Calculating the solid angle subtended by the planet

Several calculations in this section make use of two references in particular; planetary information is mainly taken from the ``Report of the IAU working group on cartographic coordinates and rotational elements of the planets and satellites'', which is periodically updated and published online and in Celestial Mechanics and Dynamical Astronomy. Precession formulae come from Lieske (1979, A & A, 73, 282).

To calculate the solid angle of the planet in question it is necessary to take account of the ellipticity of the planet, via the direction of its North Pole. The coordinates of the pole, RA_pole and Dec_pole are calculated directly from the IAU report formulae for the date given and then precessed according to Lieske's spherical coordinate formulae. If using J2000.0 for the polar coordinates, the precession parameters $\zeta_A$, z_A and $\theta_A$ (see Lieske's Fig. 1) are defined in arcseconds as

$$\zeta_A = 2306.2181 t + 0.30188 {t^2} + 0.017998 {t^3}\\$$

$$z_A = 2306.2181 t + 1.09468 {t^2} + 0.018203 {t^3}\\$$

$$\theta_A = 2004.3109 t - 0.42665 {t^2} - 0.041833 {t^3}$$

and the following quantities can be defined

$$\begin{eqnarray*} C & = & -(\sin Dec_{pole} \sin \theta_A) + (\cos Dec_{pole} \c... ...a_A) + (\cos Dec_{pole} \sin \theta_A \cos (\zeta_a + RA_{pole}) \end{eqnarray*}$$

The precessed coordinates, ${RA^\prime}_{pole}$ and ${Dec^\prime}_{pole}$ are then calculated as

$$\begin{eqnarray*} {RA^\prime}_{pole} & = & z_A + \arctan(S,C)\\ {Dec^\prime}_{pole} & = & \arcsin (D)\\ & = & \arccos (\sqrt{ S^2 + C^2}) \end{eqnarray*}$$

where (S,C) is the complex number formed from the quantities S and C. From these precessed coordinates, the planetocentric declination of the Earth can be defined as

$${Dec^\prime}_{Earth} = \arcsin {(-\sin {Dec^\prime}_{pole} \... ...} \cos {Dec^\prime}_g \cos (RA^\prime}_{pole} - {RA^\prime}_g)$$

where ${RA^\prime}_g$ and ${Dec^\prime}_g$ are the precessed geocentric coordinates of the planet. The polar inclination angle of the planet, $\alpha_{pole}$, is then given by

$$\alpha_{pole} = {Dec^\prime}_{Earth} + \frac{\pi}{2}$$

or, if Dec_Earth > 0 so the South Pole is facing Earth,

$$\alpha_{pole} = \frac{\pi}{2} - {Dec^\prime}_{Earth}.$$

With this calculation performed the semi-major axis of the planet as seen from Earth, R_semi-major is given as

$$R_{semi-major} = \frac{R_{p} (1 - \epsilon_p)}{1 - \epsilon_p \cos \alpha_{pole}}$$

where R_p is the planet's radius and $\epsilon_p$ is its ellipticity. Hence its geometrical mean radius R_gm is

$$R_{gm} = \sqrt{R_p \times R_{semi-major}}$$

and its semi-diameter as seen from the Earth, D_semi is

$$D_{semi} = \frac{R_{gm}}{d_g}$$

where d_g is the geocentric distance of the planet from the Earth as calculated in section . Finally, the solid angle subtended by the planet from the Earth, $\Omega_p$, is

$$\Omega_p = \pi {D_{semi}}^2$$