Lunisolar Precession of Earth

Let us investigate the influence of the Sun on the Earth's diurnal rotation. Consider the Earth-Sun system. See Figure 1.20. 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 $a_s= 1.496\times 10^{11}\,{\rm m}$. 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 (with respect to distant stars) 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 $x'$, $y'$, $z'$ 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 1.20. Thus, the coordinates of the Sun in the $x'$, $y'$, $z'$ system are

$$x_s' = a_s\,\cos\lambda_s,$$ (1.392)

$$y_s' =a_s\,\sin\lambda_s,$$ (1.393)

$$z_s' =0.$$ (1.394)

Note that

$$\lambda_s = \omega_s\,t,$$ (1.395)

where

$$\omega_s = \left(\frac{G\,M_s}{a_s^{\,3}}\right)^{1/2}$$ (1.396)

is the Sun's apparent orbital angular velocity about the Earth. [See Equation (1.323).] Here, $M_s=1.989\times 10^{30}\,{\rm kg}$ is the mass of the Sun.

Figure 1.20: The Earth-Sun system.

Let $\Omega$ be the Earth's angular velocity vector due to its diurnal rotation. This vector subtends 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 $\Omega$ onto the ecliptic plane subtends an angle $\phi$ with the $x'$-axis, where $\phi$ is measured in a counterclockwise (if we look from the north) sense, and is zero on the positive $x'$-axis. See Figure 1.20. 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 $\phi$. The components of $\Omega$ in the $x'$, $y'$, $z'$ system are

$${\mit\Omega}_{x'} = {\mit\Omega}\,\sin\theta\,\cos\phi,$$ (1.397)

$${\mit\Omega}_{y'} = {\mit\Omega}\,\sin\theta\,\sin\phi,$$ (1.398)

$${\mit\Omega}_{z'} = {\mit\Omega}\,\cos\theta.$$ (1.399)

Let us define a second coordinate system, centered on the Earth, such that the $z$-axis corresponds to the Earth's axis of rotation. The transformation between the $x$, $y$, $z$ system and the $x'$, $y'$, $z'$ system is

$$\left(\begin{array}{c}x\\ [0.5ex]y\\ [0.5ex]z\end{array}\right)= ... ...rray}\right)\left(\begin{array}{c}x'\\ [0.5ex]y'\\ [0.5ex]z'\end{array}\right).$$ (1.400)

(See Section A.5.) The inverse transformation is

$$\left(\begin{array}{c}x'\\ [0.5ex]y'\\ [0.5ex]z'\end{array}\right... ...d{array}\right)\left(\begin{array}{c}x\\ [0.5ex]y\\ [0.5ex]z\end{array}\right).$$ (1.401)

Of course, the previous two transformations also apply to the components of all vectors in the two coordinate systems. It is easily verified, from Equations (1.397)–(1.400), that ${\mit\Omega}={\mit\Omega}\,{\bf e}_z$ in the $x$, $y$, $z$ coordinate system. Note that ${\bf e}_y$ lies both in the Earth's equatorial plane and the ecliptic plane.

The gravitational potential of the Sun can be written

$${\mit\Phi}(x,y,z) = - \frac{G\,M_s}{\vert{\bf r}-{\bf r}_s\vert} = - \frac{G\,M_s}{[(x-x_s)^2+(y-y_s)^2+(z-z_s)^2]^{1/2}},$$ (1.402)

where

$$x_s = a_s\,(\cos\theta\,\cos\phi\,\cos\lambda_s+\cos\theta\,\sin\phi\,\sin\lambda_s),$$ (1.403)

$$y_s = a_s\,(-\sin\phi\,\cos\lambda_s+\cos\phi\,\sin\lambda_s),$$ (1.404)

$$z_s = a_s\,(\sin\theta\,\cos\phi\,\cos\lambda_s+\sin\theta\,\sin\phi\,\sin\lambda_s)$$ (1.405)

are the coordinates of the Sun in the $x$, $y$, $z$ system. [See Equations (1.392)–(1.394) and Equation (1.400).] It follows that

$${\mit\Phi}_{yz} \equiv\left.\frac{\partial^{\,2}{\mit\Phi}}{\partial y\,\partial ... ...= \frac{3}{2}\,\frac{G\,M_s}{a_s^{\,3}}\,\sin\theta\,\sin[2\,(\phi-\lambda_s)],$$ (1.406)

$${\mit\Phi}_{zx} \equiv\left.\frac{\partial^{\,2}{\mit\Phi}}{\partial z\,\partial ... ...\,G\,M_s}{2\,a_s^{\,3}}\,\cos\theta\,\sin\theta\,(1+\cos[2\,(\phi-\lambda_s)]).$$ (1.407)

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 expressions over the Sun's orbit to give

$${\mit\Phi}_{yz} =0,$$ (1.408)

$${\mit\Phi}_{zx} =-\frac{3\,G\,M_s}{2\,a_s^{\,3}}\,\cos\theta\,\sin\theta.$$ (1.409)

(This follows because the averages of $\cos[2\,(\phi-\lambda_s)]$ and $\sin[2\,(\phi-\lambda_s)]$ over a year are both zero.) Equations (1.389)–(1.391), combined with the previous two equations, reveal that the components of the gravitational torque exerted by the Sun on the Earth in the $x$, $y$, $z$ system are

$$\tau_x =0,$$ (1.410)

$$\tau_y = -\frac{3}{2}\,\frac{G\,M_s\,(I_\parallel-I_\perp)}{a_s^{\,3}}\,\cos\theta\,\sin\theta,$$ (1.411)

$$\tau_z =0,$$ (1.412)

where $I_\parallel$ and $I_\perp$ are the Earth's principal moments of inertia. [See Equations (1.385) and (1.386).] Making use of Equation (1.401), the components of the torque in the $x'$, $y'$, $z'$ system are thus

$$\tau_{x'} = \frac{3}{2}\,\frac{G\,M_s\,(I_\parallel-I_\perp)}{a_s^{\,3}}\,\cos\theta\,\sin\theta\,\sin\phi,$$ (1.413)

$$\tau_{y'} = -\frac{3}{2}\,\frac{G\,M_s\,(I_\parallel-I_\perp)}{a_s^{\,3}}\,\cos\theta\,\sin\theta\,\cos\phi,$$ (1.414)

$$\tau_{z'} =0.$$ (1.415)

Because the Earth is rotating about the principal axis of rotation whose principal moment of inertia is $I_{\parallel}$, the Earth's angular momentum can be written ${\bf L} = I_{\parallel}\,$$\Omega$. (See Section 1.7.2.) Thus, the components of ${\bf L}$ in the $x'$, $y'$, $z'$ system are

$$L_{x'} = I_\parallel\,{\mit\Omega}\,\sin\theta\,\cos\phi,$$ (1.416)

$$L_{y'} = I_\parallel\,{\mit\Omega}\,\sin\theta\,\sin\phi,$$ (1.417)

$$L_{z'} = I_\parallel\,{\mit\Omega}\,\cos\theta.$$ (1.418)

[See Equations (1.397)–(1.399).]

The angular equation of motion of the Earth is

$$\frac{d {\bf L}}{dt} = \mbox{\boldmath$\tau$} .$$ (1.419)

(See Section 1.7.1.) However, we must solve this equation in the inertial $x'$, $y'$, $z'$ coordinate system, rather than the $x$, $y$, $z$ coordinate system. The reason for this is that the Earth's angular velocity, $\Omega$, rotates about the $z'$-axis under the action of the gravitational torque exerted by the Sun on the Earth. Hence, the $x$, $y$, $z$ coordinate system, which co-moves with the Earth, accelerates with respect to the $x'$, $y'$, $z'$ system. It follows that the $x$, $y$, $z$ system is non-inertial. (See Section 1.5.4.) Making use of Equations (1.413)–(1.418), the components of the previous equation in the $x'$, $y'$, $z'$ system are

$$\frac{d}{dt}\!\left(I_\parallel\,{\mit\Omega}\,\sin\theta\,\cos\phi\right) = \frac{3}{2}\,\frac{G\,M_s\,(I_\parallel-I_\perp)}{a_s^{\,3}}\,\cos\theta\,\sin\theta\,\sin\phi,$$ (1.420)

$$\frac{d}{dt}\!\left(I_\parallel\,{\mit\Omega}\,\sin\theta\,\sin\phi\right) = -\frac{3}{2}\,\frac{G\,M_s\,(I_\parallel-I_\perp)}{a_s^{\,3}}\,\cos\theta\,\sin\theta\,\cos\phi,$$ (1.421)

$$\frac{d}{dt}\!\left(I_\parallel\,{\mit\Omega}\,\cos\theta\right) =0.$$ (1.422)

If we assume ${\mit\Omega}$ and $\theta$ are constants, but that $\phi$ varies in time, then we can solve the previous three equations to give

$$\frac{d\phi}{dt} = -{\mit\Omega}_\phi,$$ (1.423)

where

$${\mit\Omega}_\phi =\frac{3}{2}\,\frac{G\,M_s}{{\mit\Omega}\,a_s^{... ...theta = \frac{3}{2}\,\frac{\omega_s^{\,2}}{{\mit\Omega}}\,\epsilon\,\cos\theta,$$ (1.424)

and use has been made of Equations (1.379), (1.381), and (1.396).

According to Equation (1.423), the gravitational torque exerted by the Sun on the Earth, due to 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 $-{\mit\Omega}_\phi$. This precession is analogous to that of a spinning top discussed in Section 1.7.7. 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. The precession period in (sidereal) years is given by

$$T_\phi({\rm yr}) = \frac{\omega_s}{{\mit\Omega}_\phi} = \frac{2\,T_s({\rm day})}{3\,\epsilon\,\cos\theta},$$ (1.425)

where $T_s({\rm day}) = {\mit\Omega}/\omega_s = 366.26$ is the length of a sidereal year in sidereal days. (Sidereal means measured with respect to distant stars.) Thus, given that $\epsilon=3.35\times 10^{-3}$ and $\theta = 23.44^\circ$, we obtain

$$T_\phi\simeq \,\,{\rm years}.$$ (1.426)

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.

From a geocentric viewpoint, the Moon orbits the Earth counterclockwise (if we look from the north), once per month, in an approximately circular orbit of radius $a_m= 3.844\times 10^{8}\,{\rm m}$. This orbit is inclined at about $5^\circ$ to the ecliptic plane. However, in the following, we shall ignore this small inclination, and place the Moon's orbit in the ecliptic plane. Analogous analysis to that employed in the preceding part of this section reveals that the gravitational torque exerted by the Moon on the Earth (averaged over an month) gives rise to additional contribution to the precession rate ${\mit\Omega}_\phi$. In fact, by analogy with Equation (1.424), we expect

$${\mit\Omega}_\phi = \frac{3}{2\,{\mit\Omega}}\,\left(\frac{G\,M_s... ..._m^{\,3}}\right)\left(\frac{I_\parallel-I_\perp}{I_\parallel}\right)\cos\theta.$$ (1.427)

Here, $M_m=7.324\times 10^{22}\,{\rm kg}$ is the mass of the Moon. Now,

$$\omega_m = \left(\frac{G\,M_e}{a_m^{\,2}}\right)^{1/2}$$ (1.428)

is the Moon's orbital angular velocity about the Earth. [See Equation (1.323).] Here, $M_e=5.972\times 10^{24}\,{\rm kg}$ is the mass of the Earth. Making use of the previous equation, as well as Equations (1.379), (1.381), and (1.396), we obtain

$${\mit\Omega}_\phi = \frac{3}{2}\,\frac{\omega_s^{\,2}+\mu_m\,\omega_m^{\,2}}{{\mit\Omega}}\,\cos\theta,$$ (1.429)

where $\mu_m=M_m/M_e$.

According to Equations (1.423) and (1.429), the combined gravitational torque exerted by the Sun and the Moon on the Earth, due to 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 lunisolar precession in (sidereal) years is given by

$$T_\phi({\rm yr}) = \frac{\omega_s}{{\mit\Omega}_\phi} = \frac{2\,T_s({\rm day})}{3\,\epsilon\,(1+\mu_m/[T_m({\rm yr})]^2)\,\cos\theta},$$ (1.430)

where $T_m({\rm yr})= \omega_s/\omega_m=0.00748$ is the Moon's (sidereal) orbital period in years. Given that $\epsilon=3.35\times 10^{-3}$, $\theta = 23.44^\circ$, $T_s({\rm day})=366.26$, and $\mu_m=0.0123$, we obtain

$$T_\phi\simeq \,{\rm years}.$$ (1.431)

This prediction is fairly close to the observed precession period of 25,800 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 (see Table 1.4) and the Moon's orbit around the Earth, as well as the small inclination of the Moon's orbit to the Earth's.

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 1.21. 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 1.21. 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 1.21: 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.

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 (in the Earth's northern hemisphere). Likewise, the Sun reaches the autumnal equinox on about September 22, and this traditionally marks the beginning of autumn. (In fact, in our calculation, the unit vector ${\bf e}_y= -\sin\phi\,{\bf e}_{x'}+\cos\phi\,{\bf e}_{y'}$ is directed toward the autumnal equinox.) 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). 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 1.22. 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 1.22: 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.