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Next: Determination of Ecliptic Latitude Up: Planetary Latitudes Previous: Jupiter

Saturn

The ecliptic latitude of Saturn can be determined with the aid of Tables 54, 71, and 72. Table 54 allows the mean argument of latitude, $\bar{F}$, of Saturn to be calculated as a function of time. Next, Table 71 permits the deferential latitude, $\beta_0$, to be determined as a function of the true argument of latitude, $F$. Finally, Table 72 allows the quantities $\delta h_-$, $\bar{h}$, and $\delta h_+$ to be calculated as functions of the epicyclic anomaly, $\mu$. The procedure for using these tables is analogous to the previously described procedure for using the Mars tables. One example of this procedure is given below.

 
Example: May 5, 2005 CE, 00:00 UT:
 
From Cha. 8, $t-t_0=1\,950.5$ JD, $q= 2.561^\circ$, $\mu= 286.625^\circ$, $\Theta_-=0.071$, and $\Theta_+ = 0.759$. Making use of Table 54, we find:
   
$t$(JD) $\bar{F}(^\circ)$
   
+1000 $33.478$
+900 $30.130$
+50 $1.674$
+.5 $0.017$
Epoch $296.482$
  $361.781$
Modulus $1.781$
   

Thus,

\begin{displaymath}
F = \bar{F} + q =1.781+2.561 = 4.342\simeq 4^\circ.
\end{displaymath}

It follows from Table 71 that

\begin{displaymath}
\beta_0(4^\circ) = 0.173^\circ.
\end{displaymath}

Since $\mu\simeq 287^\circ$, Table 72 yields

\begin{displaymath}
\delta h_-(287^\circ) = -0.002,\mbox{\hspace{0.5cm}}\bar{h}(...
...c)=0.966, \mbox{\hspace{0.5cm}}\delta h_+(287^\circ) = -0.003,
\end{displaymath}

so

\begin{displaymath}
h = \Theta_-\,\delta h_- + \bar{h}+\Theta_+\,\delta h_+ = -0.071\times 0.002 +0.966-0.759\times 0.003 = 0.964.
\end{displaymath}

Finally,

\begin{displaymath}
\beta = h\,\beta_0 = 0.964\times 0.173 = 0.167 \simeq 0^\circ 10'.
\end{displaymath}

Thus, the ecliptic latitude of Saturn at 00:00 UT on May 5, 2005 CE was $0^\circ 10'$.


next up previous
Next: Determination of Ecliptic Latitude Up: Planetary Latitudes Previous: Jupiter
Richard Fitzpatrick 2010-07-21