Mars

According to our model, the Cartesian coordinates of a given planet in the ($x$, $y$, $z$) (see Fig. 14) system satisfy

$$P_x = a\left[C + X\,\sin\theta\,(1-\cos i)\right] - C',$$ (54)

$$P_y = a\left[S-X\,\cos\theta\,(1-\cos i)\right]-S',$$ (55)

$$P_z = a\,X\,\sin i,$$ (56)

where

$$S = \sin(x+e_c\,\sin x - e_s\,\cos x)-e_s,$$ (57)

$$C = \cos(x+e_c\,\sin x - e_s\,\cos x)-e_c,$$ (58)

$$X = S\,\cos\theta - C\,\sin\theta,$$ (59)

with $x=n\,(t-t_0)$, $n\,t_0=-(\varpi+\phi_0)$, $\varpi=\omega+\theta$, $e_s=e\,\sin\varpi$, and $e_c=e\,\cos\varpi$. Here, $n=2\pi/T$ is the mean orbital angular velocity, $T$ the mean orbital period, $a$ the semi-major radius, $e$ the ellipticity, $\omega$ the argument of the perihelion, $\theta$ the longitude of the ascending node, $i$ the inclination to the ecliptic, and $\phi _0$ the mean anomaly on day 0, of the planet in question. Furthermore,

$$S' = \sin(x'+e_c'\,\sin x' - e_s'\,\cos x')-e_s',$$ (60)

$$C' = \cos(x'+e_c'\,\sin x' - e_s'\,\cos x)-e_c',$$ (61)

where $x'=n'\,(t-t_0')$, $T'=2\pi/n'$, $n'\,t_0'=-(\omega'+\phi_0')$, $e_s'=e'\,\sin\omega'$, and $e_c'=e'\,\cos\omega'$. Here, $T'$, $e'$, $\omega'$, and $\phi_0'$ are the orbital elements of the Earth obtained in Sect. 2.6.3. The semi-major radius of the Earth is normalized to unity, for the sake of convenience.

Figure 18: The ecliptic longitude of Mars versus time. The vertical green lines indicate times of the first three oppositions. The vertical yellow line indicates the time of the maximum elongation from the guide-point.

Figure 19: The ecliptic latitude of Mars versus time. The vertical cyan lines indicates the times of the first two ascending nodes. The vertical green line indicates the time of the maximal latitude.

Figures 18 and 19 show ecliptic longitude and latitude data, respectively, for Mars, covering the years 1995-2000. In order to fit our model to the data, we need seven pieces of information. Firstly, the times of the first two ascending nodes (i.e., points at which Mars crosses the ecliptic from south to north). These are $t_1=536.726$ and $t_2=1223.644$. Secondly, the times of the first three oppositions of Mars (i.e., points at which the ecliptic longitude of Mars differs from that of the Sun by $180^\circ$), and its ecliptic longitudes at these times. These are $t_3=42.106$, $t_4=800.633$, $t_5=1574.735$ and $\lambda_3=142.905^\circ$, $\lambda_4=536.769^\circ$, $\lambda_5=934.099^\circ$, respectively. Thirdly, the approximate time of maximum elongation of Mars from its guide-point (i.e., the point at which the difference between the ecliptic longitudes of Mars and its guide-point attains its maximum value--this occurs approximately half-way between opposition and conjunction), and Mars's ecliptic longitude at this time. These are $t_6=236.0$ and $\lambda_6=201.340^\circ$, respectively. Finally, the approximate time of Mars's maximal ecliptic latitude, and its ecliptic latitude at this time. These are $t_7=40.0$ and $\beta_7=4.545^\circ$, respectively. All of the data-points are indicated in Figs. 18 and 19.

Now, an ascending node corresponds to $P_z=0$. However, it is clear, from Eq. (56), that $P_z(t+T) = P_z(t)$ for all $t$. Hence, the time difference between two successive ascending nodes must correspond to the orbital period. Thus, we obtain $T=t_2-t_1= 689.92$ days.

The remaining orbital elements are determined using an iterative procedure. Our initial guess for these elements is $e_s=e_c=i=0$, $a=1$, $t_0=t_3-\lambda_3/n$, and $\theta=n\,(t_1-t_0)$. Here, $n=2\pi/T$.

The first step in our iteration procedure involves the three opposition data-points. At opposition, $P_x/P_y=(-S_x)/(-S_y)=S'/C'$. It follows that

$$F(t,T,\lambda,\theta,i;e_s,e_c,t_0) = 0,$$ (62)

where

$$F = \left[S-X\,\cos\theta\,(1-\cos i)\right]\cos\lambda$$

$$-\left[C-X\,\cos\theta\,(1-\cos i)\right]\sin\lambda.$$ (63)

Thus, the three opposition data-points yield

$$F(t_3,T,\lambda_3,\theta,i;e_s,e_c,t_0) = 0,$$ (64)

$$F(t_4,T,\lambda_4,\theta,i;e_s,e_c,t_0) = 0,$$ (65)

$$F(t_5,T,\lambda_5,\theta,i;e_s,e_c,t_0) = 0,$$ (66)

which can be written more succinctly as

$${\bf F}({\bf x}) = {\bf0},$$ (67)

where ${\bf x} = (e_s,\,e_c,\,t_0)$. We can solve the above equation using the Newton iteration scheme described in Sect. 2.6.3. This gives us new values of $e_s$, $e_c$, and $t_0$.

The second step in our iteration procedure involves the 1st ascending node data-point. It is easily seen, from Eq. (56), that at this point (which satisfies $P_z=0$)

$$\theta \simeq x + 2\,e_c\,\sin x - 2\,e_s\,\cos x$$ (68)

to first-order in $e$, where $x=n\,(t_1-t_0)$. This gives us a new value of $\theta$.

The third step in our iteration scheme involves the maximum elongation data-point. Equations (54) and (55) yield

$$a = \left(\frac{C'\,\lambda_6-S'}{C\,\lambda_6-S}\right)_{t_6}$$ (69)

at this point. This gives us a new value of $a$. Incidentally, we chose a point of maximum elongation from the guide-point at which to perform this calculation in order to maximize its accuracy.

The fourth step in our iteration scheme involves the maximal ecliptic latitude data-point. At this point, we have

$$i = \sin^{-1}\left(\frac{\sqrt{P_x^{\,2}+P_y^{\,2}}\,\tan\beta_7}{a\,X}\right)_{t_7},$$ (70)

where the right-hand side is evaluated with the old value of $i$. This gives us a new value of $i$. Obviously, we chose a point of maximal ecliptic latitude at which to perform this calculation in order to maximize its accuracy.

Figure 20: The ecliptic latitude of Mars versus its ecliptic longitude. The blue and red curves indicate the prediction of the updated Almagest model, and the original data, respectively.

Figure 21: The residual in the ecliptic longitude of Mars versus time.

Figure 22: The residual in the ecliptic latitude of Mars versus time.

After repeating the above four steps a few times, our scheme converges to give the orbital elements for Mars listed in Tab. 1. The true elements are given in Tab. 2. It can be seen that our model determines the orbital elements of Mars to reasonable accuracy.

Figure 20 shows the ecliptic latitude versus the ecliptic longitude of Mars for part of the period 1995-2000, and compares the predictions of our updated Almagest model, made using the orbital elements given in Tab. 1, against the original data. It can be seen that the two are essentially indistinguishable. Finally, Figs. 21 and 22 give the residuals in the ecliptic longitude and latitude of Mars, respectively, versus time. It can be seen that the maximum error in longitude is about $35'$, whereas the maximum error in latitude is about $6'$. Clearly, our updated version of Ptolemy's model does a reasonably good job of accounting for Mars's apparent motion.