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Next: Useful mathematics Up: Lunar motion Previous: Historical note

Exercises

  1. Demonstrate that the lunar equation of motion, Equation (11.33), can be written in the canonical form

    $\displaystyle \ddot{\bf r} +n^{\,2}\,a^{\,3}\,\frac{{\bf r}}{r^{\,3}} =\nabla{\cal R},
$

    where

    $\displaystyle {\cal R} = \frac{n'^{\,2}\,a'^{\,3}}{r'}\left(\frac{r}{r'}\right)...
...}{r'}\left(\frac{E-M}{E+M}\right)\left(\frac{r}{r'}\right)^3 P_3(\cos\theta)
$

    is the disturbing function due to the gravitational influence of the Sun. Here, $ {\bf r}$ represents the position of the Moon relative to the Earth, $ {\bf r}'$ the position of the Sun relative to the Earth-Moon barycenter, $ \theta $ the angle subtended between $ {\bf r}$ and $ {\bf r}'$ , $ n$ the mean orbital angular velocity of the Moon around the Earth, $ n'$ the mean (apparent) orbital angular velocity of the Sun around the Earth-Moon barycenter, $ a$ the mean radius of the former orbit, $ a'$ the mean radius of the latter orbit, $ E$ the mass of the Earth, and $ M$ the mass of the moon.

  2. Consider the equation of motion of the Earth-Moon barycenter, (11.34). Approximating the orbit of the barycenter around the Sun as a circle of major radius $ a'$ , and that of the Moon and the Earth about the barycenter as a circle of major radius $ a$ , and then averaging over the motions of the Moon and the Earth, show that Equation (11.34) reduces to

    $\displaystyle \ddot{{\bf r}}+n'^{\,2}\,a'^{\,3}\,\frac{{\bf r}'}{r'^{\,3}}=-\frac{3}{4}\,\epsilon\, n'^{\,2}\,a'^{\,5}\,\frac{{\bf r}'}{r'^{\,5}},
$

    where $ n'$ is the mean orbital angular velocity of the barycenter,

    $\displaystyle \epsilon =\frac{E\,M}{(E+M)^{\,2}}\left(\frac{a}{a'}\right)^2,
$

    $ E$ is the mass of the Earth, and $ M$ is the mass of the Moon. Hence, deduce that the combined disturbing action of the Earth and the Moon causes an additional precession of the perihelion of the barycenter's orbit at the rate

    $\displaystyle \dot{\varpi}' = \frac{3\,\epsilon}{4}\,n',
$

    or $ 7.7$ arc seconds per (Julian) century. Note that this precession rate is much less than that caused by the other planets in the solar system. (See Section 5.4.)

  3. Because it is the Earth-Moon barycenter, rather than the Earth, that describes a Keplerian orbit about the Sun, it follows that, when observed from the Earth, the Sun will not apparently move in a Keplerian orbit. Let $ \delta \lambda'$ and $ \delta\beta'$ be the associated displacements of the Sun, in ecliptic longitude and latitude, respectively, with respect to a Keplerian orbit. Demonstrate that

    $\displaystyle \delta\lambda'$ $\displaystyle \simeq \left(\frac{M}{E+M}\right)\left(\frac{a}{a'}\right)\sin D,$    
    $\displaystyle \delta \beta'$ $\displaystyle \simeq I\left(\frac{M}{E+M}\right)\left(\frac{a}{a'}\right)\sin F,$    

    where $ M$ is the mass of the Moon, $ E$ the mass of the Earth, $ a$ the major radius of the lunar orbit, $ a'$ the major radius of the barycenter's orbit, $ I$ the inclination of the lunar orbit to the ecliptic plane, $ D$ the mean elongation of the Moon from the Earth, and $ F$ the lunar mean argument of latitude. Show that $ \delta\lambda' = 6.4''$ and $ \delta\beta' = 0.58''$ .

  4. Demonstrate that the lowest-order evection term,

    $\displaystyle \frac{15}{4}\,m\,e\,\sin(2\,D-{\cal M}),
$

    appearing in Equation (11.303), can be represented as the combined effect of periodic variations in the eccentricity, $ e$ , of the lunar orbit, and the mean longitude, $ \alpha$ , of the lunar perigee: in other words, $ e\rightarrow
e\,[1-(15/8)\,m\,\cos (2\, D)]$ and $ \alpha\rightarrow \alpha - (15/8)\,m\,\sin (2\, D)$ . Likewise, show that the lowest-order evection term,

    $\displaystyle \frac{3}{8}\,m\,I\,\sin(2\,D-F),
$

    appearing in Equation (11.320), can be represented as the combined effect of periodic variations in the inclination, $ I$ , of the lunar orbit, and the mean longitude, $ \gamma$ , of the lunar ascending node: in other words, $ I\rightarrow
I\,[1-(3/8)\,m\,\cos (2\, D)]$ and $ \gamma\rightarrow \gamma - (3/8)\,m\,\sin (2\, D)$ .

  5. Suppose that the major radius of the lunar orbit were reduced by a multiplicative factor $ \zeta$ ; that is, $ a\rightarrow \zeta\,a$ , where $ 0<\zeta<1$ . Assuming that the masses of the Earth and Sun, and the major radius of the terrestrial orbit, remain constant, demonstrate that the parameter $ m$ , which measures the lowest-order perturbing influence of the Sun on the lunar orbit, would be reduced by a factor $ \zeta^{\,3/2}$ ; in other words, $ m\rightarrow \zeta^{\,3/2}\,m$ . Given that $ m=0.07480$ for the true lunar orbit, how small would $ \zeta$ have to be before the (theoretical) precession rate of the lunar perigee became equal to the regression rate of the ascending node to within $ 1$ percent? What is the corresponding major radius of the lunar orbit in units of mean Earth radii? (The true major radius of the lunar orbit is $ 60.9$ mean Earth radii.)

  6. An artificial satellite orbits the Moon in a low-eccentricity orbit whose major radius is twice the lunar radius. The plane of the satellite orbit is slightly inclined to the plane of the Moon's orbit about the Earth. Given that the mass of the Earth is $ 81.3$ times that of the Moon, and the major radius of the lunar orbit is $ 221.3$ times the lunar radius, estimate the precession period of the satellite orbit's perilune (i.e., its point of closest approach to the Moon) in months due to the perturbing influence of the Earth. Likewise, estimate the regression rate of the satellite orbit's ascending node (with respect to the plane of the lunar orbit) in months. (Assume that the Moon is a perfect sphere.)

  7. The mean ecliptic longitudes (measured with respect to the vernal equinox at a fixed epoch) of the Moon and the Sun increase at the rates $ 13.176359^\circ$ per day and $ 0.98560912^\circ$ per day, respectively. However, the vernal equinox regresses in such a manner that, on average, it completes a full circuit every $ 25,772$ years. Furthermore, the lunar perigee precesses in such a manner that, on average, it completes a full circuit every $ 8.848$ years, whereas the lunar ascending mode regresses in such a manner that, on average, it completes a full circuit every $ 18.615$ years (Yoder 1995). A sidereal month is the mean period of the Moon's orbit with respect to the fixed stars, a tropical month is the mean time required for the Moon's ecliptic longitude (with respect to the true vernal equinox) to increase by $ 360^\circ$ , a synodic month is the mean period between successive new moons, an anomalistic month is the mean period between successive passages of the Moon through its perigee, and a draconic month is the mean period between successive passages of the Moon through its ascending node. Use the preceding information to demonstrate than the lengths of a sidereal, tropical, synodic, anomalistic, and draconic month are $ 27.32166$ , $ 27.32158$ , $ 29.5306$ , $ 27.5546$ , and $ 27.2123$ days, respectively.

  8. To first order in the Moon's orbital eccentricity and inclination, the geocentric ecliptic longitude and latitude of the Moon, relative to the Sun, are written

        $\displaystyle {\mit\Delta}\lambda$ $\displaystyle \simeq (n-n')\,t + 2\,e\,\sin[(n-n_p)\,t],$    
    and   $\displaystyle {\mit\Delta}\beta$ $\displaystyle \simeq I\,\sin[(n-n_n)\,t],$        

    respectively, where $ e=0.05488$ and $ I=0.9008$ radians. Here, $ n$ and $ n'$ are the mean geocentric orbital angular velocities of the Moon and Sun, respectively, $ n_p$ is the mean orbital angular velocity of the lunar perigee, and $ n_n$ is the mean orbital angular velocity of the lunar ascending node. Note that $ 2\pi/(n-n') = T_{\rm synodic}$ , $ 2\pi/(n-n_p)=T_{\rm anomalistic}$ , and $ 2\pi/(n-n_n)= T_{\rm draconic}$ , where $ T_{\rm synodic}$ , $ T_{\rm anomalistic}$ , and $ T_{\rm draconic}$ are the lengths of a synodic, anomalistic, and draconic month, respectively. At $ t=0$ , we have $ {\mit\Delta}\lambda={\mit\Delta}\beta=0^\circ$ . In other words, at $ t=0$ , the Moon and Sun have exactly the same geocentric ecliptic longitudes and latitudes, which implies that a solar eclipse occurs at this time. Suppose we can find some time period $ T$ that satisfies $ T=j_1\,T_{\rm synodic}=j_2\,T_{\rm anomalistic}=
j_3\,T_{\rm draconic}$ , where $ j_1$ , $ j_2$ , $ j_3$ are positive integers. Demonstrate that $ {\mit\Delta}\lambda={\mit\Delta}\beta=0^\circ$ at $ t=T$ . Thus, if the period $ T$ , which is known as the saros, existed then solar (and lunar) eclipses would occur in infinite sequences spaced $ j_1$ synodic months apart (Roy 2005). Show that for $ 0< j_1,\,j_2,\,j_3< 1000$ , the closest approximation to the saros is obtained when $ j_1=223$ , $ j_2=239$ , and $ j_3=242$ . Demonstrate that if $ {\mit\Delta}\lambda={\mit\Delta}\beta=0^\circ$ at $ t=0$ (i.e., if there is a solar eclipse at $ t=0$ ) then, exactly $ 223$ synodic months later, $ {\mit\Delta}\lambda = -0.3^\circ$ and $ {\mit\Delta}\beta=-0.1^\circ$ . It turns out that these values of $ {\mit \Delta }\lambda $ and $ {\mit\Delta}\beta$ are sufficiently small that the eclipse reoccurs. In fact, because $ 223$ synodic months almost satisfies the saros condition, solar (and lunar) eclipses occur in series of about 70 eclipses spaced $ 223$ synodic months, or 18 years and 11 days, apart.

  9. Let the $ x$ -, $ y$ -, and $ z$ -axes be the lunar principal axes of rotation passing through the lunar center of mass. Because the Moon is not quite spherically symmetric, its principal moments of inertia are not exactly equal to one another. Let us label the principal axes such that $ {\cal I}_{zz}>{\cal I}_{yy}>{\cal I}_{xx}$ . To a first approximation, the Moon is spinning about the $ z$ -axis, which is orientated normal to its orbital plane. Moreover, the Moon spins in such a manner that the $ x$ -axis always points approximately in the direction of the Earth. Let $ \eta$ be the (small) angle subtended between the $ x$ -axis and the line joining the centers of the Moon and the Earth. A slight generalization of the analysis in Section 8.11 reveals that

    $\displaystyle \eta = \eta_o + \eta_p,
$

    where $ \eta_o=\delta\lambda$ and $ \eta_p$ are the Moon's optical and physical libration (in ecliptic longitude), respectively, $ \delta \lambda$ is defined in Equation (11.104), and

    $\displaystyle \skew{3}\ddot{\eta}_p + n_0^{\,2}\,\eta_p =-n_0^{\,2}\,\eta_o.
$

    Here, $ n_0= [3\,({\cal I}_{yy}-{\cal I}_{xx})/{\cal I}_{zz}]^{1/2}\,n = 0.3446^\circ$ per day is the Moon's free libration rate, whereas $ n= 13.1764^\circ$ per day is the lunar mean sidereal orbital angular velocity.

    We can write

    $\displaystyle \delta\lambda \simeq \sum_{i=1,5} A_i\,\sin(n_i\,t-\gamma_i),
$

    where the $ i=1,$ $ 2$ , $ 3$ , $ 4$ , and $ 5$ terms correspond to the major inequality, the reduction to the ecliptic, variation, evection, and the annual inequality, respectively. Furthermore, $ A_1=22,640''$ , $ A_2=418''$ , $ A_3=2,370''$ , $ A_4=4,586''$ , $ A_5=666''$ , and $ n_1=n$ , $ n_2=2\,n_{\rm dr}$ , $ n_3=2\,n_{\rm sy}$ , $ n_4=2\,n_{\rm sy}-n_{\rm an}$ , $ n_5=n'$ . Here, $ n_{\rm dr}=13.2293^\circ$ per day, $ n_{\rm sy}=12.1908^\circ$ per day, $ n_{\rm an}=13.0650^\circ$ per day, are the lunar mean draconic, synodic, and anomalistic orbital angular velocities, respectively, and $ n'=0.9856^\circ$ per day is the Earth's mean sidereal orbital angular velocity. Demonstrate that

    $\displaystyle \eta_p \simeq \sum_{i=1,5} A_i'\,\sin(n_i\,t-\gamma_i),
$

    where $ A_1'=15.7''$ , $ A_2'=0.07''$ , $ A_3'=0.48''$ , $ A_4'=4.3''$ , and $ A_5'=94.0''$ are the forced libration amplitudes associated with the major inequality, reduction to the ecliptic, variation, evection, and the annual inequality, respectively. [The observed values of $ A_1'$ , $ A_3'$ , $ A_4'$ , and $ A_5'$ are $ 16.8''$ , $ 0.50''$ , $ 4.1''$ , and $ 90.7''$ , respectively. $ A_2'$ is too small to measure. (Meeus 2005).]


next up previous
Next: Useful mathematics Up: Lunar motion Previous: Historical note
Richard Fitzpatrick 2016-03-31