# Semi-Empirical Satellite-to-Ground Quantum Key Distribution Model for Realistic Receivers
## 2 Satellite-to-Ground Communication Geometry  
To determine the channel efficiency in a satellite-based QKD model, one needs to calculate the time-dependent distance \(d(t)\) between the transmitter and the receivers, as well as the satellite elevation angle \(\theta_{EI}\) above the horizon. We considered several typical satellite passages and calculated one passing through the zenith in a circular orbit. Three non-zenith actual passages of the Micius satellite were simulated for 31 October 2021, 9 March 2022, and 23 March 2022, with the maximum elevation angles of \(32.5^{\circ}\), \(57.5^{\circ}\), and \(42^{\circ}\), respectively. To compute the time dependency of the satellite's elevation angle for the (\(55^{\circ}42^{\prime}\) N, \(36^{\circ}45^{\prime}\) E) location, we used the simplified general perturbations model SGP4 [30], given the satellite's two-line elements (TLE) orbital data [31; 32]. We also considered a simple model to calculate the trace-time function of the elevation angle for the limit case corresponding to the satellite passing through the zenith point over the ground station.Here, we assumed that the Micius satellite moves in a perfectly circular orbit (the eccentricity is 0) at an altitude of 500 km and an inclination of \(97.33^{\circ}\)[13]. The angular velocity \(\omega\) of a low-Earth orbit satellite may be approximated by a constant in the Earth-centered fixed frame, according to Ref. [33]:  
\[\omega\approx\omega_{S}-\omega_{E}\cos i\,, \tag{1}\]  
where \(\omega_{S}\approx\sqrt{g/R}\) is the angular velocity of the satellite in the Earth-centred inertial coordinate system with acceleration due to gravity of \(g\) and an orbit radius of \(R\), the orbit inclination is \(i\), and \(\omega_{E}\) is the angular velocity of the Earth's rotation. The numerical valuesof the orbital parameters and angular velocity data of the Micius satellite for our model are listed in Table 1.  
We approximated the magnitude of the tangential velocity by its minimum value for a given orbit in the Earth-centered fixed frame. The absolute fluctuation of the tangential velocity magnitude for the LEO satellite is quite negligible for any latitude, according to Ref. [34]. Therefore, angular velocity is independent of the ground station latitude in this model. The elevation angle for the zenith satellite passage was calculated using a set of equations derived from simple geometric considerations (see Figure 1 and Appendix A.1), as follows:  
\[\theta_{El}=\arccos\left(\sqrt{\frac{\sin^{2}\alpha}{1+\left(\frac{R_{E}}{R} \right)^{2}-2\left(\frac{R_{E}}{R}\right)\cos\alpha}}\right), \tag{2}\]  
where the central angle can be expressed as a linear function of time as a \(\alpha=\omega t\). Next, we determined the distance between a satellite and a ground station using the elevation angle from TLE data over time for real satellite passage as well as a computed satellite pass above the zenith. Appendix A.2 has a detailed derivation of the distance, which is given by:  
\[d^{2}+d\cdot 2R_{E}\sin\theta_{El}+R_{E}^{2}-R^{2}=0. \tag{3}\]  
Figure 2 shows the dependence of the satellite elevation angle and the distance between the satellite and the ground station on time obtained from Equations (1)-(3) for the typical satellite passages, by using TLE data and the calculated zenith satellite passage. The calculations were carried out when the satellite was in the ground station's field of view, i.e., when the elevation angle was above 20 degrees above the horizon. \begin{table}
\begin{tabular}{c c c c} \hline \hline \(R\) (km) & \(i\) (degrees) & \(\omega_{E}\) (Rad/s) & \(\omega_{S}\) (Rad/s) \\ \hline
6864 & 97.3 & \(7.3\times 10^{-5}\) & \(111.4\times 10^{-5}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: The orbital parameters of the satellite. Figure 1: Schematic of the Earth-satellite geometry.