Tidal Elongation

Consider two point masses, $m$ and $m'$, executing circular orbits about their common center of mass, $C$, with angular velocity $\omega$. Let $R$ be the distance between the masses, and $\rho$ the distance between point $C$ and mass $m$--see Figure 41. We know, from Section 6.3, that

$$\omega^2 = \frac{G\,M}{R^3},$$ (927)

and

$$\rho = \frac{m'}{M}\,R,$$ (928)

where $M=m+m'$.

Figure 41: Two orbiting masses.

Let us transform to a non-inertial frame of reference which 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

$$a_g = \frac{G\,m'}{R^2}$$ (929)

directed toward the center of mass, and a centrifugal acceleration (see Chapter 7)

$$a_c = \omega^2\,\rho$$ (930)

directed away from the center of mass. However, it is easily demonstrated, using Equations (927) and (928), that

$$a_c=a_g.$$ (931)

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 42. 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 42: 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, 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 43. Hence, each constituent point experiences the same centrifugal acceleration: i.e.,

$${\bf g}_c = - \omega^2\,\rho\,{\bf e}_z.$$ (932)

It follows that

$${\bf g}_c = - \nabla\chi,$$ (933)

where

$$\chi = \omega^2\,\rho\,z$$ (934)

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

$$\chi = \frac{G\,m'}{R}\frac{r}{R}\,P_1(\cos\theta).$$ (935)

Figure 43: The center $B$ of the mass distribution $m$ orbits about the center of mass $C$ in a circle of radius $\rho$. If the mass distribution is non-rotating then a non-central point $D$ must maintain a constant spatial relationship to $B$. It follows that point $D$ orbits some point $C'$, which 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

$${\bf g}_g = -\nabla\Phi',$$ (936)

where the gravitational potential takes the form

$$\Phi' = -\frac{G\,m'}{R'}.$$ (937)

Here, $R'$ is the distance between points $A$ and $D$. Note that 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.

Now,

$${\bf R}' = {\bf R} - {\bf r},$$ (938)

where ${\bf R}'$ is the vector $\stackrel{\displaystyle \rightarrow}{DA}$, and ${\bf R}$ the vector $\stackrel{\displaystyle \rightarrow}{BA}$--see Figure 42. It follows that

$$R'^{\,-1} = \left(R^2 - 2\,{\bf R}\cdot{\bf r}+ r^2\right)^{-1/2} = \left(R^2 - 2\,R\,r\,\cos\theta+ r^2\right)^{-1/2}.$$ (939)

Expanding in powers of $r/R$, we obtain

$$R'^{\,-1} = \frac{1}{R}\sum_{n=0,\infty} \left(\frac{r}{R}\right)^n P_n(\cos\theta).$$ (940)

Hence,

$$\Phi' \simeq - \frac{G\,m'}{R}\left[1+ \frac{r}{R}\,P_1(\cos\theta) + \frac{r^2}{R^2}\,P_2(\cos\theta)\right]$$ (941)

to second-order in $r/R$.

Adding $\chi$ and $\Phi'$, we obtain

$$\chi+\Phi' \simeq - \frac{G\,m'}{R}\left[1 + \frac{r^2}{R^2}\,P_2(\cos\theta)\right]$$ (942)

to second-order in $r/R$. Note that $\chi+\Phi'$ is the potential due to the net external force acting on the mass distribution $m$. This potential is constant up to first-order in $r/R$, because the first-order variations in $\chi$ and $\Phi'$ cancel one another. The cancellation is a manifestation of the balance between the centrifugal and gravitational accelerations in the equivalent point mass problem discussed above. 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 $z$. Hence, at positive $z$, the gravitational acceleration is larger than the centrifugal, 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 (942), which acts to elongate it along the $z$-axis. This effect is known as tidal elongation.

In order to calculate the tidal elongation of the mass distribution $m$ we need to add the potential, $\chi+\Phi'$, due to the external forces, to the gravitational potential, $\Phi$, generated by the distribution itself. Assuming that the mass distribution is spheroidal with mass $m$, mean radius $a$, and ellipticity $\epsilon$, it follows from Equations (901), (911), and (942) that the total surface potential is given by

$$\chi +\Phi'+\Phi \simeq - \frac{G\,m}{a} - \frac{G\,m'}{R}$$

$$-\frac{4}{15}\,\frac{G\,m}{a}\,\epsilon\,P_2(\cos\theta) - \frac{G\,m'\,a^2}{R^3}\,P_2(\cos\theta),$$ (943)

where we have treated $\epsilon$ and $a/R$ as small quantities. As before, the condition for equilibrium is that the total potential be constant over the surface of the spheroid. Hence, we obtain

$$\epsilon = -\frac{15}{4}\,\frac{m'}{m}\left(\frac{a}{R}\right)^3$$ (944)

as our prediction for the ellipticity induced in a self-gravitating spherical mass distribution of total mass $m$ and radius $a$ by a second mass, $m'$, which is in a circular orbit of radius $R$ about the distribution. Thus, if $a_+$ is the maximum radius of the distribution, and $a_-$ the minimum radius (see Figure 44), then

$$\frac{a_+-a_-}{a} = -\epsilon = \frac{15}{4}\,\frac{m'}{m}\left(\frac{a}{R}\right)^3.$$ (945)

Figure 44: Tidal elongation.

Consider the tidal elongation of the Earth due to the Moon. In this case, we have $a=6.37\times 10^6\,{\rm m}$, $R=3.84\times 10^8\,{\rm m}$, $m=5.97\times 10^{24}\,{\rm kg}$, and $m'=7.35\times 10^{22}\,{\rm kg}$. Hence, we calculate that $-\epsilon=2.1\times 10^{-7}$, or

$$\Delta a = a_+-a_- = -\epsilon\,a = 1.34\,{\rm m}.$$ (946)

We, thus, predict that tidal forces due to the Moon cause the Earth to elongate along the axis joining its center to the Moon by about $1.3$ meters. Since water is obviously more fluid than rock (especially on relatively short time-scales) most of this elongation takes place in the oceans rather than in 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. Since the Earth is rotating, whilst the tidal bulge of the oceans remains relatively stationary, the Moon's tidal force causes the ocean at a given point on the Earth's surface to rise and fall, by about a meter, twice daily, giving rise to the phenomenon known as the tides.

Consider the tidal elongation of the Earth due to the Sun. In this case, we have $a=6.37\times 10^6\,{\rm m}$, $R=1.50\times 10^{11}\,{\rm m}$, $m=5.97\times 10^{24}\,{\rm kg}$, and $m'=1.99\times 10^{30}\,{\rm kg}$. Hence, we calculate that $-\epsilon=9.6\times 10^{-8}$, or

$$\Delta a = a_+-a_- = -\epsilon\,a = 0.61\,{\rm m}.$$ (947)

Thus, the tidal elongation due to the Sun is about half that due to the Moon. It follows that the 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 (note that the name has nothing to do with the season). Conversely, the 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.

In reality, the amplitude of the tides varies significantly from place to place on the Earth's surface, due to the presence of the continents, which impede the flow of the oceanic tidal bulge around the Earth. Moreover, there is a time-lag of approximately 12 minutes between the Moon being directly overhead (or directly below) and high tide, because of the finite inertia of the oceans. Similarly, the time-lag between a spring tide and a full moon, or a new moon, can be up to 2 days.