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Next: Evection in latitude Up: Lunar motion Previous: Parallactic inequality

Evection

Next, let us consider terms in the solution of the lunar equations of motions that depend on the lunar eccentricity, $ e$ , but are independent of the lunar inclination, $ I$ .

According to Equations (11.133) and (11.145),

    $\displaystyle a_6$ $\displaystyle = \frac{3}{4}\,x_1-\frac{3}{4}\,y_1-3\,x_1\,x_4+\frac{3}{2}\,y_1\,y_4,$ (11.261)
and   $\displaystyle b_6$ $\displaystyle =-\frac{3}{4}\,x_1+\frac{3}{4}\,y_1+\frac{3}{2}\,x_1\,y_4-\frac{3}{2}\,y_1\,x_4.$     (11.262)

It follows from Equations (11.192), (11.193), (11.199), (11.217), and (11.218) that

    $\displaystyle x_6$ $\displaystyle =\frac{3\,(6\,x_1-17\,y_1)}{64} +\left( \frac{-1146\,x_1+72\,c\,x_1-1985\,y_1-204\,c\,y_1}{512}\right)m,$ (11.263)
and   $\displaystyle y_6$ $\displaystyle =\frac{3\,(-6\,x_1+17\,y_1)}{32}+\left(\frac{522\,x_1-72\,c\,x_1+2225\,y_1+204\,c\,y_1}{256}\right)m.$ (11.264)

According to Equations (11.134) and (11.146),

    $\displaystyle a_7$ $\displaystyle = \frac{3}{4}\,x_1+\frac{3}{4}\,y_1-3\,x_1\,x_4-\frac{3}{2}\,y_1\,y_4,$ (11.265)
and   $\displaystyle b_7$ $\displaystyle =-\frac{3}{4}\,x_1-\frac{3}{4}\,y_1+\frac{3}{2}\,x_1\,y_4+\frac{3}{2}\,y_1\,x_4.$     (11.266)

It follows from Equations (11.192), (11.193), (11.200), (11.217), and (11.218) that

    $\displaystyle x_7$ $\displaystyle =\left(\frac{-46\,x_1-3\,y_1}{128}\right),$ (11.267)
and   $\displaystyle y_7$ $\displaystyle =\left(\frac{6\,x_1+17\,y_1}{64}\right).$ (11.268)

According to Equations (11.128) and (11.140),

    $\displaystyle a_1$ $\displaystyle =\frac{3}{4}\,x_6 -\frac{3}{4}\,y_6-3\,x_4\,x_6+\frac{3}{2}\,y_4\,y_6,$ (11.269)
and   $\displaystyle b_1$ $\displaystyle = -\frac{3}{4}\,x_6+\frac{3}{4}\,y_6-\frac{3}{2}\,x_4\,y_6+\frac{3}{2}\,y_4\,x_6$   $\displaystyle .
$ (11.270)

It follows from Equation (11.217) and (11.218), as well as the previous expressions for $ x_6$ and $ y_6$ , that

    $\displaystyle a_1$ $\displaystyle = a_{1\,x}\,x_1+ a_{1\,y}\,y_1,$ (11.271)
and   $\displaystyle b_1$ $\displaystyle = b_{1\,x}\,x_1+a_{1\,y}\,y_1,$     (11.272)

where

    $\displaystyle a_{1\,x}$ $\displaystyle = \frac{81}{256},$ (11.273)
    $\displaystyle a_{1\,y}$ $\displaystyle = -\frac{459}{512},$ (11.274)
    $\displaystyle b_{1\,x}$ $\displaystyle = -\frac{459}{512},$ (11.275)
and   $\displaystyle b_{1\,y}$ $\displaystyle = \frac{2601}{1024}.$     (11.276)

Thus, Equations (11.190) and (11.191) reduce to

$\displaystyle \left( \begin{array}{ll} A_{xx}, & A_{xy}\\ [0.5ex] A_{yx},&A_{yy...
...end{array} \right) = \left( \begin{array}{c} 0\\ [0.5ex] 0 \end{array} \right),$ (11.277)

where

    $\displaystyle A_{xx}$ $\displaystyle = \omega_1^{\,2}+3\left(1+\frac{1}{2}\,m^{\,2}\right) + a_{1\,x}\,m^{\,3},$ (11.278)
    $\displaystyle A_{xy}$ $\displaystyle = 2\,\omega_1+a_{1\,y}\,m^{\,3},$ (11.279)
    $\displaystyle A_{yx}$ $\displaystyle = 2\,\omega_1+b_{1\,x}\,m^{\,3},$ (11.280)
and   $\displaystyle A_{yy}$ $\displaystyle = \omega_1^{\,2}+b_{1\,y}\,m^{\,3}.$ (11.281)

The non-trivial solution of the homogenous matrix equation (11.277) is obtained by setting the determinant of the matrix to zero (Riley 1974d). In other words,

$\displaystyle {\mit\Delta} \equiv A_{xx}\,A_{yy}- A_{xy}\,A_{y\,x}= 0.$ (11.282)

It follows from Equations (11.194), (11.273)-(11.276), and (11.278)-(11.281) that

$\displaystyle {\mit\Delta} = \left(\frac{3-4\,c}{2}\right)m^{\,2} + \frac{255}{16}\,m^{\,3} +{\cal O}(m^{\,4}).$ (11.283)

Thus, setting $ {\mit\Delta}=0$ , we get

$\displaystyle c = \frac{3}{4}+ \frac{255}{32}\,m+{\cal O}(m^{\,2}).$ (11.284)

Moreover, Equations (11.278)-(11.281) reduce to

    $\displaystyle A_{xx}$ $\displaystyle = 4+{\cal O}(m^{\,3}),$ (11.285)
    $\displaystyle A_{xy}$ $\displaystyle =2-\frac{3}{2}\,m^{\,2}+{\cal O}(m^{\,3}),$ (11.286)
    $\displaystyle A_{yx}$ $\displaystyle =2-\frac{3}{2}\,m^{\,2}+{\cal O}(m^{\,3}),$ (11.287)
and   $\displaystyle A_{yy}$ $\displaystyle = 1-\frac{3}{2}\,m^{\,2}+{\cal O}(m^{\,3}),$     (11.288)

which implies, from Equation (11.277), that

$\displaystyle \frac{y_1}{x_1} =-2\left(1+\frac{3}{4}\,m^{\,2}\right).$ (11.289)

The arbitrary parameter $ y_1$ is chosen such that the parameter $ l_1$ , appearing in Equation (11.123), is the same as in the undisturbed motion. Thus, making use of Equations (11.172),

    $\displaystyle x_1$ $\displaystyle = -1 + \frac{11}{12}\,m^{\,2},$ (11.290)
and   $\displaystyle y_1$ $\displaystyle = 2 - \frac{1}{3}\,m^{\,.2}.$     (11.291)

Hence, Equations (11.263), (11.264), (11.267), and (11.268) reduce to

    $\displaystyle x_6$ $\displaystyle = -\frac{15}{8} -\frac{199}{32}\,m,$ (11.292)
    $\displaystyle y_6$ $\displaystyle = \frac{15}{4} + \frac{67}{4}\,m,$ (11.293)
    $\displaystyle x_7$ $\displaystyle = \frac{5}{16},$ (11.294)
and   $\displaystyle x_8$ $\displaystyle =\frac{7}{16}.$     (11.295)

According to Equation (11.126),

$\displaystyle a_{02} = -\frac{3}{2}\,x_1^{\,2}+\frac{3}{4}\,y_1^{\,2} =\frac{3}{2},$ (11.296)

where use has been made of the previous expressions for $ x_1$ and $ y_1$ . Equation (11.189) yields

$\displaystyle x_{02} = -\frac{1}{2}.$ (11.297)

According to Equations (11.129) and (11.141),

    $\displaystyle a_2$ $\displaystyle = -\frac{3}{2}\,x_1^{\,2} -\frac{3}{4}\,y_1^{\,2}=-\frac{9}{2},$ (11.298)
and   $\displaystyle b_2$ $\displaystyle =\frac{3}{2}\,x_1\,y_1=-3,$     (11.299)

where use has been made of the previous expressions for $ x_1$ and $ y_1$ . Hence, Equations (11.192), (11.193), and (11.195) yield

    $\displaystyle x_2$ $\displaystyle =\frac{1}{4},$ (11.300)
and   $\displaystyle y_2$ $\displaystyle =\frac{1}{4}.$ (11.301)

It follows from Equations (11.122)-(11.124), (11.158), (11.160) (11.161), (11.165), (11.166), (11.172), (11.173), (11.177), and (11.178), as well as the previous expressions for $ x_{02}$ , $ x_1$ , $ x_2$ , $ x_6$ , $ x_7$ , $ y_1$ , $ y_2$ , $ y_6$ , and $ y_{7}$ , that the net perturbation of the lunar orbit due to terms in the solution of the lunar equations of motion that depend on $ e$ , but are independent of $ I$ , is

    $\displaystyle \delta R$ $\displaystyle =-\left(e-\frac{11}{12}\,m^{\,2}\,e\right)\,\cos{\cal M} +\frac{1}{2}\,e^{\,2}-\frac{1}{2}\,e^{\,2}\,\cos(2\,{\cal M})$        
      $\displaystyle \phantom{=}-\left(\frac{15}{8}\,m\,e +\frac{155}{32}\,m^{\,2}\,e\right)\cos (2\,D-{\cal M}) -\frac{17}{16}\,m^{\,2}\,e\,\cos(2\,D+{\cal M}),$ (11.302)
    $\displaystyle \delta \lambda$ $\displaystyle =2\,e\,\sin{\cal M} +\frac{5}{4}\,e^{\,2}\,\sin(2\,{\cal M})$        
      $\displaystyle \phantom{=}+\left(\frac{15}{4}\,m\,e +\frac{263}{16}\,m^{\,2}\,e\right)\sin (2\,D-{\cal M}) +\frac{17}{8}\,m^{\,2}\,e\,\cos(2\,D+{\cal M}) ,$ (11.303)
and   $\displaystyle \delta \beta$ $\displaystyle = 0.$     (11.304)

The previous expressions are accurate to $ {\cal O}(e^{\,2})$ and $ {\cal O}(m^{\,2}\,e)$ .

The first two terms on the right-hand side of expression (11.303) are Keplerian in origin (i.e., they are independent of the perturbing action of the Sun). In fact, the first is due to the eccentricity of the lunar orbit (i.e., the fact that the geometric center of the orbit is slightly shifted from the center of the Earth), and the second is due to the ellipticity of the orbit (i.e., the fact that the orbit is slightly noncircular). These terms are usually called the major inequality

The third term on the right-hand side of expression (11.303) corresponds to evection, and is due to the combined action of the Sun and the eccentricity of the lunar orbit. In fact, evection can be thought of as causing a slight reduction in the eccentricity of the lunar orbit around the times of the new moon and the full moon (i.e., $ D=0^\circ$ and $ D=180^\circ$ ), and a corresponding slight increase in the eccentricity around the times of the first and last quarter moons (i.e., $ D=90^\circ$ and $ D=270^\circ$ ). (See Exercise 4.) This follows because the evection term in Equation (11.303) augments the eccentricity term, $ 2\,e\,\sin {\cal M}$ , when $ \cos (2\,D)=-1$ , and reduces the term when $ \cos (2\,D) = +1$ . Evection generates a perturbation in the lunar ecliptic longitude that oscillates sinusoidally with a period of $ 31.8$ days. This oscillation period is in good agreement with observations. However, the amplitude of the oscillation (calculated using $ m=0.07480$ and $ e=0.05488$ ) is $ 4,216$ arc seconds, which is somewhat less than the observed amplitude of $ 4,586$ arc seconds (Chapront-Touzé and Chapront 1988). Again, this discrepancy between theory and observation is due to the fact that we have only calculated the lowest-order (in $ m$ and $ e$ ) contributions to evection.


next up previous
Next: Evection in latitude Up: Lunar motion Previous: Parallactic inequality
Richard Fitzpatrick 2016-03-31