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Jupiter

The ecliptic latitude of Jupiter can be determined with the aid of Tables 50, 69, and 70. Table 50 allows the mean argument of latitude, $\bar{F}$, of Jupiter to be calculated as a function of time. Next, Table 69 permits the deferential latitude, $\beta_0$, to be determined as a function of the true argument of latitude, $F$. Finally, Table 70 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, $q= -0.091^\circ$, $\mu= 208.192^\circ$, $\Theta_-=-0.469$, and $\Theta_+ = -0.121$. Making use of Table 50, we find:
   
$t$(JD) $\bar{F}(^\circ)$
   
+1000 $83.081$
+900 $74.773$
+50 $4.154$
+.5 $0.042$
Epoch $293.660$
  $455.710$
Modulus $95.710$
   

Thus,

\begin{displaymath}
F = \bar{F} + q =95.710-0.091 = 95.619\simeq 96^\circ.
\end{displaymath}

It follows from Table 69 that

\begin{displaymath}
\beta_0(96^\circ) = 1.297^\circ.
\end{displaymath}

Since $\mu\simeq 208^\circ$, Table 70 yields

\begin{displaymath}
\delta h_-(208^\circ) = 0.014,\mbox{\hspace{0.5cm}}\bar{h}(2...
...rc)=1.197, \mbox{\hspace{0.5cm}}\delta h_+(208^\circ) = 0.016,
\end{displaymath}

so

\begin{displaymath}
h = \Theta_-\,\delta h_- + \bar{h}+\Theta_+\,\delta h_+ = -0.469\times 0.014 +1.197-0.121\times 0.016 = 1.188.
\end{displaymath}

Finally,

\begin{displaymath}
\beta = h\,\beta_0 = 1.188\times 1.297 = 1.541 \simeq 1^\circ 32'.
\end{displaymath}

Thus, the ecliptic latitude of Jupiter at 00:00 UT on May 5, 2005 CE was $1^\circ 32'$.


next up previous
Next: Saturn Up: Planetary Latitudes Previous: Mars
Richard Fitzpatrick 2010-07-21