of gravity we have the following two relationships:
mr1 = -GMEmr/r2 - GMmd/d2
MEr2 = -GMME R/R2.
Dividing Eq. 2 by m and Eq. 3 by ME we get
r1 = -GME r/r2 - GM d/d2
r2 = -GM R/R2.
Substituting Eq. 4 and Eq. 5 into Eq. 1 we get the force per unit mass acting on m:
r = -GME r/r2 - GM d/d2 - (-GM R/R2),
which factors to
r = -GME r/r2 - GM( d/d2 - R/R2 ) º aE + aM.
Note that aE is the force per unit mass caused by the earth, which seeks to keep the earth in a spherical shape, whereas aM is the force per unit mass caused by the moon, known as the tide. It is this tidal force that causes the shape of the earth to deviate from that of a perfect sphere. Note that aM is the difference between the earth's overall attraction to the moon, and the mass element m's attraction to the moon.
Consider aM = - GM( d/d2 - R/R2 )on the earth's equator, where q is 90° (see Fig. 1 and Fig. 2).
To get aM in terms of i and j, we eliminate d and R. From Fig. 2,
d = -[R - r cos f]i + r sin f j,
and
d = [r2 + R2 - 2rR cos f]1/2,
so that
d = = .
Since R = -i, we have
aM = -GM + i + j .
You will analyze this field in the problem set that follows.