next up previous
Next: Determination of Ecliptic Latitude Up: The Moon Previous: Determination of Ecliptic Longitude

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

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

Thus,

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


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


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

Table 37 yields

\begin{displaymath}
q_1(a_1)=5.555^\circ,~~q_2(a_2)= -0.663^\circ,~~q_3(a_3) = -0.631^\circ,
\end{displaymath}


\begin{displaymath}q_4(a_4)=-0.139^\circ,~~q_5(a_5)= 0.086^\circ.
\end{displaymath}

Hence,

\begin{displaymath}
\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,
\end{displaymath}

or

\begin{displaymath}
\lambda =3.091\simeq 3^\circ 05'.
\end{displaymath}

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

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

Thus,

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


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


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

Table 37 yields

\begin{displaymath}
q_1(a_1)=5.562^\circ,~~q_2(a_2)= 1.311^\circ,~~q_3(a_3) = -0.394^\circ,
\end{displaymath}


\begin{displaymath}q_4(a_4)=0.017^\circ,~~q_5(a_5)= -0.021^\circ.
\end{displaymath}

Hence,

\begin{displaymath}
\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,
\end{displaymath}

or

\begin{displaymath}
\lambda =25.950\simeq 25^\circ 57'.
\end{displaymath}

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.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{epsfiles/moon.eps}}
\end{figure}


next up previous
Next: Determination of Ecliptic Latitude Up: The Moon Previous: Determination of Ecliptic Longitude
Richard Fitzpatrick 2010-07-21