Precession and Forced Nutation of the Earth

Consider the Earth-Sun system--see Figure 45. From a geocentric viewpoint, the Sun orbits the Earth counter-clockwise (looking from the north), once per year, in an approximately circular orbit of radius $a_s= 1.50\times 10^{11}\,{\rm m}$. In astronomy, the plane of the Sun's apparent orbit relative to the Earth is known as the ecliptic plane. Let us define non-rotating Cartesian coordinates, centered on the Earth, which are such that the $x$- and $y$-axes lie in the ecliptic plane, and the $z$-axis is normal to this plane (in the sense that the Earth's north pole lies at positive $z$). It follows that the $z$-axis is directed toward a point in the sky (located in the constellation Draco) known as the north ecliptic pole. In the following, we shall treat the $Oxyz$ system as inertial. This is a reasonable approximation because the orbital acceleration of the Earth is much smaller than the acceleration due to its diurnal rotation. It is convenient to parameterize the instantaneous position of the Sun in terms of a counter-clockwise (looking from the north) azimuthal angle $\lambda_s$ that is zero on the positive $x$-axis--see Figure 45.

Figure 45: The Earth-Sun system.

Let $\mbox{\boldmath$\omega$}$ be the Earth's angular velocity vector due to its daily rotation. This vector makes an angle $\theta$ with the $z$-axis, where $\theta = 23.44^\circ$ is the mean inclination of the ecliptic to the Earth's equatorial plane. Suppose that the projection of $\mbox{\boldmath$\omega$}$ onto the ecliptic plane subtends an angle $\phi$ with the $y$-axis, where $\phi$ is measured in a counter-clockwise (looking from the north) sense--see Figure 45. The orientation of the Earth's axis of rotation (which is, of course, parallel to $\mbox{\boldmath$\omega$}$) is thus determined by the two angles $\theta$ and $\phi$. Note, however, that these two angles are also Euler angles, in the sense given in Chapter 8. Let us examine the Earth-Sun system at an instant in time, $t=0$, when $\phi =0$: i.e., when $\mbox{\boldmath$\omega$}$ lies in the $y$-$z$ plane. At this particular instant, the $x$-axis points towards the so-called vernal equinox, which is defined as the point in the sky where the ecliptic plane crosses the projection of the Earth's equator (i.e., the plane normal to $\mbox{\boldmath$\omega$}$) from south to north. A counter-clockwise (looking from the north) angle in the ecliptic plane that is zero at the vernal equinox is generally known as an ecliptic longitude. Thus, $\lambda_s$ is the Sun's ecliptic longitude.

According to Equation (926), the potential energy of the Earth-Sun system is written

$$U = M_s\,\Phi = - \frac{G\,M_s\,M}{a_s} + \frac{G\,M_s\,(I_\parallel-I_\perp)}{a_s^{\,3}}\,P_2[\cos(\gamma_s)],$$ (957)

where $M_s$ is the mass of the Sun, $M$ the mass of the Earth, $I_\parallel$ the Earth's moment of inertia about its axis of rotation, $I_\perp$ the Earth's moment of inertia about an axis lying in its equatorial plane, and $P_2(x)=(1/2)\,(3\,x^2-1)$. Furthermore, $\gamma_s$ is the angle subtended between $\mbox{\boldmath$\omega$}$ and ${\bf r}_s$, where ${\bf r}_s$ is the position vector of the Sun relative to the Earth.

It is easily demonstrated that (with $\phi =0$)

$$\mbox{\boldmath$\omega$}= \omega\,(0,\,\sin\theta,\,\cos\theta),$$ (958)

and

$${\bf r}_s = a_s\,(\cos\lambda_s,\,\sin\lambda_s,\,0).$$ (959)

Hence,

$$\cos\gamma_s =\frac{\mbox{\boldmath$\omega$}\cdot{\bf r}_s}{... ...omega$}\vert\,\vert{\bf r}_s\vert}= \sin\theta\,\sin\lambda_s,$$ (960)

giving

$$U = - \frac{G\,M_s\,M}{a_s} + \frac{G\,M_s\,(I_\parallel-I_\perp)}{2\,a_s^{\,3}}\,(3\,\sin^2\theta\,\sin^2\lambda_s -1).$$ (961)

Now, we are primarily interested in the motion of the Earth's axis of rotation over time-scales that are much longer than a year, so we can average the above expression over the Sun's orbit to give

$$U = - \frac{G\,M_s\,M}{a_s} + \frac{G\,M_s\,(I_\parallel-I_\perp)}{2\,a_s^{\,3}}\,\left(\frac{3}{2}\,\sin^2\theta -1\right)$$ (962)

(since the average of $\sin^2\lambda_s$ over a year is $1/2$). Thus, we obtain

$$U = U_0 - \epsilon\,\alpha_s\,\cos(2\,\theta),$$ (963)

where $U_0$ is a constant, and

$$\alpha_s = \frac{3}{8}\,I_\parallel\,n_s^{\,2}.$$ (964)

Here,

$$\epsilon = \frac{I_{\parallel}-I_\perp}{I_\parallel}=0.00335$$ (965)

is the Earth's ellipticity, and

$$n_s =\frac{d\lambda_s}{dt}= \left(\frac{G\,M_s}{a_s^{\,3}}\right)^{1/2}$$ (966)

the Sun's apparent orbital angular velocity.

According to Section 8.9, the rotational kinetic energy of the Earth can be written

$$K = \frac{1}{2}\left(I_\perp\,\dot{\theta}^2 + I_\perp\,\sin^2\theta\,\dot{\phi}^2 + I_\parallel\,\omega^2\right),$$ (967)

where the Earth's angular velocity

$$\omega = \cos\theta\,\dot{\phi} + \dot{\psi}$$ (968)

is a constant of the motion. Here, $\psi$ is the third Euler angle. Hence, the Earth's Lagrangian takes the form

$${\cal L} = K - U = \frac{1}{2}\left(I_\perp\,\dot{\theta}^2 ... ...parallel\,\omega^2\right) +\epsilon\,\alpha_s\,\cos(2\,\theta)$$ (969)

where any constant terms have been neglected. One equation of motion which can immediately be derived from this Lagrangian is

$$\frac{d}{dt}\!\left(\frac{\partial{\cal L}}{\partial \dot{\theta}}\right)- \frac{\partial {\cal L}}{\partial\theta} = 0,$$ (970)

which reduces to

$$I_\perp\,\ddot{\theta} - \frac{\partial {\cal L}}{\partial \theta} = 0.$$ (971)

Consider steady precession of the Earth's rotational axis, which is characterized by $\dot{\theta}=0$, with both $\dot{\phi}$ and $\dot{\psi}$ constant. It follows, from the above equation, that such motion must satisfy the constraint

$$\frac{\partial {\cal L}}{\partial\theta} = 0.$$ (972)

Thus, we obtain

$$\frac{1}{2}\,I_\perp\,\sin(2\,\theta)\,\dot{\phi}^{\,2} - I_... ...mega\,\dot{\phi} - 2\,\epsilon\,\alpha_s\,\sin(2\,\theta) = 0,$$ (973)

where use has been made of Equations (968) and (969). Now, as can easily be verified after the fact, $\vert\dot{\phi}\vert\ll \omega$, so the above equation reduces to

$$\dot{\phi} \simeq -\frac{4\,\epsilon\,\alpha_s\,\cos\theta}{I_{\parallel}\,\omega} = \Omega_\phi,$$ (974)

which can be integrated to give

$$\phi \simeq- \Omega_\phi\,t,$$ (975)

where

$$\Omega_\phi = \frac{3}{2}\,\frac{\epsilon\,n_s^{\,2}}{\omega}\,\cos\theta,$$ (976)

and use has been made of Equation (964). According to the above expression, the mutual interaction between the Sun and the quadrupole gravitational field generated by the Earth's slight oblateness causes the Earth's axis of rotation to precess steadily about the normal to the ecliptic plane at the rate $-\Omega_\phi$. The fact that $-\Omega_\phi$ is negative implies that the precession is in the opposite direction to the direction of the Earth's rotation and the Sun's apparent orbit about the Earth. Incidentally, the interaction causes a precession of the Earth's rotational axis, rather than the plane of the Sun's orbit, because the Earth's axial moment of inertia is much less than the Sun's orbital moment of inertia. The precession period in years is given by

$$T_\phi({\rm yr}) = \frac{n_s}{\Omega_\phi} = \frac{2\,T_s({\rm day})}{3\,\epsilon\,\cos\theta},$$ (977)

where $T_s({\rm day}) = \omega/n_s = 365.24$ is the Sun's orbital period in days. Thus, given that $\epsilon=0.00335$ and $\theta = 23.44^\circ$, we obtain

$$T_\phi\simeq 79,200\,\,{\rm years}.$$ (978)

Unfortunately, the observed precession period of the Earth's axis of rotation about the normal to the ecliptic plane is approximately 25,800 years, so something is clearly missing from our model. It turns out that the missing factor is the influence of the Moon.

Using analogous arguments to those given above, the potential energy of the Earth-Moon system can be written

$$U = - \frac{G\,M_m\,M}{a_m} + \frac{G\,M_m\,(I_\parallel-I_\perp)}{a_m^{\,3}}\,P_2[\cos(\gamma_m)],$$ (979)

where $M_m$ is the lunar mass, and $a_m$ the radius of the Moon's (approximately circular) orbit. Furthermore, $\gamma_m$ is the angle subtended between $\mbox{\boldmath$\omega$}$ and ${\bf r}_m$, where

$$\mbox{\boldmath$\omega$}= \omega\,\left(-\sin\theta\,\sin\phi,\,\sin\theta\,\cos\phi,\,\cos\theta\right)$$ (980)

is the Earth's angular velocity vector, and ${\bf r}_m$ is the position vector of the Moon relative to the Earth. Here, for the moment, we have retained the $\phi$ dependence in our expression for $\mbox{\boldmath$\omega$}$ (since we shall presently differentiate by $\phi$, before setting $\phi =0$). Now, the Moon's orbital plane is actually slightly inclined to the ecliptic plane, the angle of inclination being $\iota_m=5.16^\circ$. Hence, we can write

$${\bf r}_m \simeq a_m\,\left(\cos\lambda_m,\,\sin\lambda_m,\,\iota_m\,\sin(\lambda_m-\varpi_n)\right),$$ (981)

to first order in $\iota_m$, where $\lambda_m$ is the Moon's ecliptic longitude, and $\varpi_n$ is the ecliptic longitude of the lunar ascending node, which is defined as the point on the lunar orbit where the Moon crosses the ecliptic plane from south to north. Of course, $\lambda_m$ increases at the rate $n_m$, where

$$n_m = \frac{d\lambda_m}{dt}\simeq \left(\frac{G\,M}{a_m^{\,3}}\right)^{1/2}$$ (982)

is the Moon's orbital angular velocity. It turns out that the lunar ascending node precesses steadily, in the opposite direction to the Moon's orbital rotation, in such a manner that it completes a full circuit every $18.6$ years. This precession is caused by the perturbing influence of the Sun--see Chapter 14. It follows that

$$\frac{d\varpi_n}{dt}= -\Omega_n,$$ (983)

where $2\pi/\Omega_n=18.6\,{\rm years}$. Now, from (980) and (981),

$$\cos\gamma_m = \frac{\mbox{\boldmath$\omega$}\cdot{\bf r}_m}... ...\lambda_m-\phi)+\iota_m\,\cos\theta\,\sin(\lambda_m-\varpi_n),$$ (984)

so (979) yields

$$U \simeq - \frac{G\,M_m\,M}{a_m} + \frac{G\,M_m\,(I_\parallel-I_\perp)}{2\,a_m^{\,3}}\left[3\,\sin^2\theta\,\sin^2(\lambda_m-\phi)\right.$$

$$\left. + 3\,\iota_m\,\sin(2\,\theta)\,\sin(\lambda_m-\phi)\,\sin(\lambda_m-\varpi_n)-1\right]$$ (985)

to first order in $\iota_m$. Given that we are interested in the motion of the Earth's axis of rotation on time-scales that are much longer than a month, we can average the above expression over the Moon's orbit to give

$$U \simeq U_0' - \epsilon\,\alpha_m\,\cos(2\,\theta) + \epsilon\,\beta_m\,\sin(2\,\theta)\,\cos(\varpi_n-\phi),$$ (986)

[since the average of $\sin^2(\lambda_m-\phi)$ over a month is $1/2$, whereas that of $\sin(\lambda_m-\phi)\,\sin(\lambda_m-\varpi_m)$ is $(1/2)\,\cos(\varpi_m-\phi)$]. Here, $U_0'$ is a constant,

$$\alpha_m = \frac{3}{8}\,I_\parallel\,\mu_m\,n_m^{\,2},$$ (987)

$$\beta_m = \frac{3}{4}\,I_\parallel\,\iota_m\,\mu_m\,n_m^{\,2},$$ (988)

and

$$\mu_m=\frac{M_m}{M} = 0.0123$$ (989)

is the ratio of the lunar to the terrestrial mass. Now, gravity is a superposable force, so the total potential energy of the Earth-Moon-Sun system is the sum of Equations (963) and (986). In other words,

$$U = U_0'' -\epsilon\,\alpha\,\cos(2\,\theta) +\epsilon\,\beta_m\,\sin(2\,\theta)\,\cos(\varpi_n-\phi),$$ (990)

where $U_0''$ is a constant, and

$$\alpha=\alpha_s+\alpha_m.$$ (991)

Finally, making use of (967), the Lagrangian of the Earth is written

$${\cal L}= \frac{1}{2}\left(I_\perp\,\dot{\theta}^2 + I_\perp... ...eta)- \epsilon\,\beta_m\,\sin(2\,\theta)\,\cos(\varpi_n-\phi),$$ (992)

where any constant terms have been neglected. Recall that $\omega$ is given by (968), and is a constant of the motion.

Two equations of motion that can immediately be derived from the above Lagrangian are

$$\frac{d}{dt}\!\left(\frac{\partial {\cal L}}{\partial \dot{\theta}}\right)-\frac{\partial{\cal L}}{\partial \theta} = 0,$$ (993)

$$\frac{d}{dt}\!\left(\frac{\partial {\cal L}}{\partial \dot{\phi}}\right)-\frac{\partial{\cal L}}{\partial \phi} = 0.$$ (994)

(The third equation, involving $\psi$, merely confirms that $\omega$ is a constant of the motion.) The above two equations yield

$$0 = I_\perp\,\ddot{\theta} - \frac{1}{2}\,I_\perp\,\sin(2\,\theta)\,\... ...parallel}\,\sin\theta\,\omega\,\dot{\phi} +2\,\epsilon\,\alpha\,\sin(2\,\theta)$$

$$+ 2\,\epsilon\,\beta_m\,\cos(2\,\theta)\,\cos(\varpi_n-\phi),$$ (995)

$$0 = \frac{d}{dt}\!\left(I_\perp\,\sin^2\theta\,\dot{\phi} + I_{\paral... ...theta\,\omega\right) + \epsilon\,\beta_m\,\sin(2\,\theta)\,\sin(\varpi_n-\phi),$$ (996)

respectively. Let

$$\theta(t) = \theta_0+ \epsilon\,\theta_1(t),$$ (997)

$$\phi(t) = \epsilon\,\phi_1(t),$$ (998)

where $\theta_0=23.44^\circ$ is the mean inclination of the ecliptic to the Earth's equatorial plane. To first order in $\epsilon$, Equations (995) and (996) reduce to

$$0 \simeq I_\perp\,\ddot{\theta}_1 + I_{\parallel}\,\sin\theta_0\,\omega\,\... ...\,\alpha\,\sin(2\,\theta_0) + 2\,\beta_m\,\cos(2\,\theta_0)\,\cos(\Omega_n\,t),$$ (999)

$$0 \simeq I_\perp\,\sin^2\theta_0\,\ddot{\phi}_1 - I_{\parallel}\,\sin\theta_0\,\omega\,\dot{\theta}_1 - \beta_m\,\sin(2\,\theta_0)\,\sin(\Omega_n\,t),$$ (1000)

respectively, where use has been made of Equation (983). However, as can easily be verified after the fact, $d/dt\ll \omega$, so we obtain

$$\dot{\phi}_1 \simeq -\frac{4\,\alpha\,\cos\theta_0}{I_\parallel\,\omega}-\frac{2\,\beta_m\,\cos(2\,\theta_0)}{I_\parallel\,\omega\,\sin\theta_0}\,\cos(\Omega_n\,t),$$ (1001)

$$\dot{\theta}_1 \simeq -\frac{2\,\beta_m\,\cos\theta_0}{I_\parallel\,\omega}\,\sin(\Omega_n\,t).$$ (1002)

The above equations can be integrated, and then combined with Equations (997) and (998), to give

$$\phi(t) = - \Omega_\phi\,t - \delta\phi\,\sin(\Omega_n\,t),$$ (1003)

$$\theta(t) = \theta_0 + \delta\theta\,\cos(\Omega_n\,t),$$ (1004)

where

$$\Omega_\phi = \frac{3}{2}\,\frac{\epsilon\,(n_s^{\,2}+\mu_m\,n_m^{\,2})}{\omega}\,\cos\theta_0,$$ (1005)

$$\delta\phi = \frac{3}{2}\,\frac{\epsilon\,\iota_m\,\mu_m\,n_m^{\,2}}{\omega\,\Omega_n}\,\frac{\cos(2\,\theta_0)}{\sin\theta_0},$$ (1006)

$$\delta\theta = \frac{3}{2}\,\frac{\epsilon\,\iota_m\,\mu_m\,n_m^{\,2}}{\omega\,\Omega_n}\,\cos\theta_0.$$ (1007)

Incidentally, in the above, we have assumed that the lunar ascending node coincides with the vernal equinox at time $t=0$ (i.e., $\varpi_m=0$ at $t=0$), in accordance with our previous assumption that $\phi =0$ at $t=0$.

According to Equation (1003), the combined gravitational interaction of the Sun and the Moon with the quadrupole field generated by the Earth's slight oblateness causes the Earth's axis of rotation to precess steadily about the normal to the ecliptic plane at the rate $-\Omega_\phi$. As before, the negative sign indicates that the precession is in the opposite direction to the (apparent) orbital motion of the sun and moon. The period of the precession in years is given by

$$T_\phi({\rm yr}) = \frac{n_s}{\Omega_\phi} = \frac{2\,T_s({\rm day})}{\epsilon\,(1+\mu_m/[T_m({\rm yr})]^2)\,\cos\theta_0},$$ (1008)

where $T_m({\rm yr})= n_s/n_m=0.081$ is the Moon's (synodic) orbital period in years. Given that $\epsilon=0.00335$, $\theta_0=23.44^\circ$, $T_s({\rm day})=365.24$, and $\mu_m=0.0123$, we obtain

$$T_\phi\simeq 27,600\,{\rm years}.$$ (1009)

This prediction is fairly close to the observed precession period of $25,800\,{\rm years}$. The main reason that our estimate is slightly inaccurate is because we have neglected to take into account the small eccentricities of the Earth's orbit around the Sun, and the Moon's orbit around the Earth.

The point in the sky toward which the Earth's axis of rotation points is known as the north celestial pole. Currently, this point lies within about a degree of the fairly bright star Polaris, which is consequently sometimes known as the north star or the pole star. It follows that Polaris appears to be almost stationary in the sky, always lying due north, and can thus be used for navigational purposes. Indeed, mariners have relied on the north star for many hundreds of years to determine direction at sea. Unfortunately, because of the precession of the Earth's axis of rotation, the north celestial pole is not a fixed point in the sky, but instead traces out a circle, of angular radius $23.44^\circ$, about the north ecliptic pole, with a period of 25,800 years. Hence, a few thousand years from now, the north celestial pole will no longer coincide with Polaris, and there will be no convenient way of telling direction from the stars.

The projection of the ecliptic plane onto the sky is called the ecliptic, and coincides with the apparent path of the Sun against the backdrop of the stars. Furthermore, the projection of the Earth's equator onto the sky is known as the celestial equator. As has been previously mentioned, the ecliptic is inclined at $23.44^\circ$ to the celestial equator. The two points in the sky at which the ecliptic crosses the celestial equator are called the equinoxes, since night and day are equally long when the Sun lies at these points. Thus, the Sun reaches the vernal equinox on about March 21st, and this traditionally marks the beginning of spring. Likewise, the Sun reaches the autumn equinox on about September 22nd, and this traditionally marks the beginning of autumn. However, the precession of the Earth's axis of rotation causes the celestial equator (which is always normal to this axis) to precess in the sky, and thus also causes the equinoxes to precess along the ecliptic. This effect is known as the precession of the equinoxes. The precession is in the opposite direction to the Sun's apparent motion around the ecliptic, and is of magnitude $1.4^\circ$ per century. Amazingly, this miniscule effect was discovered by the Ancient Greeks (with the help of ancient Babylonian observations). In about 2000 BC, when the science of astronomy originated in ancient Egypt and Babylonia, the vernal equinox lay in the constellation Aries. Indeed, the vernal equinox is still sometimes called the first point of Aries in astronomical texts. About 90 BC, the vernal equinox moved into the constellation Pisces, where it still remains. The equinox will move into the constellation Aquarius (marking the beginning of the much heralded ``Age of Aquarius'') in about 2600 AD. Incidentally, the position of the vernal equinox in the sky is of great significance in astronomy, since it is used as the zero of celestial longitude (much as Greenwich is used as the zero of terrestrial longitude).

Equations (1003) and (1004) indicate that the small inclination of the lunar orbit to the ecliptic plane, combined with the precession of the lunar ascending node, causes the Earth's axis of rotation to wobble sightly. This wobble is known as nutation, and is superimposed on the aforementioned precession. In the absence of precession, nutation would cause the north celestial pole to periodically trace out a small ellipse on the sky, the sense of rotation being counter-clockwise. The nutation period is 18.6 years: i.e., the same as the precession period of the lunar ascending node. The nutation amplitudes in the polar and azimuthal angles $\theta$ and $\phi$ are

$$\delta\theta = \frac{3}{2}\,\frac{\epsilon\,\iota_m\,\mu_m\,T_n({\rm yr})}{T_s({\rm day})\,[T_m({\rm yr})]^2}\,\cos\theta_0,$$ (1010)

$$\delta\phi = \frac{3}{2}\,\frac{\epsilon\,\iota_m\,\mu_m\,T_n({\rm yr})}{T_s({\rm day})\,[T_m({\rm yr})]^2}\,\frac{\cos(2\,\theta_0)}{\sin\theta_0},$$ (1011)

respectively, where $T_n({\rm yr}) = n_s/\Omega_n = 18.6$. Given that $\epsilon=0.00335$, $\theta_0=23.44^\circ$, $\iota_m=5.16^\circ$, $T_s({\rm day})=365.24$, $T_m({\rm yr})=0.081$, and $\mu_m=0.0123$, we obtain

$$\delta\theta = 8.2^{''},$$ (1012)

$$\delta\phi = 15.3^{''}.$$ (1013)

The observed nutation amplitudes are $9.2^{''}$ and $17.2^{''}$, respectively. Hence, our estimates are quite close to the mark. Any inaccuracy is mainly due to the fact that we have neglected to take into account the small eccentricities of the Earth's orbit around the Sun, and the Moon's orbit around the Earth. The nutation of the Earth was discovered in 1728 by the English astronomer James Bradley, and was explained theoretically about 20 years later by d'Alembert and L. Euler. Nutation is important because the corresponding gyration of the Earth's rotation axis appears to be transferred to celestial objects when they are viewed using terrestrial telescopes. This effect causes the celestial longitudes and latitudes of heavenly objects to oscillate sinusoidally by up to $20^{''}$ (i.e., about the maximum angular size of Saturn) with a period of 18.6years. It is necessary to correct for this oscillation in order to accurately guide terrestrial telescopes to particular objects.

Note, finally, that the type of forced nutation discussed above, which is driven by an external torque, is quite distinct from the free nutation described in Section 8.9.