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Free precession of Earth

This appendix extends the treatment of the free precession of the Earth's axis of rotation, presented in Section 8.8, to take into account the fact that the Earth is not a completely rigid body.

The Earth is modeled as a nearly spherical body, of uniform mass density $ \gamma$ , whose boundary lies at

$\displaystyle r = R + \delta(\theta,\phi),$ (E.1)

where $ \vert\delta\vert\ll R$ . Here, $ r$ , $ \theta $ , $ \phi $ are standard spherical coordinates in the terrestrial body frame. (See Section 8.6.) Moreover, $ R$ is the Earth's mean radius. The terrestrial mass density field is well approximated as

$\displaystyle \rho({\bf r}) = \rho_0(r)+\gamma\,\delta(\theta,\phi)\,\delta(r-R),$ (E.2)

where $ \delta(r-R)$ is a Dirac delta function (Riley 1974b), and

$\displaystyle \rho_0(r)=\left\{ \begin{array}{lll} \gamma&\mbox{\hspace{0.5cm}}&r\leq R\\ [0.5ex] 0&&r>R \end{array} \right. .$ (E.3)


$\displaystyle \delta(\theta,\phi)$ $\displaystyle = - \frac{2}{3}\,\epsilon\,R\,P_2(\cos\theta) + \skew{3}\tilde{\delta}(\theta,\phi)$    
  $\displaystyle = -\frac{2}{3}\sqrt{\frac{4\pi}{5}}\,\epsilon\,R\,Y_2^{\,0}+ \skew{3}\tilde{\delta},$ (E.4)

where the first term on the right-hand side describes the Earth's rotational flattening (see Section 6.5), whereas the second is the additional displacement of the Earth's surface caused by its lack of complete rigidity. The dimensionless parameter $ \epsilon$ , which measures the degree of rotational flattening, is termed the ellipticity. (See Section 3.6.) Here, the $ Y_l^{\,m}(\theta,\phi)$ are spherical harmonics. (See Section A.12.)

Let $ x'=r\,\sin\theta\,\cos\phi$ , $ y'=r\,\sin\theta\,\sin\phi$ , $ z'=r\,\cos\theta$ be Cartesian coordinates in the Earth's body frame. In particular, let the $ z'$ -axis correspond to the symmetry axis passing through the Earth's geographic poles. Suppose that the Earth's angular velocity vector, $ \Omega$ , is slightly inclined to its symmetry axis, such that

$\displaystyle \frac{\mbox{\boldmath$\Omega$}}{\mit\Omega} \simeq l\,{\bf e}_{x'} + m\,{\bf e}_{y'}+{\bf e}_{z'},$ (E.5)

where $ \vert l\vert$ , $ \vert m\vert\ll 1$ . In the following, we shall neglect terms that are second order in the small quantities $ l$ and $ m$ .

The Earth rotational flattening is a response to the conventional centrifugal potential

$\displaystyle \chi_0 = -\frac{1}{2}\,{\mit\Omega}^{\,2}\,(x'^{\,2}+y'^{\,2}).$ (E.6)

(See Section 6.5.) However, the previous potential is calculated on the assumption that the Earth's axis of rotation coincides exactly with its symmetry axis. In fact, because the two axes are slightly inclined to one another, the true centrifugal potential is

$\displaystyle \chi = -\frac{1}{2}\,\vert$$\displaystyle \mbox{\boldmath$\Omega$}$$\displaystyle \times {\bf r}\vert^{\,2} \simeq - \frac{1}{2}\,{\mit\Omega}^{\,2}\left(x'^{\,2}+y'^{\,2} - 2\,l\,x'\,y'-2\,m\,y'\,z'\right).$ (E.7)

Thus, there is a residual potential,

$\displaystyle \delta\chi = \chi-\chi_0$ $\displaystyle = {\mit\Omega}^{\,2}\,(l\,x'+m\,y')\,z'$    
  $\displaystyle ={\mit\Omega}^{\,2}\,r^{\,2}\,(l\,\cos\phi+m\,\sin\phi)\,\sin\theta\,\cos\theta$    
  $\displaystyle = \sqrt{\frac{2\pi}{15}}\,{\mit\Omega}^{\,2}\,r^{\,2}\left[(l+{\rm i}\,m)\,Y_2^{\,-1}-(l-{\rm i}\,m)\,Y_2^{\,1}\right],$ (E.8)

that is not compensated by the Earth's rotational flattening, and to which the Earth responds elastically. Note that this potential is a solid harmonic of degree 2. (See Section C.2.) According to the analysis of Section C.4, the elastic response of the Earth to the residual potential is such that

$\displaystyle \skew{3}\tilde{\delta}(\theta,\phi) = - \frac{h_2}{g}\,\delta\chi(R,\theta,\phi),$ (E.9)

where $ g=G\,M/R^{\,2}$ is the Earth's surface gravity, and $ M=(4\pi/3)\,\gamma\,R^{\,3}$ its mass. Here, the dimensionless parameter $ h_2$ is a measure of terrestrial rigidity. In fact, $ h_2=5/2$ in the limit of zero rigidity, and $ h_2=0$ in the opposite limit of complete rigidity. [See Equation (C.50).] It follows that the net radial displacement of the Earth's surface, in response to the overall centrifugal potential, can be written

$\displaystyle \delta(\theta,\phi) =- \frac{2}{3}\sqrt{\frac{4\pi}{5}}\,\epsilon...
...h_2\,\zeta\,R\left[(l-{\rm i}\,m)\,Y_2^{\,1}-(l+{\rm i}\,m)\,Y_2^{\,-1}\right],$ (E.10)


$\displaystyle \zeta = \frac{{\mit\Omega}^{\,2}\,R^{\,3}}{3\,G\,M}$ (E.11)

is a dimensionless parameter that measures the strength of the centrifugal potential. (See Section 6.5.)

Now, it is easily shown that

    $\displaystyle x'^{\,2}+y'^{\,2}$ $\displaystyle = \frac{4\sqrt{\pi}}{3}\,r^{\,2}\left(Y_0^{\,0} - \frac{1}{\sqrt{5}}\,Y_2^{\,0}\right),$ (E.12)
    $\displaystyle x'^{\,2}+z'^{\,2}$ $\displaystyle = \sqrt{\pi}\,r^{\,2}\left(\frac{4}{3}\,Y_0^{\,0} + \sqrt{\frac{2...{2}{3}\sqrt{\frac{1}{5}}\,Y_2^{\,0} + \sqrt{\frac{2}{15}}\,Y_2^{\,-2}\right),$ (E.13)
    $\displaystyle y'^{\,2}+z'^{\,2}$ $\displaystyle = \sqrt{\pi}\,r^{\,2}\left(\frac{4}{3}\,Y_0^{\,0} -\sqrt{\frac{2}...{2}{3}\sqrt{\frac{1}{5}}\,Y_2^{\,0} - \sqrt{\frac{2}{15}}\,Y_2^{\,-2}\right),$ (E.14)
    $\displaystyle x'\,y'$ $\displaystyle = - {\rm i}\sqrt{\frac{2\pi}{15}}\,r^{\,2}\left(Y_2^{\,2}-Y_2^{\,-2}\right),$ (E.15)
    $\displaystyle x'\,z'$ $\displaystyle = \sqrt{\frac{2\pi}{15}}\,r^{\,2}\left(-Y_2^{\,1}+Y_2^{\,-1}\right),$ (E.16)
and   $\displaystyle y'\,z'$ $\displaystyle = {\rm i}\,\sqrt{\frac{2\pi}{15}}\,r^{\,2}\left(Y_2^{\,1}+Y_2^{\,-1}\right).$     (E.17)

[See Equations (A.164)-(A.172).] Thus,

$\displaystyle {\cal I}_{z'z'}$ $\displaystyle = \int_0^\infty \oint \rho({\bf r})\,(x'^{\,2}+y'^{\,2})\,r^{\,2}\,d{\mit\Omega}$    
  $\displaystyle =\gamma\int_0^R\oint (x'^{\,2}+y'^{\,2})\,r^{\,2}\,d{\mit\Omega} ...
...oint(x'^{\,2}+y'^{\,2})\,\delta(\theta,\phi)\,d{\mit\Omega}\,\right\vert _{r=R}$    
  $\displaystyle = \frac{2}{5}\,M\,R^{\,2} + \gamma\,R^{\,4}\,\frac{4\sqrt{\pi}}{3}\oint\left(Y_0^{\,0}-\frac{1}{\sqrt{5}}\,Y_2^{\,0}\right)\delta\,d{\mit\Omega},$ (E.18)

where $ d{\Omega}=\sin\theta\,d\theta\,d\phi$ is an element of solid angle, and use has been made of Equations (E.2), (E.3), and (E.12). Similarly, the other elements of the moment of inertia tensor (see Section 8.3) are written

    $\displaystyle {\cal I}_{x'x'}$ $\displaystyle = \frac{2}{5}\,M\,R^{\,2} + \gamma\,R^{\,4}\sqrt{\pi}\oint\left(\...
...}{5}}\,Y_2^{\,0} - \sqrt{\frac{2}{15}}\,Y_2^{\,-2}\right)\delta\,d{\mit\Omega},$ (E.19)
    $\displaystyle {\cal I}_{y'y'}$ $\displaystyle = \frac{2}{5}\,M\,R^{\,2} + \gamma\,R^{\,4}\sqrt{\pi}\oint\left(\...
...}{5}}\,Y_2^{\,0} + \sqrt{\frac{2}{15}}\,Y_2^{\,-2}\right)\delta\,d{\mit\Omega},$ (E.20)
    $\displaystyle {\cal I}_{x'y'}$ $\displaystyle = \gamma\,R^{\,4}\,{\rm i}\sqrt{\frac{2\pi}{15}}\oint\left(Y_2^{\,2}-Y_2^{\,-2}\right)\delta\, d{\mit\Omega},$ (E.21)
    $\displaystyle {\cal I}_{x'z'}$ $\displaystyle =- \gamma\,R^{\,4}\sqrt{\frac{2\pi}{15}}\oint\left(-Y_2^{\,1}+Y_2^{-1}\right)\delta\,d{\mit\Omega},$ (E.22)
and   $\displaystyle {\cal I}_{y'z'}$ $\displaystyle = -\gamma\,R^{\,4}\,{\rm i}\,\sqrt{\frac{2\pi}{15}}\oint\left(Y_2^{\,1}+Y_2^{-1}\right)\delta\,d{\mit\Omega}.$ (E.23)

According to Equations (A.162) and (A.163),

$\displaystyle \oint Y_l^{\,m} \,Y_{l'}^{\,-m'}\,d{\mit\Omega} = (-1)^m\,\delta_{ll'}\,\delta_{mm'}.$ (E.24)

Hence, combining Equations (E.10) with Equations (E.18)-(E.23), we deduce that

    $\displaystyle {\cal I}_{z'z'}$ $\displaystyle = {\cal I}_0\left(1+\frac{2\,\epsilon}{3}\right)= {\cal I}_\parallel,$ (E.25)
    $\displaystyle {\cal I}_{x'x'}$ $\displaystyle = {\cal I}_0\left(1-\frac{\epsilon}{3}\right)={\cal I}_\perp,$ (E.26)
    $\displaystyle {\cal I}_{y'y'}$ $\displaystyle = {\cal I}_0\left(1-\frac{\epsilon}{3}\right)={\cal I}_\perp,$ (E.27)
    $\displaystyle {\cal I}_{x'y'}$ $\displaystyle = 0,$ (E.28)
    $\displaystyle {\cal I}_{x'z'}$ $\displaystyle = \frac{3}{2}\,h_2\,\zeta\,{\cal I}_0\,l,$ (E.29)
and   $\displaystyle {\cal I}_{y'z'}$ $\displaystyle = \frac{3}{2}\,h_2\,\zeta\,{\cal I}_0\,m,$     (E.30)

where $ {\cal I}_0 = (2/5)\,M\,R^{\,2}$ . It can be seen that the elastic yielding of the Earth introduces small off-diagonal elements into its moment of inertia tensor, but does not affect the diagonal elements.

The components of the Earth's angular momentum vector, $ {\bf L}$ , take the form (see Section 8.3),

    $\displaystyle L_{x'}$ $\displaystyle = {\cal I}_{x'x'}\,{\mit \Omega}_{x'} + {\cal I}_{x'y'}\,{\mit\Om...
...z'} \simeq {\mit\Omega}\,{\cal I}_\perp\left(1+\frac{3}{2}\,h_2\,\zeta\right)l,$ (E.31)
    $\displaystyle L_{y'}$ $\displaystyle = {\cal I}_{x'y'}\,{\mit \Omega}_{x'} + {\cal I}_{y'y'}\,{\mit\Om...
...{z'}\simeq {\mit\Omega}\,{\cal I}_\perp\left(1+\frac{3}{2}\,h_2\,\zeta\right)m,$ (E.32)
and   $\displaystyle L_{z'}$ $\displaystyle = {\cal I}_{x'z'}\,{\mit \Omega}_{x'} + {\cal I}_{y'z'}\,{\mit\Om...
...{y'}+{\cal I}_{z'z'}\,{\mit\Omega}_{z'}\simeq {\mit\Omega}\,{\cal I}_\parallel.$     (E.33)

According to Equation (8.25), in the absence of an external torque, the Earth's rotational equation of motion is written

$\displaystyle \frac{d{\bf L}}{dt'} = {\bf L}\times$   $\displaystyle \mbox{\boldmath$\Omega$}$$\displaystyle ,$ (E.34)

where $ d/dt'$ denotes a time derivative in the body frame. Assuming that the direction cosines $ l$ and $ m$ are the only quantities that evolve in time, the $ x'$ - and $ y'$ -components of the previous equation yield

    $\displaystyle \frac{dl}{dt'} + \omega\,m$ $\displaystyle = 0,$ (E.35)
and   $\displaystyle \frac{dm}{dt'}-\omega\,l$ $\displaystyle = 0,$     (E.36)

respectively, where

$\displaystyle \omega =\left[\frac{{\cal I}_\parallel-(1+3\,h_2\,\zeta/2)\,{\cal I}_\perp}{(1+3\,h_2\,\zeta/2)\,{\cal I}_\perp}\right]{\mit\Omega}.$ (E.37)

It follows that, in the body frame,

    $\displaystyle l$ $\displaystyle =\alpha\,\cos(\omega\,t),$ (E.38)
and   $\displaystyle m$ $\displaystyle =\alpha\,\sin(\omega\,t),$     (E.39)

where $ \alpha$ is the arbitrary angle of inclination between the Earth's axis of rotation and its symmetry axis. Of course, for consistency with our previous analysis, this angle must be small. The previous two equations indicate that, in the body frame, the Earth's axis of rotation precesses about its symmetry axis at the angular frequency $ \omega$ . The period of this precession in sidereal days is

$\displaystyle T = \frac{(1+3\,h_2\,\zeta/2)\,{\cal I}_\perp}{{\cal I}_\parallel...
...\tilde{\epsilon}+ 3\,h_2\,\zeta/2}{\skew{3}\tilde{\epsilon} - 3\,h_2\,\zeta/2},$ (E.40)


$\displaystyle \skew{3}\tilde{\epsilon}= \frac{{\cal I}_\parallel-{\cal I}_\perp}{{\cal I}_\parallel}$ (E.41)

is the Earth's dynamical ellipticity. (See Section 8.10 and Appendix F.) Note that, were the Earth completely rigid (i.e., $ h_2=0$ ), the precession period would be

$\displaystyle T _0\simeq \frac{1-\skew{3}\tilde{\epsilon}}{\skew{3}\tilde{\epsilon} }.$ (E.42)

Approximating the Earth as a uniform elastic body of uniform rigidity $ \mu$ , we find that $ \skew{3}\tilde{\epsilon}=\epsilon$ [i.e., the Earth's dynamical ellipticity is equal to the actual ellipticity due to its slightly flattened shape; see Equations (E.25)-(E.27)] and

$\displaystyle h_2 = \frac{5/2}{1+(38\pi/3)\,(\mu\,R^{\,4}/G\,M^{\,2})}.$ (E.43)

[See Equation (C.50).] Using $ \epsilon=3.35\times 10^{-3}$ , $ {\mit\Omega}=7.29\times 10^{-5}\,{\rm rad./s}$ , $ R=6.37\times 10^6\,{\rm m}$ , $ M = 5.97\times 10^{24}\,{\rm kg}$ , and $ \mu=1\times 10^{11}\,{\rm N\,m}^{-2}$ (see Section 6.5), we deduce that

    $\displaystyle \skew{3}\tilde{\epsilon}$ $\displaystyle = 3.35\times 10^{-3},$ (E.44)
and   $\displaystyle h_2$ $\displaystyle = 0.67.$     (E.45)

Hence, our estimate for the free precession period of the Earth's axis of rotation (in its body frame) becomes

$\displaystyle T = 453\,\,{\rm days}.$ (E.46)

If the Earth were modeled as completely rigid then the corresponding precession period would be

$\displaystyle T_0 = 297\,{\rm days}.$ (E.47)

Now, the observed precession period is $ 434$ days (Yoder 1995). Thus, our estimate for the precession period that takes the Earth's lack of complete rigidity into account is far closer to the correct answer than our estimate that treats the Earth as a completely rigid body.

We can improve our estimate for the Earth's free precession period by taking into account the fact that the Earth is not actually a uniform body. This is possible because Equation (E.40) turns out to be completely general (i.e., although it was derived on the assumption that the Earth is a uniform body, it remains valid for a non-uniform Earth) (Love 1909). The Earth's actual dynamical ellipticity, which is deduced from the measured rate of luni-solar precession, turns out to be

$\displaystyle \skew{3}\tilde{\epsilon} = 3.273\times 10^{-3}.$ (E.48)

(See Appendix F.) Furthermore, the value of the Love number $ h_2$ , which is deduced from the measured yielding of the Earth to the Moon's tidal gravitational field, is

$\displaystyle h_2=0.60.$ (E.49)

(Love 1909.) Hence, our improved estimate for the Earth's free precession period becomes

$\displaystyle T = 445\,\,{\rm days},$ (E.50)

which lies within 3 percent of the correct answer. The corresponding precession period for a completely rigid Earth is

$\displaystyle T = 304\,\,{\rm days}.$ (E.51)

(See Section 8.8.) The remaining difference between our improved estimate for the Earth's free precession period and the observed precession period can be explained by taking into account the effect of the Earth's fluid core, as well as the fact that much of its surface is covered with fluid oceans (Smith and Dahlen 1981).
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Next: Forced precession and nutation Up: Celestialhtml Previous: Simple application of Darwin-Radau
Richard Fitzpatrick 2016-03-31