Example Longitude Calculations

Example 1: May 5, 2005 CE, 00:00 UT:
 
From Sect. 5.1, $t-t_0=1950.5$ JD, $\lambda_S = 44.602^\circ$, and $M_S= 120.001^\circ$. Making use of Table 36, we find:

$t$(JD)	$\bar{\lambda}(^\circ)$	$M(^\circ)$	$\bar{F}(^\circ)$
+1000	$216.396$	$104.993$	$269.350$
+900	$338.757$	$238.494$	$26.415$
+50	$298.820$	$293.250$	$301.468$
+.5	$6.588$	$6.532$	$6.615$
Epoch	$218.322$	$134.916$	$93.284$
	$1078.883$	$778.185$	$697.132$
Modulus	$358.883$	$58.185$	$337.132$

It follows that

$$\tilde{D}=\bar{\lambda}-\lambda_S = 358.883-44.602 = 314.281^\circ.$$

Thus,

$$a_1=M\simeq 58^\circ,~~~a_2=2\tilde{D}-M = 2\times 314.281-58.185=570.377\simeq 210^\circ,$$

$$a_3=\tilde{D}\simeq 314^\circ,~~~a_4 = M_S\simeq 120^\circ,$$

$$a_5=2\bar{F} = 2\times 337.132=674.264\simeq 314^\circ.$$

Table 37 yields

$$q_1(a_1)=5.555^\circ,~~q_2(a_2)= -0.663^\circ,~~q_3(a_3) = -0.631^\circ,$$

$$q_4(a_4)=-0.139^\circ,~~q_5(a_5)= 0.086^\circ.$$

Hence,

$$\lambda = \bar{\lambda} + q_1+q_2+q_3+q_4+q_5=358.883+5.555-0.663-0.631-0.139+0.086= 363.091^\circ,$$

or

$$\lambda =3.091\simeq 3^\circ 05'.$$

Thus, the ecliptic longitude of the moon at 00:00 UT on May 5, 2005 CE was 3AR05.

 
Example 2: December 25, 1800 CE, 00:00 UT:
 
From Sect. 5.1, $t-t_0=-72\,690.5$ JD, $\lambda_S = 273.055^\circ$, and $M_S= 353.814^\circ$. Making use of Table 36, we find:

$t$(JD)	$\bar{\lambda}(^\circ)$	$M(^\circ)$	$\bar{F}(^\circ)$
-70,000	$-27.752$	$-149.506$	$-134.519$
-2,000	$-72.793$	$-209.986$	$-178.701$
-600	$-345.838$	$-278.996$	$-17.610$
-90	$-105.876$	$-95.849$	$-110.642$
-.5	$-6.588$	$-6.532$	$-6.615$
Epoch	$218.322$	$134.916$	$93.284$
	$-340.525$	$-605.953$	$-354.803$
Modulus	$19.475$	$114.047$	$5.197$

It follows that

$$\tilde{D}=\bar{\lambda}-\lambda_S = 19.475-273.055 = -253.580^\circ.$$

Thus,

$$a_1=M\simeq 114^\circ,~~~a_2=2\tilde{D}-M = -2\times 253.580-114.047=-621.207 \simeq 99^\circ,$$

$$a_3=\tilde{D}\simeq 106^\circ,~~~a_4 = M_S\simeq 354^\circ,$$

$$a_5=2\bar{F} = 2\times 5.197=10.394\simeq 10^\circ.$$

Table 37 yields

$$q_1(a_1)=5.562^\circ,~~q_2(a_2)= 1.311^\circ,~~q_3(a_3) = -0.394^\circ,$$

$$q_4(a_4)=0.017^\circ,~~q_5(a_5)= -0.021^\circ.$$

Hence,

$$\lambda = \bar{\lambda} + q_1+q_2+q_3+q_4+q_5=19.475+5.562+1.311-0.394+0.017-0.021= 25.950^\circ,$$

or

$$\lambda =25.950\simeq 25^\circ 57'.$$

Thus, the ecliptic longitude of the moon at 00:00 UT on December 25, 1800 CE was 25AR57.

Figure 23: The orbit of the moon about the earth. Here, $G$, $L$, $N$, $N'$, $\Omega$, $F$, and $\Upsilon$ represent the earth, moon, ascending node, descending node, longitude of the ascending node, argument of latitude, and vernal equinox, respectively. View is from northern ecliptic pole. The moon orbits counterclockwise.