Reduction to ecliptic

Finally, let us calculate the remaining terms in the solution of the lunar equations of motion.

According to Equation (11.153),

$$c_2 = \frac{3}{2}\,x_1\,z_1.$$ (11.321)

It follows from Equations (11.209), (11.211), (11.290), and (11.315) that

$$z_2 = -\frac{3}{2}$$ (11.322)

According to Equation (11.154),

$$c_3 = \frac{3}{2}\,x_1\,z_1.$$ (11.323)

It follows from Equations (11.209), (11.212), (11.290), and (11.315) that

$$z_3 = \frac{1}{2}$$ (11.324)

According to Equation (11.127),

$$a_{03} = \frac{3}{4}\,z_1^{\,2}=\frac{3}{4},$$ (11.325)

where use has been made of Equation (11.315). Equation (11.189) yields

$$x_{03} =-\frac{1}{4}.$$ (11.326)

Finally, according to Equations (11.130) and (11.142),

$$a_3 =-\frac{3}{4}\,z_1 = -\frac{3}{4},$$ (11.327)

$$b_3 = 0,$$ (11.328)

where use has been made of Equation (11.315). Hence, Equations (11.192), (11.193), and (11.196) yield

$$x_3 =\frac{1}{4},$$ (11.329)

$$y_3 =-\frac{1}{4}.$$ (11.330)

It follows from Equations (11.122)–(11.124), (11.159), (11.162), (11.174), (11.185), and (11.186), as well as the previous expressions for $x_{03}$, $x_3$, $y_3$, $z_2$, and $z_3$, that the net perturbation to the lunar orbit due to the remaining terms in the solution of the lunar equations of motion is

$$\delta R =0$$ (11.331)

$$\delta \lambda =-\frac{1}{4}\,I^{\,2}\,\sin(2\,F),$$ (11.332)

$$\delta \beta =-e\,I\,\sin(F-{\cal M}) + e\,I\,\sin(F+{\cal M}).$$ (11.333)

These expressions are accurate to ${\cal O}(I^{\,2})$ and ${\cal O}(e\,I)$.

All of the terms on the right-hand sides of Equations (11.332) and (11.333) are Keplerian in origin (i.e., they are independent of the perturbing influence of the Sun). The term on the right-hand side of Equation (11.332) is due to the slight inclination of the lunar orbit to the ecliptic plane, and is known as the reduction to the ecliptic.