Cassini's laws

Consider Figure 8.13. Here, $M$, $E$, and $N$, represent the center of the Moon, the center of the Earth, and the north ecliptic pole, respectively. Moreover, $MZ$ is the instantaneous normal to the Moon's equatorial plane, and $MP$ the instantaneous normal to the Moon's orbital plane. Let ${\bf e}$, $\zeta$, $\nu$, and ${\bf p}$ be unit vectors parallel to $ME$, $MZ$, $MN$, and $MP$, respectively. The fixed angle, $I=5.16^\circ$, subtended between the directions of ${\bf p}$ and $\nu$, represents the fixed inclination of the Moon's orbital plane to the ecliptic plane. Furthermore, as is well known, because of the perturbing action of the Sun, the normal to the Moon's orbital plane, ${\bf p}$, precesses about the normal to the ecliptic plane, $\nu$, in the opposite sense to the Moon's orbital motion, such that it completes a full circuit every $18.6$ years. (See Chapter 11.) Recall that precession in the opposite sense to orbital motion is usually termed regression.

In 1693, the astronomer Gian Domenico Cassini (1625–1712) formulated a set of empirical laws that succinctly describe the Moon's axial rotation. According to these laws:

1. The Moon spins at a uniform rate that matches its mean orbital rotation rate.
2. The normal to the Moon's equatorial plane subtends a fixed angle, $\iota=1.59^\circ$, with the normal to the ecliptic plane.
3. The normal to the Moon's equatorial plane, the normal to the Moon's orbital plane, and the normal to the ecliptic plane, are coplanar vectors that are orientated such that the latter vector lies between the other two.

Law 1 effectively states that the Moon is locked in a $1$:$1$ spin-orbit resonance. (See Section 8.11.) Let the $x$-, $y$-, and $z$-axes be the Moon's principal axes of rotation, and let ${\cal I}_{xx}$, ${\cal I}_{yy}$, and ${\cal I}_{zz}$ be the corresponding principal moments of inertia. Furthermore, let us label the principal axes such that the Moon's equatorial plane corresponds to the $x$-$y$ plane, the normal to the equatorial plane corresponds to the $z$-axis, and ${\cal I}_{yy}>{\cal I}_{xx}$. In this case, as we saw in the previous section, a $1$:$1$ spin-orbit resonant state is such that the Moon's $x$-axis always points approximately in the direction of the Earth; that is, ${\bf e}$ is almost parallel to the $x$-axis.

Law 2 states that the angle, $\iota$, subtended between $\zeta$ and $\nu$ is fixed. Moreover, because the angles $I$ and $\iota$ are both small (when expressed in radians), we deduce that the vectors $\nu$ and ${\bf p}$ are almost parallel to $\zeta$.

Law 3 states that the vectors $\zeta$, $\nu$, and ${\bf p}$ all lie in the same plane, with $\zeta$ and ${\bf p}$ on opposite sides of $\nu$. In other words, as the normal to the Moon's orbital plane, ${\bf p}$, regresses about the normal to the ecliptic plane, $\nu$, the normal to the Moon's equatorial plane, $\zeta$, regresses at the same rate, such that $\zeta$ is always directly opposite ${\bf p}$ with respect to $\nu$.

Figure 8.13: Geometry of Cassini's laws.

Cassini's first law was accounted for in the previous section. The ultimate aim of this section is to account for Cassini's second and third laws. Our approach is largely based on that of Danby (1992). In order to simplify the analysis, we shall assume that the Moon orbits around the Earth, at the uniform angular velocity, $n$, in a circular orbit of major radius $a$. When expressed in terms of the $x$, $y$, $z$ coordinate system, $\zeta$$= (0,\,0,\,1)$. Furthermore, because the unit vectors $\nu$ and ${\bf p}$ are almost parallel to $\zeta$, we can write

$$\mbox{\boldmath$\nu$} \simeq(\nu_x,\,\nu_y,\,1),$$ (8.190)

$${\bf p} \simeq (p_x,\,p_y,\,1),$$ (8.191)

where $\vert\nu_x\vert$, $\vert\nu_y\vert$, $\vert p_x\vert$, $\vert p_y\vert\ll 1$. Similarly, because the unit vector ${\bf e}$ is almost parallel to the $x$-axis, we have

$${\bf e} \simeq (1,\,e_y,\,e_z),$$ (8.192)

where $\vert e_y\vert$, $\vert e_z\vert\ll 1$. The position vector, ${\bf r}=(x,\, y,\, z)$, of the center of the Earth with respect to the center of the Moon is written

$${\bf r} = a\,{\bf e}.$$ (8.193)

Finally, given Cassini's first law, and assuming that the Moon's spin axis is almost parallel to the $z$-axis, the Moon's spin angular velocity takes the form

$$\mbox{\boldmath$\omega$} = n\,{\bf w},$$ (8.194)

Here, ${\bf w}$ is a unit vector such that

$${\bf w}\simeq (w_x,\,w_y,\,1),$$ (8.195)

where $\vert w_x\vert$, $\vert w_y\vert\ll 1$.

According to Equations (8.150), (8.151), (8.153), and (8.154),

$$\skew{5}\dot{\omega}_x + \sigma\,\omega_y\,\omega_z =3\,n^{\,2}\,\sigma\,\frac{y\,z}{a^{\,2}},$$ (8.196)

$$\skew{5}\dot{\omega}_y -\tau\,\omega_x\,\omega_z =-3\,n^{\,2}\,\tau\,\frac{x\,z}{a^{\,2}},$$ (8.197)

because $r=a$ and $G\,m_p=n^{\,2}\,a^{\,3}$. Here,

$$\sigma =\frac{{\cal I}_{zz}-{\cal I}_{yy}}{{\cal I}_{xx}},$$ (8.198)

$$\tau =\frac{{\cal I}_{zz}-{\cal I}_{xx}}{{\cal I}_{yy}}.$$ (8.199)

Note that $\vert\sigma\vert$, $\vert\tau\vert\ll 1$ because the Moon is almost spherically symmetric. To second order in small quantities, Equations (8.196) and (8.197) yield

$$n^{-1}\,\dot{w}_x + \sigma\,w_y \simeq 0,$$ (8.200)

$$n^{-1}\,\dot{w}_y -\tau\,w_x \simeq -3\,\tau\,e_z,$$ (8.201)

where use has been made of Equations (8.192)–(8.195).

The unit vector $\nu$ is stationary in an inertial frame whose coordinate axes are fixed with respect to distant stars. Hence, in the $x$, $y$, $z$ body frame, which rotates with respect to the aforementioned fixed frame at the angular velocity $\omega$, we have (see Section 6.2)

$$\frac{d\mbox{\boldmath$\nu$}}{dt} + \mbox{\boldmath$\omega$}\times \mbox{\boldmath$\nu$}= {\bf0}.$$ (8.202)

It follows, from Equations (8.190), (8.194), and (8.195), that

$$w_x \simeq n^{-1}\skew{3}\dot{\nu}_y +\nu_x,$$ (8.203)

$$w_y \simeq-n^{-1}\,\skew{3}\dot{\nu}_x+\nu_y.$$ (8.204)

These expressions can be combined with Equations (8.200) and (8.201) to give

$$n^{-2}\,\skew{3}\ddot{\nu}_x -(1-\tau)\,n^{-1}\,\skew{3}\dot{\nu}_y+\tau\,\nu_x \simeq 3\,\tau\,e_z,$$ (8.205)

$$n^{-2}\,\skew{3}\ddot{\nu}_y+(1-\sigma)\,n^{-1}\,\skew{3}\dot{\nu}_x+\sigma\,\nu_y \simeq 0.$$ (8.206)

It now remains to express $e_z$ in terms of $\nu_x$ and $\nu_y$.

By definition, ${\bf p}$ is normal to ${\bf e}$, as the vector ${\bf e}$ lies in the plane of the Moon's orbit. Hence, according to Equations (8.191) and (8.192),

$${\bf p}\cdot{\bf e} \simeq p_x+ e_z= 0,$$ (8.207)

which implies that

$$e_z \simeq - p_x.$$ (8.208)

Let $A$ be the ascending node of the Earth's apparent orbit about the Moon (which implies that $A$ is the descending node of the Moon's actual orbit about the Earth), and let ${\bf a}$ be a unit vector parallel to $MA$. See Figure 8.13 and Section 4.12. By definition, ${\bf a}$ is normal to both ${\bf p}$ and $\nu$. In fact, we can write

$${\bf a} = \frac{\mbox{\boldmath$\nu$}\times {\bf p}}{\sin I},$$ (8.209)

where $I$ is the angle subtended between the vectors ${\bf p}$ and $\nu$. It follows from Equations (8.190), (8.191), and the fact that $I$ is small, that

$${\bf a} \simeq I^{\,-1}\left(\nu_y-p_y,\,\,p_x-\nu_x,\,\,\nu_x\,p_y-\nu_y\,p_x\right).$$ (8.210)

Now,

$${\bf a}\cdot{\bf e} =\cos\psi,$$ (8.211)

$${\bf a}\times {\bf e}\cdot{\bf p} =\sin\psi,$$ (8.212)

where $\psi$ is the angle subtended between ${\bf a}$ and ${\bf e}$. See Figure 8.13. Thus, Equations (8.191), (8.192), and (8.210) yield

$$\nu_x = p_x+ I\,\sin\psi,$$ (8.213)

$$\nu_y = p_y+I\,\cos\psi,$$ (8.214)

and

$${\bf a} \simeq \left(\cos\psi,\,\,-\sin\psi,\,\,\nu_y\,\sin\psi-\nu_x\,\cos\psi\right).$$ (8.215)

In fact, $\psi$ is the longitude of the Earth relative to the ascending node of its apparent orbit around the Moon. It follows that

$$\psi = (n+g)\,t,$$ (8.216)

where $g$ is the uniform regression rate of the Earth's apparent ascending node (which is the same as the regression rate of the true ascending node of the Moon's orbit around the Earth). Here, for the sake of simplicity, we have assumed that the Earth passes through its apparent ascending node at time $t=0$. Hence, Equations (8.205), (8.206), (8.208), (8.213), and (8.216) can be combined to give

$$n^{-2}\,\skew{3}\ddot{\nu}_x - (1-\tau)\,n^{-1}\,\skew{3}\dot{\nu}_y + 4\,\tau\,\nu_x \simeq 3\,\tau\,I\,\sin[(n+g)\,t],$$ (8.217)

$$n^{-2}\,\skew{3}\ddot{\nu}_y+(1-\sigma)\,n^{-1}\,\skew{3}\dot{\nu}_x+\sigma\,\nu_y \simeq 0.$$ (8.218)

The previous two equations govern the Moon's physical libration in latitude. As is the case for libration in longitude, there are both free and forced modes. The free modes satisfy

$$n^{-2}\,\skew{3}\ddot{\nu}_x - (1-\tau)\,n^{-1}\,\skew{3}\dot{\nu}_y + 4\,\tau\,\nu_x \simeq 0,$$ (8.219)

$$n^{-2}\,\skew{3}\ddot{\nu}_y+(1-\sigma)\,n^{-1}\,\skew{3}\dot{\nu}_x+\sigma\,\nu_y \simeq 0.$$ (8.220)

Let us search for solutions of the form

$$\nu_x(t) =\skew{3}\hat{\nu}_x\,\sin(s\,n\,t-\phi),$$ (8.221)

$$\nu_y(t) =\skew{3}\hat{\nu}_y\,\cos(s\,n\,t-\phi),$$ (8.222)

where $\skew{3}\hat{\nu}_x$, $\skew{3}\hat{\nu}_y$, $\phi$ are constants. It follows that

$$\frac{\skew{3}\hat{\nu}_y}{\skew{3}\hat{\nu}_x}= \frac{s^{\,2}-4\,\tau}{(1-\tau)\,s}=\frac{(1-\sigma)\,s}{s^{\,2}-\sigma}.$$ (8.223)

Given that $\vert\sigma\vert$ and $\vert\tau\vert$ are both small compared to unity, two independent free libration modes can be derived from the preceding expression. The first mode is such that $s\simeq 1+3\,\tau/2$ and $\skew{3}\hat{\nu}_y/\skew{3}\hat{\nu}_x\simeq 1-3\,\tau/2$, whereas the second is such that $s\simeq 2\sqrt{\sigma\,\tau}$ and $\skew{3}\hat{\nu}_y/\skew{3}\hat{\nu}_x \simeq -2\sqrt{\tau/\sigma}$. In the Moon's body frame, these modes cause the normal to the ecliptic plane, $\nu$$=(\nu_x,\,\nu_y,\,1)$, to precess about the normal to the Moon's equatorial plane, $\zeta$$= (0,\,0,\,1)$, in such a manner that

$$\nu_x \simeq A_1\,\sin(\omega_1\,t-\phi_1)+ A_2\,\sin(\omega_2\,t-\phi_2),$$ (8.224)

$$\nu_y \simeq K_1\,A_1\,\cos(\omega_1\,t-\phi_1)+ K_2\,A_2\,\cos(\omega_2\,t-\phi_2),$$ (8.225)

where $\omega_1 \simeq n\,(1+3\,\tau/2)$, $\omega_2\simeq 2\,n\sqrt{\sigma\,\tau}$, $K_1\simeq 1-3\,\tau/2$, $K_2\simeq -2\sqrt{\tau/\sigma}$, and the constants $A_1$, $A_2$, $\phi_1$, $\phi_2$ are arbitrary. The observed values of $n$, $\sigma$, and $\tau$ are $13.1764^\circ$ per day, $4.0362\times 10^{-4}$, and $6.3149\times 10^{-4}$, respectively (Konopliv et al. 1998; Dickey et al. 1994). [Actually, $\sigma$ and $\tau$ are measured by fitting observations of lunar libration obtained from laser ranging to the theory described here.] Thus, it follows that $\omega_1=13.1889^\circ$ per day, $\omega_2 = 1.3304\times 10^{-2\,\circ}$ per day, $K_1=0.9991$, and $K_2=-2.5017$. In the body frame, the first mode causes $\nu$ to regress about $\zeta$ with a period of $27.2957$ days, whereas the second mode causes $\nu$ to precess about $\zeta$ with a period of $74.1$ years. Both these modes of libration have been detected by means of lunar laser ranging. The measured amplitude of the first mode is $A_1=0.37''$, whereas that of the second mode is $A_2=3.25''$ (Jin and Li 1996). Incidentally, the second mode is very similar in nature to the Chandler wobble of the Earth. (See Section 8.8.) Note that if $\sigma$ and $\tau$ were of opposite sign—that is, if ${\cal I}_{zz}$ were intermediate between ${\cal I}_{xx}$ and ${\cal I}_{yy}$—the second mode of libration would grow exponentially in time, rather than oscillate at a constant amplitude; in other words, the Moon's spin state would be unstable. In fact, the Moon's principal axes of rotation are orientated such that ${\cal I}_{zz}>{\cal I}_{yy}>{\cal I}_{xx}$, which ensures that the Moon spins in a stable manner.

Let us now search for forced solutions of Equations (8.217) and (8.218) of the form

$$\nu_x =\skew{3}\hat{\nu}_x\,\sin[(n+g)\,t],$$ (8.226)

$$\nu_y =\skew{3}\hat{\nu}_y\,\cos[(n+g)\,t],$$ (8.227)

where $\skew{3}\hat{\nu}_x$, $\skew{3}\hat{\nu}_y$ are constants. It follows that

$$\left[4\,\tau-(1+g/n)^{\,2}\right]\skew{3}\hat{\nu}_x + (1-\tau)\,(1+g/n)\,\skew{3}\hat{\nu}_y \simeq 3\,\tau\,I,$$ (8.228)

$$(1-\sigma)\,(1+g/n)\,\skew{3}\hat{\nu}_x+\left[\sigma-(1+g/n)^{\,2}\right]\skew{3}\hat{\nu}_y \simeq 0.$$ (8.229)

Hence, recalling that $\sigma$, $\tau$, and $g/n$ are all small compared to unity, we obtain the following mode of forced libration:

$$\frac{\skew{3}\hat{\nu}_y}{\skew{3}\hat{\nu}_x} \simeq 1-\frac{g}{n},$$ (8.230)

$$\skew{3}\hat{\nu}_x \simeq \frac{3\,\tau\,I}{3\,\tau-2\,g/n}.$$ (8.231)

In the Moon's body frame, this mode causes the vectors $\nu$ and ${\bf p}$ to regress about $\zeta$ (i.e., the $z$-axis) in such a manner that

$$\mbox{\boldmath$\nu$} \simeq (-\iota\,\sin\psi,\,\,-\iota\,\cos\psi,\,\,1),$$ (8.232)

$${\bf p} \simeq \left(-(I+\iota)\,\sin\psi,\,\,-(I+\iota)\,\cos\psi,\,\,1\right),$$ (8.233)

where

$$\iota = \frac{3\,\tau\,I}{2\,g/n-3\,\tau},$$ (8.234)

and use has been made of Equations (8.213), (8.214), and (8.216). Because the observed values of $I$, $\tau$, and $g/n$ are $I=5.16^\circ$, $\tau=6.3149\times 10^{-4}$, and $g/n=4.0185\times 10^{-3}$ (Konopliv et al. 1998; Dickey et al. 1994; Yoder 1995), we deduce that

$$\iota = 1.59^\circ.$$ (8.235)

According to Equation (8.232), $\nu$ regresses around $\zeta$, with a period of $27.2122$ days (i.e., a draconic month), in such a manner that $\nu$ subtends a fixed angle of $\iota=1.59^\circ$ with respect to $\zeta$. This accounts for Cassini's second law. According to Equation (8.233), ${\bf p}$ simultaneously regress around $\zeta$, with the same period, in such a manner that $\zeta$$\cdot$   $\nu$$\times {\bf p}=0$. In other words, the three vectors $\zeta$, $\nu$, and ${\bf p}$ always lie in the same plane. Moreover, it is clear that $\nu$ is intermediate between $\zeta$ and ${\bf p}$. This accounts for Cassini's third law. The angle $I'=I+\iota$, subtended between $\zeta$ and ${\bf p}$, which is also the angle of inclination between the Moon's equatorial and orbital planes, takes the fixed value

$$I' = \frac{2\,(g/n)\,I}{2\,g/n-3\,\tau}=6.75^\circ.$$ (8.236)

Note that this angle would be zero in the absence of any regression of the Moon's orbital ascending node (i.e., if $g/n$ were zero). In other words, the nonzero angle of inclination between the Moon's equatorial and orbital planes is a direct consequence of this regression, which is ultimately due to the perturbing action of the Sun. Because the regression of the Moon's orbital ascending node is also responsible for the forced nutation of the Earth's axis of rotation (see Section 8.10), it follows that this nutation is closely related to the forced inclination between the Moon's equatorial and orbital planes.