next up previous
Next: Mercury Up: Planetary Latitudes Previous: Determination of Ecliptic Latitude

Venus

The ecliptic latitude of Venus can be determined with the aid of Tables 58, 73, and 74. Table 58 allows the mean argument of latitude, $\bar{F}$, of Venus to be calculated as a function of time. Next, Table 73 permits the epicyclic latitude, $\beta_0$, to be determined as a function of the true argument of latitude, $F$. Finally, Table 74 allows the quantities $\delta h_-$, $\bar{h}$, and $\delta h_+$ to be calculated as functions of the epicyclic anomaly, $\mu$.

The procedure for using the tables is as follows:

  1. Determine the fractional Julian day number, $t$, corresponding to the date and time at which the ecliptic latitude is to be calculated with the aid of Tables 27-29. Form $\Delta t = t-t_0$, where $t_0= 2\,451\,545.0$ is the epoch.
  2. Calculate the planetary equation of center, $q$, ecliptic anomaly, $\mu$, and interpolation parameters $\Theta_+$ and $\Theta_-$ using the procedure set out in Cha. 9.
  3. Enter Table 58 with the digit for each power of 10 in ${\Delta} t$ and take out the corresponding values of $\Delta\bar{F}$. If $\Delta t$ is negative then the corresponding values are also negative. The value of the mean argument of latitude, $\bar{F}$, is the sum of all the $\Delta\bar{F}$ values plus the value of $\bar{F}$ at the epoch.
  4. Form the true argument of latitude, $F=\bar{F} + q$. Add as many multiples of $360^\circ$ to $F$ as is required to make it fall in the range $0^\circ $ to $360^\circ$. Round $F$ to the nearest degree.
  5. Enter Table 73 with the value of $F$ and take out the corresponding value of the epicyclic latitude, $\beta_0$. It is necessary to interpolate if $F$ is odd.
  6. Enter Table 74 with the value of $\mu$ and take out the corresponding values of $\delta h_-$, $\bar{h}$, and $\delta h_+$. If $\mu > 180^\circ$ then it is necessary to make use of the identities $\delta h_\pm(360^\circ - \mu) = \delta h_\pm(\mu)$ and $\bar{h}(360^\circ - \mu) = \bar{h}(\mu)$.
  7. Form the deferential latitude correction factor, $h = \Theta_-\,\delta h_-+ \bar{h} + \Theta_+\,\delta h_+$.
  8. The ecliptic latitude, $\beta$, is the product of the epicyclic latitude, $\beta_0$, and the deferential latitude correction factor, $h$. The decimal fraction can be converted into arc minutes using Table 31. Round to the nearest arc minute.
One example of this procedure is given below.

 
Example: May 5, 2005 CE, 00:00 UT:
 
From Cha. 9, $t-t_0=1\,950.5$ JD, $q= -0.712^\circ$, $\mu= 21.689^\circ$, $\Theta_-=-0.355$, and $\Theta_+ = -0.125$. Making use of Table 58, we find:
   
$t$(JD) $\bar{F}(^\circ)$
   
+1000 $162.138$
+900 $1.924$
+50 $80.107$
+.5 $0.801$
Epoch $105.253$
  $350.223$
Modulus $350.223$
   

Thus,

\begin{displaymath} F = \bar{F} + q = 350.223-0.712 = 349.511\simeq 350^\circ. \end{displaymath}

It follows from Table 73 that

\begin{displaymath} \beta_0(350^\circ) = -0.423^\circ. \end{displaymath}

Since $\mu\simeq 22^\circ$, Table 74 yields

\begin{displaymath} \delta h_-(22^\circ) = 0.008,\mbox{\hspace{0.5cm}}\bar{h}(22^\circ)=0.591, \mbox{\hspace{0.5cm}}\delta h_+(22^\circ) = 0.008, \end{displaymath}

so

\begin{displaymath} h = \Theta_-\,\delta h_- + \bar{h}+\Theta_+\,\delta h_+ = -0.355\times 0.008+0.591-0.125\times 0.008 = 0.587. \end{displaymath}

Finally,

\begin{displaymath} \beta = h\,\beta_0 =- 0.587\times 0.423 = -0.248\simeq -0^\circ 15'. \end{displaymath}

Thus, the ecliptic latitude of Venus at 00:00 UT on May 5, 2005 CE was $-0^\circ 15'$.

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