Tidal Torques

The fact that there is a time-lag between the Moon passing overhead and the occurrence of a high tide suggests the physical scenario illustrated in Figure 20. According to this scenario, the Moon, which is of mass $m'$, and which is treated as a point particle, orbits the Earth (it actually orbits the center of mass of the Earth-Moon system, but this amounts to almost the same thing) in an approximately circular orbit of radius $R$. Moreover, the orbital angular velocity of the Moon is [see Equation (322)]

$$\omega =\frac{(G\,M)^{\,1/2}}{R^{\,3/2}},$$ (343)

where $M=m+m'$ is the total mass of the Earth-Moon system. The Earth is treated as a sphere of mass $m$, and radius $a$, that rotates daily about its axis (which is approximately normal to the orbital plane of the Moon) at the angular velocity $\Omega$. Note, incidentally, that the Earth rotates in the same sense that the Moon orbits, as indicated in the figure. Now, as we saw in the previous section, spatial gradients in the Moon's gravitational field produce a slight tidal elongation of the Earth. However, because of the finite inertia of the oceans, this elongation does not quite line up along the axis joining the centers of the Earth and Moon. In fact, since $\Omega>\omega$, the tidal elongation is carried ahead (in the sense defined by the Earth's rotation) of this axis by some small angle $\delta$ (say), as shown in the figure.

Figure 20: Origin of tidal torque.

Defining a spherical coordinate system, $r$, $\theta$, $\phi$, whose origin is the center of the Earth, and which is orientated such that the Earth-Moon axis always corresponds to $\theta=0$ (see Figure 17), the Earth's external gravitational potential is [cf., Equation (140)]

$$\Phi(r,\theta) = -\frac{G\,m}{r} +\frac{2}{5}\,\frac{G\,m\,a^2}{r^3}\,\epsilon \left[3\,\cos^2(\theta-\delta)-1\right],$$ (344)

where $\epsilon$ is the ellipticity induced by the tidal field of the Moon. Note that the second term on the right-hand side of the above expression is the contribution of the Earth's tidal bulge, which attains its maximum amplitude at $\theta=\delta$, rather than $\theta=0$, because of the aforementioned misalignment between the bulge and the Earth-Moon axis. Equations (339) and (344) can be combined to give

$$\Phi(r,\theta) = -\frac{G\,m}{r} -\frac{3}{4}\,\frac{G\,m'\,... ...(\frac{a}{R}\right)^3 \left[3\,\cos^2(\theta-\delta)-1\right].$$ (345)

Now, from (84), the torque that the Earth's gravitational field exerts on the Moon is

$$\tau = \left.- m'\,\frac{\partial\Phi}{\partial\theta}\right... ...}{2}\,\frac{G\,m'^{\,2}}{a}\left(\frac{a}{R}\right)^6\,\delta,$$ (346)

where use has been made of Equation (345), as well as the fact that $\delta$ is a small angle. Note that there is zero torque in the absence of a misalignment between the Earth's tidal bulge and the Earth-Moon axis. The torque $\tau$ acts to increase the Moon's orbital angular momentum. By conservation of angular momentum, an equal and opposite torque, $-\tau$, is applied to the Earth, and acts to decrease its rotational angular momentum. Note that if the Moon were sufficiently close to the Earth that its orbital angular velocity exceeded the Earth's rotational angular velocity (i.e., if $\omega >\Omega$) then the phase-lag between the tides and the Moon would cause the tidal bulge to fall slightly behind the Earth-Moon axis (i.e., $\delta<0$). In this case, the gravitational torque would act to reduce the Moon's orbital angular momentum, and to increase the Earth's rotational angular momentum.

The Earth's rotational equation of motion is

$$I\,\dot{\Omega} = -\tau,$$ (347)

where $I$ is its moment of inertia. Very crudely approximating the Earth as a uniform sphere, we have $I=(2/5)\,m\,a^2$. Hence, the previous two equations can be combined to give

$$\frac{\dot\Omega}{\Omega} \simeq -\frac{45}{4}\,\frac{\omega... ...\left(\frac{m'}{m}\right)^2\left(\frac{a}{R}\right)^3\,\delta,$$ (348)

where use has been made of Equation (343), as well as the fact that $m\gg m'$. Now, a time-lag of 12 minutes between the Moon being overhead and the occurrence of a high-tide implies a phase-lag of $\delta\sim 0.05$ radians (i.e., $\delta \sim 3^\circ$). Hence, employing the observed values $m=5.97\times 10^{24}\,{\rm kg}$, $m'=7.35\times 10^{22}\,{\rm kg}$, $a=6.37\times 10^6\,{\rm m}$, $R=3.84\times 10^8\,{\rm m}$, $\Omega = 7.27\times 10^{-5}\,{\rm rad./s}$, and $\omega=2.67\times 10^{-6}\,{\rm rad./s}$, we find that

$$\frac{\dot{\Omega}}{\Omega} \simeq -3.8\times 10^{-17}\,{\rm s}^{-1}.$$ (349)

It follows that, under the influence of the tidal torque, the Earth's rotation is gradually decelerating. Indeed, according to the above estimate, the length of a day should be increasing at the rate of about 10 milliseconds a century. (In fact, as a consequence of the severe approximations made in the our calculation, this estimate is about a factor of 5 too high.) The time-scale for the tidal torque to significantly reduce the Earth's rotational angular velocity is estimated to be

$$T \simeq \frac{\Omega}{\vert\dot{\Omega}\vert} \sim 8\times 10^8\,{\rm years}.$$ (350)

This time-scale is comparable with the Earth's age, which is thought to be $4.5\times 10^9$ years. Hence, we conclude that, whilst the Earth is certainly old enough for the tidal torque to have significantly reduced its rotational angular velocity, it is plausible (especially when we take into account that our estimate for $\vert\dot{\Omega}\vert$ is 5 times too high) that it is not sufficiently old for the torque to have driven the Earth-Moon system to a final steady-state. Such a state, in which the Earth's rotational angular velocity matches the Moon's orbital angular velocity, is termed synchronous. In a synchronous state, the Moon would appear stationary to an observer on the Earth's surface, and, hence, there would be no tides (from the observer's perspective), no phase-lag, and no tidal torque.

Up to now, we have concentrated on the effect of the tidal torque on the rotation of the Earth. Let us now examine its effect on the orbit of the Moon. Now, the total angular momentum of the Earth-Moon system is

$$L \simeq \frac{2}{5}\,m\,a^2\,\Omega + m'\,R^{\,2}\,\omega,$$ (351)

where the first term on the right-hand side is the rotational angular momentum of the Earth, and the second the orbital angular momentum of the Moon. Of course, $L$ is a conserved quantity. Moreover, $\omega$ and $R$ are related according to Equation (343). It follows that

$$\frac{\dot{R}}{R} \simeq - \frac{4}{5}\,\frac{m}{m'}\left(\f... ...{\dot{\Omega}}{\omega}\simeq 1.9\times 10^{-17}\,{\rm s}^{-1},$$ (352)

where use has been made of (349). In other words, the tidal torque causes the radius of the Moon's orbit to gradually increase. According to the above estimate, this increase should take place at the rate of about $22\,{\rm cm}$ a year. (Again, this estimate is a factor of 5 too large.) We also have

$$\frac{\dot{\omega}}{\omega} = -\frac{3}{2}\,\frac{\dot{R}}{R} \sim -2.8\times 10^{-17}\,{\rm s}^{-1}.$$ (353)

In other words, the tidal torque produces a gradual angular deceleration in the Moon's orbital motion. According to the above estimate, this deceleration should take place at the rate of 150 arc seconds per century squared. (As before, this estimate is a factor of 5 too large.)

The net rate at which the tidal torques acting on the Moon and the Earth do work is

$$\dot{E} = \tau\,(\omega-\Omega).$$ (354)

Note that $\dot{E}<0$, since $\tau>0$ and $\Omega>\omega$. This implies that the deceleration of the Earth's rotation, and the Moon's orbital motion, that are induced by the tidal torques are necessarily associated with the dissipation of energy. Most of this dissipation occurs in a turbulent boundary layer at the bottoms of shallow seas such as the European shelf around the British Isles, the Patagonian shelf off Argentina, and the Bering Sea. In the absence of dissipation, there can be no steady phase-lag between the Moon and the Earth's tidal bulge, and, therefore, no tidal torques.

Of course, we would expect spatial gradients in the gravitational field of the Earth to generate a tidal bulge in the Moon. We would also expect dissipative effects to generate a phase-lag between this bulge and the Earth. This would allow the Earth to exert a gravitational torque that acts to drive the Moon toward a synchronous state in which its rotational angular velocity matches its orbital angular velocity. By analogy with the previous analysis, the de-spinning rate of the Moon is estimated to be

$$\frac{\dot{\Omega}' }{\Omega'} \simeq -\frac{45}{4}\,\frac{\... ...'}{R}\right)^3\,\delta \sim -2.3\times 10^{-13}\,{\rm s}^{-1},$$ (355)

where $\Omega'$ is the Moon's rotational angular velocity, $a'=1.74\times 10^6\,{\rm m}$ the lunar radius, and $\delta$ the phase-lag. The above numerical estimate is made with the guesses $\Omega'=10\,\omega$ and $\delta=0.01$. Thus, the time required for the Moon to achieve a synchronous state is

$$T' \simeq \frac{\Omega'}{\vert\dot{\Omega}'\vert}\sim 10^5\,{\rm years}.$$ (356)

This is considerably less than the age of the Moon. Hence, it is not surprising that the Moon has actually achieved a synchronous state, as evidenced by the fact that the same side of the Moon is always visible from the Earth.