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),

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

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

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

$$x_6 =\frac{3\,(6\,x_1-17\,y_1)}{64} +\left( \frac{-1146\,x_1+72\,c\,x_1-1985\,y_1-204\,c\,y_1}{512}\right)m,$$ (11.263)

$$y_6 =\frac{3\,(-6\,x_1+17\,y_1)}{32}+\left(\frac{522\,x_1-72\,c\,x_1+2225\,y_1+204\,c\,y_1}{256}\right)m.$$ (11.264)

According to Equations (11.134) and (11.146),

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

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

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

$$x_7 =\left(\frac{-46\,x_1-3\,y_1}{128}\right),$$ (11.267)

$$y_7 =\left(\frac{6\,x_1+17\,y_1}{64}\right).$$ (11.268)

According to Equations (11.128) and (11.140),

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

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

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

$$a_1 = a_{1\,x}\,x_1+ a_{1\,y}\,y_1,$$ (11.271)

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

where

$$a_{1\,x} = \frac{81}{256},$$ (11.273)

$$a_{1\,y} = -\frac{459}{512},$$ (11.274)

$$b_{1\,x} = -\frac{459}{512},$$ (11.275)

$$b_{1\,y} = \frac{2601}{1024}.$$ (11.276)

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

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

where

$$A_{xx} = \omega_1^{\,2}+3\left(1+\frac{1}{2}\,m^{\,2}\right) + a_{1\,x}\,m^{\,3},$$ (11.278)

$$A_{xy} = 2\,\omega_1+a_{1\,y}\,m^{\,3},$$ (11.279)

$$A_{yx} = 2\,\omega_1+b_{1\,x}\,m^{\,3},$$ (11.280)

$$A_{yy} = \omega_1^{\,2}+b_{1\,y}\,m^{\,3}.$$ (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,

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

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

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

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

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

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

$$A_{xx} = 4+{\cal O}(m^{\,3}),$$ (11.285)

$$A_{xy} =2-\frac{3}{2}\,m^{\,2}+{\cal O}(m^{\,3}),$$ (11.286)

$$A_{yx} =2-\frac{3}{2}\,m^{\,2}+{\cal O}(m^{\,3}),$$ (11.287)

$$A_{yy} = 1-\frac{3}{2}\,m^{\,2}+{\cal O}(m^{\,3}),$$ (11.288)

which implies, from Equation (11.277), that

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

$$x_1 = -1 + \frac{11}{12}\,m^{\,2},$$ (11.290)

$$y_1 = 2 - \frac{1}{3}\,m^{\,.2}.$$ (11.291)

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

$$x_6 = -\frac{15}{8} -\frac{199}{32}\,m,$$ (11.292)

$$y_6 = \frac{15}{4} + \frac{67}{4}\,m,$$ (11.293)

$$x_7 = \frac{5}{16},$$ (11.294)

$$x_8 =\frac{7}{16}.$$ (11.295)

According to Equation (11.126),

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

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

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

According to Equations (11.129) and (11.141),

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

$$b_2 =\frac{3}{2}\,x_1\,y_1=-3,$$ (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

$$x_2 =\frac{1}{4},$$ (11.300)

$$y_2 =\frac{1}{4}.$$ (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

$$\delta R =-\left(e-\frac{11}{12}\,m^{\,2}\,e\right)\,\cos{\cal M} +\frac{1}{2}\,e^{\,2}-\frac{1}{2}\,e^{\,2}\,\cos(2\,{\cal M})$$

$$\phantom{=}-\left(\frac{15}{8}\,m\,e +\frac{155}{32}\,m^{\,2}\,e\right)\cos (2\,D-{\cal M}) -\frac{17}{16}\,m^{\,2}\,e\,\cos(2\,D+{\cal M}),$$ (11.302)

$$\delta \lambda =2\,e\,\sin{\cal M} +\frac{5}{4}\,e^{\,2}\,\sin(2\,{\cal M})$$

$$\phantom{=}+\left(\frac{15}{4}\,m\,e +\frac{263}{16}\,m^{\,2}\,e\right)\sin (2\,D-{\cal M}) +\frac{17}{8}\,m^{\,2}\,e\,\cos(2\,D+{\cal M}) ,$$ (11.303)

$$\delta \beta = 0.$$ (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 Section 11.18, 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.