Tidal Elongation of Earth

Consider two point objects, of masses $m$ and $m'$, executing circular orbits about their common center of mass, $C$, with angular velocity $\omega$. Let $a$ be the distance between the masses, and $\rho$ the distance between point $C$ and mass $m$. See Figure 1.24. We know from Section 1.10.8 that

$$\omega^{2} = \frac{G\,M}{a^{3}},$$ (1.450)

and

$$\rho = \frac{m'}{M}\,a,$$ (1.451)

where $M=m+m'$.

Figure 1.24: Two orbiting masses.

Let us transform to a non-inertial frame of reference that rotates, about an axis perpendicular to the orbital plane and passing through $C$, at the angular velocity $\omega$. In this reference frame, both masses appear to be stationary. Consider mass $m$. In the rotating frame, this mass experiences a gravitational acceleration

$$a_g = \frac{G\,m'}{a^{2}}$$ (1.452)

directed toward the center of mass, and a centrifugal acceleration (see Section 1.10.2)

$$a_c = \omega^{2}\,\rho$$ (1.453)

directed away from the center of mass. However, it is easily demonstrated, using Equations (1.450) and (1.451), that

$$a_c=a_g.$$ (1.454)

In other words, the gravitational and centrifugal accelerations balance, as must be the case if mass $m$ is to remain stationary in the rotating frame. Let us investigate how this balance is affected if the masses $m$ and $m'$ have finite spatial extents.

Let the center of the mass distribution $m'$ lie at $A$, the center of the mass distribution $m$ at $B$, and the center of mass at $C$. See Figure 1.25. We wish to calculate the centrifugal and gravitational accelerations at some point $D$ in the vicinity of point $B$. It is convenient to adopt spherical coordinates, centered on point $B$, and aligned such that the $z$-axis coincides with the line $BA$.

Figure 1.25: Calculation of tidal forces.

Let us assume that the mass distribution $m$ is orbiting around $C$, but is not rotating about an axis passing through its center of mass, in order to exclude rotational flattening from our analysis. If this is the case then it is easily seen that each constituent point of $m$ executes circular motion of angular velocity $\omega$ and radius $\rho$. See Figure 1.26. Hence, each point experiences the same centrifugal acceleration:

$${\bf g}_c = - \omega^{2}\,\rho\,{\bf e}_z.$$ (1.455)

It follows that

$${\bf g}_c = - \nabla\chi',$$ (1.456)

where

$$\chi' = \omega^{2}\,\rho\,z$$ (1.457)

is the centrifugal potential and $z=r\,\cos\theta$. The centrifugal potential can also be written

$$\chi' = \frac{G\,m'}{a}\frac{r}{a}\,P_1(\cos\theta),$$ (1.458)

where

$$P_1(x)= x$$ (1.459)

is a Legendre polynomial of degree 1.

Figure: 1.26 The center $B$ of mass distribution $m$ orbits about the center of mass $C$ in a circle of radius $\rho$. If $m$ is non-rotating then a non-central point $D$ maintains a constant spatial relationship to $B$, such that $D$ orbits some point $C'$ that has the same spatial relationship to $C$ that $D$ has to $B$, in a circle of radius $\rho$.

The gravitational acceleration at point $D$ due to mass $m'$ is given by

$${\bf g}_g = -\nabla{\mit\Phi}',$$ (1.460)

where the gravitational potential takes the form

$${\mit\Phi}' = -\frac{G\,m'}{a'}.$$ (1.461)

(See Section 1.8.5.) Here, $a'$ is the distance between points $A$ and $D$. The gravitational potential generated by the mass distribution $m'$ is the same as that generated by an equivalent point mass at $A$, as long as the distribution is spherically symmetric, which we shall assume to be the case. (See Section 1.8.3.)

Now,

$${\bf a}' = {\bf a} - {\bf r},$$ (1.462)

where ${\bf a}'$ is the vector $\stackrel{\displaystyle \rightarrow}{DA}$, and ${\bf a}$ the vector $\stackrel{\displaystyle \rightarrow}{BA}$. See Figure 1.25. It follows that

$$a'^{\,-1} = \left(a^{2} - 2\,{\bf a}\cdot{\bf r}+ r^{2}\right)^{-1/2} = \left(a^{2} - 2\,a\,r\,\cos\theta+ r^{2}\right)^{-1/2}.$$ (1.463)

Expanding in powers of $r/a$, it is easily demonstrated that

$${\mit\Phi}' \simeq - \frac{G\,m'}{a}\left[1+ \frac{r}{a}\,P_1(\cos\theta) + \frac{r^{2}}{a^{2}}\,P_2(\cos\theta)\right],$$ (1.464)

to second order in $r/a$, where the Legendre polynomials $P_2(x)$ and $P_1(x)$ are defined in Equations (1.329) and (1.459), respectively.

Adding $\chi'$ and ${\mit\Phi}'$, we find that

$$\chi=\chi'+{\mit\Phi}' \simeq - \frac{G\,m'}{a}\left[1 + \frac{r^{2}}{a^{2}}\,P_2(\cos\theta)\right],$$ (1.465)

to second order in $r/a$. Note that $\chi$ is the potential due to the net externally generated force acting on the mass distribution $m$ in the rotating frame. This potential is constant up to first order in $r/a$, because the first-order variations in $\chi'$ and ${\mit\Phi}'$ cancel each other. The cancellation is a manifestation of the balance between the centrifugal and gravitational accelerations in the equivalent point mass problem discussed previously. However, this balance is only exact at the center of the mass distribution $m$. Away from the center, the centrifugal acceleration remains constant, whereas the gravitational acceleration increases with increasing $z$. At positive $z$, the gravitational acceleration is larger than the centrifugal acceleration, giving rise to a net acceleration in the $+z$-direction. Likewise, at negative $z$, the centrifugal acceleration is larger than the gravitational, giving rise to a net acceleration in the $-z$-direction. It follows that the mass distribution $m$ is subject to a residual acceleration, represented by the second-order variation in Equation (1.465), that acts to elongate it along the $z$-axis. This effect is known as tidal elongation.

Suppose that the mass distribution $m$ is a uniform fluid sphere of radius $R$. Let us estimate the elongation of this distribution due to the tidal potential specified in Equation (1.465), which (neglecting constant terms) can be written

$$\chi(r,\theta)=\frac{G\,m}{R}\,\zeta\left(\frac{r}{R}\right)^2 P_2(\cos\theta).$$ (1.466)

Here, the dimensionless parameter

$$\zeta = - \frac{m'}{m}\left(\frac{R}{a}\right)^3$$ (1.467)

is (minus) the typical ratio of the tidal acceleration to the gravitational acceleration at $r\simeq R$. Let us assume that $\vert\zeta\vert\ll 1$. By analogy with the analysis in Section 1.10.2, in the presence of the tidal potential, the distribution becomes slightly spheroidal in shape, such that its outer boundary satisfies Equation (1.328). Moreover, the induced ellipticity, $\epsilon$, of the distribution is related to the normalized amplitude, $\zeta$, of the tidal potential according to

$$\epsilon = \frac{15}{4}\,\zeta.$$ (1.468)

[See Equation (1.364).]

Consider the tidal elongation of the Earth due to the Moon. In this case, we have $R=6.371\times 10^6\,{\rm m}$, $a=3.844\times 10^8\,{\rm m}$, $m=5.972\times 10^{24}\,{\rm kg}$, and $m'=7.324\times 10^{22}\,{\rm kg}$. Hence, we find that

$$\zeta = -5.58\times 10^{-8}.$$ (1.469)

Thus, according to Equation (1.468), the ellipticity of the Earth induced by the tidal effect of the Moon is

$$\epsilon = \frac{15}{4}\,\zeta\simeq -2.09\times 10^{-7}.$$ (1.470)

The fact that $\epsilon$ is negative implies that the Earth is elongated along the $z$-axis; that is, along the axis joining its center to that of the Moon. [See Equation (1.328).] If $R_+$ and $R_-$ are the greatest and least radii of the Earth, respectively, due to this elongation, then

$${\mit\Delta} R = R_+-R_- = -\epsilon\,R = 1.33\,{\rm m}.$$ (1.471)

Thus, we predict that the tidal effect of the Moon (which is actually due to spatial gradients in the Moon's gravitational field) causes the Earth to elongate along the axis joining its center to that of the Moon by about $133$ centimeters. This turns out to be an overestimate because the tidal potential of the Moon is not strong enough to force the rocks that make up the Earth to respond to it as a fluid.

Consider the tidal elongation of the Earth due to the Sun. In this case, we have $R=6.371\times 10^6\,{\rm m}$, $a=1.496\times 10^{11}\,{\rm m}$, $m=5.972\times 10^{24}\,{\rm kg}$, and $m'=1.989\times 10^{30}\,{\rm kg}$. Hence, we find that

$$\zeta = -2.65\times 10^{-8},$$ (1.472)

and

$$\epsilon = \frac{15}{4}\,\zeta =-9.95\times 10^{-8},$$ (1.473)

with

$${\mit\Delta} R = R_+-R_- = -\epsilon\,R = 0.63\,{\rm m}.$$ (1.474)

Again, this turns out to be an overestimate because the tidal potential of the Sun is not strong enough to force the rocks that make up the Earth to respond to it as a fluid. Nevertheless, we can conclude that the tidal elongation of the Earth due to the Sun is about half that due to the Moon.

Because the Earth's oceans are liquid, their tidal elongation is significantly larger than that of the underlying land. Hence, the oceans rise, relative to the land, in the region of the Earth closest to the Moon, and also in the region furthest away. Because the Earth is rotating, while the tidal bulge of the oceans remains relatively stationary, the Moon's tidal effect causes the ocean at a given point on the Earth's surface to rise and fall twice daily, giving rise to the phenomenon known as the tides. There is also an oceanic tidal bulge due to the Sun that is about half as large as that due to the Moon. Consequently, ocean tides are particularly high when the Sun, the Earth, and the Moon lie approximately in a straight line, so that the tidal effects of the Sun and the Moon reinforce one another. This occurs at a new moon, or at a full moon. These type of tides are called spring tides (the name has nothing to do with the season). Conversely, ocean tides are particularly low when the Sun, the Earth, and the Moon form a right angle, so that the tidal effects of the Sun and the Moon partially cancel one another. These type of tides are called neap tides. Generally speaking, we would expect two spring tides and two neap tides per month.