Venus

The ecliptic longitude of Venus can be determined with the aid of Tables 58-60. Table 58 allows the mean longitude, $\bar{\lambda}$, and the mean anomaly, $M$, of Venus to be calculated as functions of time. Next, Table 59 permits the equation of center, $q$, and the radial anomaly, $\zeta$, to be determined as functions of the mean anomaly. Finally, Table 60 allows the quantities $\delta\theta_-$, $\bar{\theta}$, and $\delta\theta_+$ 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 longitude 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 ecliptic longitude, $\lambda_S$, and radial anomaly, $\zeta_S$, of the sun using the procedure set out in Sect. 5.1.
3. Enter Table 58 with the digit for each power of 10 in ${\Delta} t$ and take out the corresponding values of $\Delta\bar{\lambda}$ and $\Delta M$. If $\Delta t$ is negative then the corresponding values are also negative. The value of the mean longitude, $\bar{\lambda}$, is the sum of all the $\Delta\bar{\lambda}$ values plus value of $\bar{\lambda}$ at the epoch. Likewise, the value of the mean anomaly, $M$, is the sum of all the $\Delta M$ values plus the value of $M$ at the epoch. Add as many multiples of $360^\circ$ to $\bar{\lambda}$ and $M$ as is required to make them both fall in the range $0^\circ$ to $360^\circ$. Round $M$ to the nearest degree.
4. Enter Table 59 with the value of $M$ and take out the corresponding value of the equation of center, $q$, and the radial anomaly, $\zeta$. It is necessary to interpolate if $M$ is odd.
5. Form the epicyclic anomaly, $\mu = \bar{\lambda}+q-\lambda_S$. Add as many multiples of $360^\circ$ to $\mu$ as is required to make it fall in the range $0^\circ$ to $360^\circ$. Round $\mu$ to the nearest degree.
6. Enter Table 60 with the value of $\mu$ and take out the corresponding values of $\delta\theta_-$, $\bar{\theta}$, and $\delta\theta_+$. If $\mu > 180^\circ$ then it is necessary to make use of the identities $\delta\theta_\pm(360^\circ - \mu) =-\delta\theta_\pm(\mu)$ and $\bar{\theta}(360^\circ - \mu) =-\bar{\theta}(\mu)$.
7. Form $z = (1-\zeta_S)/(1-\zeta)$.
8. Obtain the values of $\bar{z}$ and $\delta z$ from Table 5. Form $\xi = (\bar{z}-z)/\delta z$.
9. Enter Table 17 with the value of $\xi$ and take out the corresponding values of $\Theta_-$ and $\Theta_+$. If $\xi<0$ then it is necessary to use the identities $\Theta_+(\xi)=-\Theta_-(-\xi)$ and $\Theta_-(\xi)=-\Theta_+(-\xi)$.
10. Form the equation of the epicycle, $\theta = \Theta_-\,\delta\theta_-+ \bar{\theta} + \Theta_+\,\delta\theta_+$.
11. The ecliptic longitude, $\lambda$, is the sum of the ecliptic longitude of the sun, $\lambda_S$, and the equation of the epicycle, $\theta$. If necessary convert $\lambda$ into an angle in the range $0^\circ$ to $360^\circ$. The decimal fraction can be converted into arc minutes using Table 31. Round to the nearest arc minute. The final result can be written in terms of the signs of the zodiac using the table in Sect. 2.6.

Two examples of this procedure are given below.

 
Example 1: May 5, 2005 CE, 00:00 UT:
 
From Cha. 8, $t-t_0=1\,950.5$ JD, $\lambda_S = 44.602^\circ$, and $\zeta_S= -8.56\times 10^{-3}$. Making use of Table 58, we find:

$t$(JD)	$\bar{\lambda}(^\circ)$	$M(^\circ)$
+1000	$162.169$	$162.130$
+900	$1.952$	$1.917$
+50	$80.108$	$80.107$
+.5	$0.801$	$0.801$
Epoch	$181.973$	$49.237$
	$427.003$	$294.192$
Modulus	$67.003$	$294.192$

Given that $M\simeq 294^\circ$, Table 59 yields

$$q(294^\circ)= -0.712^\circ, \mbox{\hspace{0.5cm}}\zeta(294^\circ)=2.72\times 10^{-3},$$

so

$$\mu=\bar{\lambda}+q-\bar{\lambda}_S = 67.003-0.712-44.602= 21.689\simeq 22^\circ.$$

It follows from Table 60 that

$$\delta\theta_-(22^\circ) = 0.126^\circ,\mbox{\hspace{0.5cm}}... ..., \mbox{\hspace{0.5cm}}\delta\theta_+(22^\circ) = 0.129^\circ.$$

Now,

$$z= (1-\zeta_S)/(1-\zeta) = (1+8.56\times 10^{-3})/(1-2.72\times 10^{-3}) = 1.01131.$$

However, from Table 5, $\bar{z}= 1.00016$ and $\delta z = 0.02349$, so

$$\xi = (\bar{z}-z)/\delta z = (1.00016-1.01131)/0.02349 \simeq -0.48.$$

According to Table 17,

$$\Theta_-(-0.48) = -0.355, \mbox{\hspace{0.5cm}}\Theta_+(-0.48) = -0.125,$$

so

$$\theta = \Theta_-\,\delta\theta_- + \bar{\theta}+\Theta_+\,\... ... = -0.355\times 0.126 + 9.212-0.125\times 0.129 = 9.151^\circ.$$

Finally,

$$\lambda=\bar{\lambda}_S + \theta= 44.602+9.151=53.753 \simeq 53^\circ 45'.$$

Thus, the ecliptic longitude of Venus at 00:00 UT on May 5, 2005 CE was 23TA45.

 
Example 2: December 25, 1800 CE, 00:00 UT:
 
From Cha. 8, $t-t_0=-72\,690.5$ JD, $\lambda_S = 273.055^\circ$, and $\zeta_S= 1.662\times 10^{-2}$. Making use of Table 58, we find:

$t$(JD)	$\bar{\lambda}(^\circ)$	$M(^\circ)$
-70,000	$-191.810$	$-189.128$
-2,000	$-324.337$	$-324.261$
-600	$-241.301$	$-241.278$
-90	$-144.195$	$-144.192$
-.5	$-0.801$	$-0.801$
Epoch	$181.973$	$49.237$
	$-720.471$	$-850.423$
Modulus	$359.529$	$229.577$

Given that $M\simeq 230^\circ$, Table 59 yields

$$q(230^\circ)= -0.592^\circ,\mbox{\hspace{0.5cm}}\zeta(230^\circ)=-4.38\times 10^{-3},$$

so

$$\mu=\bar{\lambda}+q-\bar{\lambda}_S =359.529-0.592-273.055= 85.882\simeq 86^\circ.$$

It follows from Table 60 that

$$\delta\theta_-(86^\circ) = 0.589^\circ, \mbox{\hspace{0.5cm}... ..., \mbox{\hspace{0.5cm}}\delta\theta_+(86^\circ) = 0.607^\circ.$$

Now,

$$z= (1-\zeta_S)/(1-\zeta) = (1-1.662\times 10^{-2})/(1+4.38\times 10^{-3}) = 0.97909,$$

so

$$\xi = (\bar{z}-z)/\delta z = (1.00016-0.97909)/0.02349 \simeq 0.90.$$

According to Table 17,

$$\Theta_-(0.90) = 0.045,\mbox{\hspace{0.5cm}}\Theta_+(0.90) = 0.855,$$

so

$$\theta = \Theta_-\,\delta\theta_- + \bar{\theta}+\Theta_+\,\... ...+ = 0.045\times 0.589+34.482+0.855\times 0.607 = 35.027^\circ.$$

Finally,

$$\lambda=\bar{\lambda}_S + \theta= 273.055+35.027= 308.082 \simeq 308^\circ 5'.$$

Thus, the ecliptic longitude of Venus at 00:00 UT on December 25, 1800 CE was 8AQ5.