Forced precession and nutation of Earth

Consider the Earth-Sun system. See Figure 8.4. From a geocentric viewpoint, the Sun orbits the Earth counterclockwise (if we look from the north), once per year, in an approximately circular orbit of radius $a_s= 1.50\times 10^{11}\,{\rm m}$ (Yoder 1995). 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. See Figure 8.5. In the following, we shall treat the $x$, $y$, $z$ coordinate 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 counterclockwise (if we look from the north) azimuthal angle $\lambda_s$ that is zero on the positive $x$-axis. See Figure 8.4.

Figure 8.4: The Earth-Sun system.

Let $\omega$ 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 (Yoder 1995). Suppose that the projection of $\omega$ onto the ecliptic plane subtends an angle $\phi$ with the $y$-axis, where $\phi$ is measured in a counterclockwise (if we look from the north) sense. See Figure 8.4. The orientation of the Earth's axis of rotation (which is, of course, parallel to $\omega$) 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 Section 8.7. Let us examine the Earth-Sun system at an instant in time, $t=0$, when $\phi =0$; that is, when $\omega$ lies in the $y$-$z$ plane. At this particular instant, the $x$-axis points toward the so-called vernal equinox, which is defined as the point in the sky where the Sun's apparent orbit crosses the projection of the Earth's equator (i.e., the plane normal to $\omega$) from south to north. A counterclockwise (if we look 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 (8.69), the potential energy of the Earth-Sun system is written

$$U = M_s\,{\mit\Phi} = - \frac{G\,M_s\,M}{a_s} + \frac{G\,M_s\,({\cal I}_\parallel-{\cal I}_\perp)}{a_s^{\,3}}\,P_2[\cos(\gamma_s)],$$ (8.87)

where $M_s$ is the mass of the Sun, $M$ the mass of the Earth, ${\cal I}_\parallel$ the Earth's moment of inertia about its axis of rotation, ${\cal I}_\perp$ 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 $\omega$ 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 $\phi =0$)

$$\mbox{\boldmath$\omega$} = \omega\,(0,\,\sin\theta,\,\cos\theta),$$ (8.88)

and

$${\bf r}_s = a_s\,(\cos\lambda_s,\,\sin\lambda_s,\,0).$$ (8.89)

Hence,

$$\cos\gamma_s =\frac{\mbox{\boldmath$\omega$}\cdot{\bf r}_s}{\vert\mbox{\boldmath$\omega$}\vert\,\vert{\bf r}_s\vert}= \sin\theta\,\sin\lambda_s,$$ (8.90)

giving

$$U = - \frac{G\,M_s\,M}{a_s} + \frac{G\,M_s\,({\cal I}_\parallel-{\cal I}_\perp)}{2\,a_s^{\,3}}\,(3\,\sin^2\theta\,\sin^2\lambda_s -1).$$ (8.91)

Because we are primarily interested in the motion of the Earth's axis of rotation on timescales that are much longer than a year, we can average the preceding expression over the Sun's orbit to give

$$U = - \frac{G\,M_s\,M}{a_s} + \frac{G\,M_s\,({\cal I}_\parallel-{\cal I}_\perp)}{2\,a_s^{\,3}}\,\left(\frac{3}{2}\,\sin^2\theta -1\right)$$ (8.92)

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

$$U = U_0 - \skew{3}\tilde{\epsilon}\,\alpha_s\,\cos(2\,\theta),$$ (8.93)

where $U_0$ is a constant, and

$$\alpha_s = \frac{3}{8}\,{\cal I}_\parallel\,n_s^{\,2}.$$ (8.94)

Here,

$$\skew{3}\tilde{\epsilon} = \frac{{\cal I}_{\parallel}-{\cal I}_\perp}{{\cal I}_\parallel}=0.00335$$ (8.95)

is the Earth's dynamical ellipticity [the quoted value is determined from the Earth's observed flattening, on the simplistic assumption that it is a homogeneous body (Yoder 1995)], and

$$n_s =\frac{d\lambda_s}{dt}= \left(\frac{G\,M_s}{a_s^{\,3}}\right)^{1/2}$$ (8.96)

is the Sun's apparent orbital angular velocity.

The rotational kinetic energy of the Earth can be written (see Sections 8.4 and 8.6)

$$K = \frac{1}{2}\left({\cal I}_{\perp}\,\omega_{x'}^{\,2} + {\cal I}_\perp\,\omega_{y'}^{\,2}+{\cal I}_\parallel\,\omega_{z'}^{\,2}\right),$$ (8.97)

which reduces to

$$K = \frac{1}{2}\left({\cal I}_\perp\,\skew{5}\dot{\theta}^{\,2} +... ...,\sin^2\theta\,\,\skew{5}\dot{\phi}^{\,2} + {\cal I}_\parallel\,\omega^2\right)$$ (8.98)

with the aid of Equations (8.54)–(8.56). Here,

$$\omega = \cos\theta\,\,\skew{5}\dot{\phi} + \skew{5}\dot{\psi},$$ (8.99)

and $\psi$ is the third Euler angle. Hence, the Earth's Lagrangian takes the form

$${\cal L} = K - U = \frac{1}{2}\left({\cal I}_\perp\,\skew{5}\dot{... ...\parallel\,\omega^2\right) +\skew{3}\tilde{\epsilon}\,\alpha_s\,\cos(2\,\theta)$$ (8.100)

where any constant terms have been neglected. The Lagrangian does not depend explicitly on the angular coordinate $\psi$. It follows that the conjugate momentum is a constant of the motion (see Section 7.5). In other words,

$$p_\psi = \frac{\partial{\cal L}}{\partial\skew{5}\dot{\psi}} = {\cal I}_\parallel \,\omega$$ (8.101)

is a constant of the motion, implying that $\omega$ is also a constant of the motion. Note that $\omega$ is effectively the Earth's angular velocity of rotation about its axis [because $\vert\omega_{x'}\vert, \vert\omega_{y'}\vert\ll \omega_{z'}= \omega$, which follows because $\vert\skew{5}\dot{\phi}\vert, \vert\skew{5}\dot{\theta}\vert\ll \skew{5}\dot{\psi}$; see Equations (8.54)–(8.56)]. Another equation of motion that can be derived from the Lagrangian is

$$\frac{d}{dt}\!\left(\frac{\partial{\cal L}}{\partial \skew{5}\dot{\theta}}\right)- \frac{\partial {\cal L}}{\partial\theta} = 0,$$ (8.102)

which reduces to

$${\cal I}_\perp\,\skew{5}\ddot{\theta} - \frac{\partial {\cal L}}{\partial \theta} = 0.$$ (8.103)

Consider steady precession of the Earth's rotational axis, which is characterized by $\skew{5}\dot{\theta}=0$, with both $\skew{5}\dot{\phi}$ and $\skew{5}\dot{\psi}$ constant. It follows, from Equation (8.103), that such motion must satisfy the constraint

$$\frac{\partial {\cal L}}{\partial\theta} = 0.$$ (8.104)

Thus, we obtain

$$\frac{1}{2}\,{\cal I}_\perp\,\sin(2\,\theta)\,\skew{5}\dot{\phi}^... ...skew{5}\dot{\phi} - 2\,\skew{3}\tilde{\epsilon}\,\alpha_s\,\sin(2\,\theta) = 0,$$ (8.105)

where use has been made of Equations (8.99) and (8.100). As can easily be verified after the fact, $\vert\skew{5}\dot{\phi}\vert\ll \omega$, so Equation (8.105) reduces to

$$\skew{5}\dot{\phi} \simeq -\frac{4\,\skew{3}\tilde{\epsilon}\,\alpha_s\,\cos\theta}{{\cal I}_{\parallel}\,\omega} = -{\mit\Omega}_\phi,$$ (8.106)

which can be integrated to give

$$\phi \simeq- {\mit\Omega}_\phi\,t,$$ (8.107)

where

$${\mit\Omega}_\phi = \frac{3}{2}\,\frac{\skew{3}\tilde{\epsilon}\,n_s^{\,2}}{\omega}\,\cos\theta,$$ (8.108)

and use has been made of Equation (8.94). According to the preceding expressions, 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 $-{\mit\Omega}_\phi$. The fact that $-{\mit\Omega}_\phi$ is negative implies that the precession is in the opposite sense to that of the Earth's diurnal 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 (sidereal) years is given by

$$T_\phi({\rm yr}) = \frac{n_s}{{\mit\Omega}_\phi} = \frac{2\,T_s({\rm day})}{3\,\skew{3}\tilde{\epsilon}\,\cos\theta},$$ (8.109)

where $T_s({\rm day}) = \omega/n_s = 366.26$ is the length of a sidereal year in stellar (sidereal) days. Thus, given that $\skew{3}\tilde{\epsilon}=0.00335$ and $\theta = 23.44^\circ$, we obtain

$$T_\phi\simeq \,\,{\rm years}.$$ (8.110)

Unfortunately, the observed precession period of the Earth's axis of rotation about the normal to the ecliptic plane is approximately 25,800 years (Yoder 1995), 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 previously, the potential energy of the Earth-Moon system can be written

$$U = - \frac{G\,M_m\,M}{a_m} + \frac{G\,M_m\,({\cal I}_\parallel-{\cal I}_\perp)}{a_m^{\,3}}\,P_2[\cos(\gamma_m)],$$ (8.111)

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 $\omega$ and ${\bf r}_m$, where

$$\mbox{\boldmath$\omega$} = \omega\,\left(-\sin\theta\,\sin\phi,\,\sin\theta\,\cos\phi,\,\cos\theta\right)$$ (8.112)

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 $\omega$ (because we shall presently differentiate by $t$, before setting $\phi =0$). The Moon's orbital plane is actually slightly inclined to the ecliptic plane, the (mean) angle of inclination being $I_m=5.16^\circ$ (Yoder 1995). Hence, we can write

$${\bf r}_m \simeq a_m\,\left(\cos\lambda_m,\,\sin\lambda_m,\,I_m\,\sin(\lambda_m-\alpha_n)\right),$$ (8.113)

to first order in $I_m$, where $\lambda_m$ is the Moon's ecliptic longitude, and $\alpha_n$ 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

$$n_m = \frac{d\lambda_m}{dt}\simeq \left(\frac{G\,M}{a_m^{\,3}}\right)^{1/2}$$ (8.114)

is the Moon's mean 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 (Yoder 1995). This precession is caused by the perturbing influence of the Sun. (See Chapter 11.) It follows that

$$\frac{d\alpha_n}{dt}= -{\mit\Omega}_n,$$ (8.115)

where $2\pi/{\mit\Omega}_n=18.6\,{\rm years}$. From Equations (8.112) and (8.113),

$$\cos\gamma_m = \frac{\mbox{\boldmath$\omega$}\cdot{\bf r}_m}{\ver... ...t}= \sin\theta\,\sin(\lambda_m-\phi)+I_m\,\cos\theta\,\sin(\lambda_m-\alpha_n),$$ (8.116)

so Equation (8.111) yields

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

$$\phantom{=}\left. + 3\,I_m\,\sin(2\,\theta)\,\sin(\lambda_m-\phi)\,\sin(\lambda_m-\alpha_n)-1\right]$$ (8.117)

to first order in $I_m$. Because we are interested in the motion of the Earth's axis of rotation on timescales that are much longer than a month, we can average this expression over the Moon's orbit to give

$$U \simeq U_0' - \skew{3}\tilde{\epsilon}\,\alpha_m\,\cos(2\,\theta) + \skew{3}\tilde{\epsilon}\,\beta_m\,\sin(2\,\theta)\,\cos(\alpha_n-\phi),$$ (8.118)

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

$$\alpha_m = \frac{3}{8}\,{\cal I}_\parallel\,\mu_m\,n_m^{\,2},$$ (8.119)

$$\beta_m = \frac{3}{4}\,{\cal I}_\parallel\,I_m\,\mu_m\,n_m^{\,2},$$ (8.120)

and

$$\mu_m=\frac{M_m}{M} = 0.0123$$ (8.121)

is the ratio of the lunar to the terrestrial mass (Yoder 1995). Gravity is a superposable force, so the total potential energy of the Earth-Moon-Sun system is the sum of Equations (8.93) and (8.118). In other words,

$$U = U_0'' -\skew{3}\tilde{\epsilon}\,\alpha\,\cos(2\,\theta) +\skew{3}\tilde{\epsilon}\,\beta_m\,\sin(2\,\theta)\,\cos(\alpha_n-\phi),$$ (8.122)

where $U_0''$ is a constant, and

$$\alpha=\alpha_s+\alpha_m.$$ (8.123)

Finally, making use of Equation (8.98), the Lagrangian of the Earth is written

$${\cal L}= \frac{1}{2}\left({\cal I}_\perp\,\skew{5}\dot{\theta}^{... ...heta)- \skew{3}\tilde{\epsilon}\,\beta_m\,\sin(2\,\theta)\,\cos(\alpha_n-\phi),$$ (8.124)

where any constant terms have been neglected. Recall that $\omega$ is given by Equation (8.99), and is a constant of the motion.

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

$$\frac{d}{dt}\!\left(\frac{\partial {\cal L}}{\partial\skew{5}\dot{\theta}}\right)- \frac{\partial {\cal L}}{\partial \theta} = 0,$$ (8.125)

$$\frac{d}{dt}\!\left(\frac{\partial {\cal L}}{\partial \skew{5}\dot{\phi}}\right)-\frac{\partial{\cal L}}{\partial \phi} = 0.$$ (8.126)

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

$$={\cal I}_\perp\,\skew{5}\ddot{\theta} - \frac{1}{2}\,{\cal I}_\p... ...omega\,\skew{5}\dot{\phi} +2\,\skew{3}\tilde{\epsilon}\,\alpha\,\sin(2\,\theta)$$ 0

$$\phantom{=}+ 2\,\skew{3}\tilde{\epsilon}\,\beta_m\,\cos(2\,\theta)\,\cos(\alpha_n-\phi),$$ (8.127)

$$=\frac{d}{dt}\!\left({\cal I}_\perp\,\sin^2\theta\,\,\skew{5}\dot... ...ght) + \skew{3}\tilde{\epsilon}\,\beta_m\,\sin(2\,\theta)\,\sin(\alpha_n-\phi),$$ 0 (8.128)

respectively. Let

$$\theta(t) =\theta_0+ \skew{3}\tilde{\epsilon}\,\theta_1(t),$$ (8.129)

$$\phi(t) = \skew{3}\tilde{\epsilon}\,\phi_1(t),$$ (8.130)

where $\theta_0=23.44^\circ$ is the mean inclination of the ecliptic to the Earth's equatorial plane. To first order in $\skew{3}\tilde{\epsilon}$, Equations (8.127) and (8.128) reduce to

$$\simeq {\cal I}_\perp\,\skew{5}\ddot{\theta}_1 + {\cal I}_{\paral... ...ha\,\sin(2\,\theta_0) + 2\,\beta_m\,\cos(2\,\theta_0)\,\cos({\mit\Omega}_n\,t),$$ 0 (8.131)

$$\simeq{\cal I}_\perp\,\sin^2\theta_0\,\,\skew{5}\ddot{\phi}_1 - {... ...\,\skew{5}\dot{\theta}_1 - \beta_m\,\sin(2\,\theta_0)\,\sin({\mit\Omega}_n\,t),$$ 0 (8.132)

respectively, where use has been made of Equation (8.115). However, as can easily be verified after the fact, $d/dt\ll \omega$, so we obtain

$$\skew{5}\dot{\phi}_1 \simeq -\frac{4\,\alpha\,\cos\theta_0}{{\cal I}_\parallel\,\omega... ...,\theta_0)}{{\cal I}_\parallel\,\omega\,\sin\theta_0}\,\cos({\mit\Omega}_n\,t),$$ (8.133)

$$\skew{5}\dot{\theta}_1 \simeq -\frac{2\,\beta_m\,\cos\theta_0}{{\cal I}_\parallel\,\omega}\,\sin({\mit\Omega}_n\,t).$$ (8.134)

These equations can be integrated and then combined with Equations (8.129) and (8.130) to give

$$\phi(t) = - {\mit\Omega}_\phi\,t - \delta\phi\,\sin({\mit\Omega}_n\,t),$$ (8.135)

$$\theta(t) = \theta_0 + \delta\theta\,\cos({\mit\Omega}_n\,t),$$ (8.136)

where

$${\mit\Omega}_\phi = \frac{3}{2}\,\frac{\skew{3}\tilde{\epsilon}\,(n_s^{\,2}+\mu_m\,n_m^{\,2})}{\omega}\,\cos\theta_0,$$ (8.137)

$$\delta\phi =\frac{3}{2}\,\frac{\skew{3}\tilde{\epsilon}\,I_m\,\mu_m\,n_m^{\,2}}{\omega\,{\mit\Omega}_n}\,\frac{\cos(2\,\theta_0)}{\sin\theta_0},$$ (8.138)

$$\delta\theta = \frac{3}{2}\,\frac{\skew{3}\tilde{\epsilon}\,I_m\,\mu_m\,n_m^{\,2}}{\omega\,{\mit\Omega}_n}\,\cos\theta_0.$$ (8.139)

Incidentally, in these expressions, we have assumed that the lunar ascending node coincides with the vernal equinox at time $t=0$ (i.e., $\alpha_n=0$ at $t=0$), in accordance with our previous assumption that $\phi =0$ at $t=0$. There is also an implicit assumption that ${\mit\Omega}_n\gg {\mit\Omega}_\phi$.

According to Equation (8.135), the combined gravitational interaction of the Sun and the Moon with the gravitational 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 $-{\mit\Omega}_\phi$. As before, the negative sign indicates that the precession is in the opposite direction to the (apparent) orbital motion of the Sun and the Moon. The period of this so-called luni-solar precession in (sidereal) years is given by

$$T_\phi({\rm yr}) = \frac{n_s}{{\mit\Omega}_\phi} = \frac{2\,T_s({... ...day})}{3\,\skew{3}\tilde{\epsilon}\,(1+\mu_m/[T_m({\rm yr})]^2)\,\cos\theta_0},$$ (8.140)

where $T_m({\rm yr})= n_s/n_m=0.00748$ is the Moon's (sidereal) orbital period in years. Given that $\skew{3}\tilde{\epsilon}=0.00335$, $\theta_0=23.44^\circ$, $T_s({\rm day})=366.26$, and $\mu_m=0.0123$, we obtain

$$T_\phi\simeq \,{\rm years}.$$ (8.141)

This prediction is fairly close to the observed precession period of 25,800 years (Yoder 1995). 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, and have also treated the Earth as a homogeneous body. (See Appendix F.)

The point in the sky toward which the Earth's axis of rotation is directed 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 pole star. See Figure 8.5. 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. See Figure 8.5. 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.

Figure: 8.5 Path of the north celestial pole against the backdrop of the stars as consequence of the precession of the equinoxes (calculated assuming constant precessional speed and obliquity). Numbers indicate years relative to start of common era. Stellar positions and magnitudes from Hoffleit and Warren (1991).

The projection of the ecliptic plane onto the sky is called the ecliptic circle, and coincides with the apparent path of the Sun against the backdrop of the stars. 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, because night and day are equally long when the Sun lies at these points. Thus, the Sun reaches the vernal equinox on about March 20, and this traditionally marks the beginning of spring. Likewise, the Sun reaches the autumnal equinox on about September 22, 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; it 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; Heath 1991). In about 2000 BCE, when the science of astronomy originated in ancient Egypt and Babylonia, the vernal equinox lay in the constellation Aries. See Figure 8.6. Indeed, the vernal equinox is still sometimes called the first point of Aries in astronomical texts. About 90 BCE, 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 CE. Incidentally, the position of the vernal equinox in the sky is of great significance in astronomy, because it is used as the zero of celestial longitude (much as the Greenwich meridian is used as the zero of terrestrial longitude).

Figure: 8.6 Path of the vernal equinox against the backdrop of the stars as a consequence of the precession of the equinoxes (calculated assuming constant precessional speed and obliquity). Numbers indicate years relative to start of common era. Stellar positions and magnitudes from Hoffleit and Warren (1991).

Equations (8.135) and (8.136) 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 (from the Latin nutare, to nod), 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 counterclockwise. The nutation period is 18.6 years; the same as the precession period of the lunar ascending node. The nutation amplitudes in the polar and azimuthal angles $\theta$ and $\phi$ are

$$\delta\theta = \frac{3}{2}\,\frac{\skew{3}\tilde{\epsilon}\,I_m\,\mu_m\,T_n({\rm yr})}{T_s({\rm day})\,[T_m({\rm yr})]^2}\,\cos\theta_0,$$ (8.142)

$$\delta\phi =\frac{3}{2}\,\frac{\skew{3}\tilde{\epsilon}\,I_m\,\mu_m\,T_n({\r... ...})}{T_s({\rm day})\,[T_m({\rm yr})]^2}\,\frac{\cos(2\,\theta_0)}{\sin\theta_0},$$ (8.143)

respectively, where $T_n({\rm yr}) = n_s/{\mit\Omega}_n = 18.6$. Given that $\skew{3}\tilde{\epsilon}=0.00335$, $\theta_0=23.44^\circ$, $I_m=5.16^\circ$, $T_s({\rm day})=366.26$, $T_m({\rm yr})=0.0748$, and $\mu_m=0.0123$, we obtain

$$\delta\theta = 9.6'',$$ (8.144)

$$\delta\phi = 17.9''.$$ (8.145)

The observed nutation amplitudes are $9.2''$ and $17.2''$, respectively (Meeus 2005). 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, and have also treated the Earth as a homogeneous body. (See Appendix F.) The nutation of the Earth was discovered in 1728 by the English astronomer James Bradley (1693–1748); it was explained theoretically about 20 years later by d'Alembert and 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 apparent angular size of Saturn) with a period of 18.6 years. It is necessary to correct for this oscillation (as well as for the precession of the equinoxes) to accurately guide terrestrial telescopes to particular celestial objects.