# Caching-Aware Intelligent Handover Strategy for LEO Satellite Networks
## 2 System Model
### Remaining Service Time  
Due to the high mobility, a LEO satellite stays in the visible range of the user for about 10 min generally, during which the satellite can communicate with the user. This period is called the maximum service time. The maximum remaining service time refers to the time that the communication link can maintain before the serving satellite moves out of the user's visual range. In Figure 2, due to the user's speed is far less than the speed of the satellite, we assume that the user terminal is stationary relative to the satellite. For every user and its serving satellite, the elevation angle between the user and the satellite will gradually increase from the minimum value to the maximum value and then, finally, decrease to the minimum value. Specifically, at \(T_{0}\), the user terminal enters the satellite coverage area, and the elevation angle from the user terminal to the satellite is the smallest. At this point, the satellite begins to provide communication services for users, and the remaining service time is the longest. With the movement of the satellite, the elevation angle increases gradually, and the user terminal gets the maximum elevation angle at \(T_{1}\). The position of the satellite footprint \(Q_{1}\) coincides with the point \(H\), and \(H\) stands for the closest point from the location of the user terminal to the satellite footprint trajectory. After \(T_{1}-T_{0}\), the elevation angle of the user terminal is minimum again in \(T_{2}\) slot, and the user leaves the satellite service coverage area. According to the geometric relationship, we can obtain the maximum service time \(T_{\text{max}}\)[20] as follows:  
\[T_{\text{max}}=\frac{2\Gamma(t_{0})}{\omega}, \tag{1}\]  
where \(\Gamma(t_{0})\) is the radian of \(\angle Q_{0}OH\), \(O\) is the center of the Earth, and \(\omega\) is the angular velocity of the satellite. \(\Gamma(t_{0})\) can be calculated with \(\text{Y}(t_{0})\) and \(\text{Y}_{\text{min}}\)[20] as follows:  
\[\cos\Gamma(t_{0})\cos(\text{Y}_{\text{min}})=\cos\text{Y}(t_{0}), \tag{2}\]  
where \(\text{Y}_{\text{min}}\) is the radian of \(\angle Q_{1}OU\), \(O\) is the center of the Earth, and \(Q_{1}U\) represents the shortest distance between the user terminal and the satellite footprint trajectory. \(\text{Y}(t_{0})\) is the radian of \(\angle Q_{0}OU\), and \(Q_{0}U\) represents the longest distance between the user terminal  
Figure 1: Handover scenario in LEO satellite system. and the satellite footprint trajectory. According to the geometric relationship, we can calculate the maximum service time with the total radian of the satellite footprint trajectory \(2\Gamma(t_{0})\). First, \(\mathrm{Y}(t_{0})\)[20] can be obtained as shown in the Figure 3. It can be expressed as  
\[(R_{e}+h)\cos(\theta_{\mathrm{min}}+\mathrm{Y}(t_{0}))=R_{e}\cos\theta_{ \mathrm{min}}, \tag{3}\]  
where \(R_{e}\) is the radius of Earth, \(h\) is the altitude of satellite orbit, and \(\theta_{\mathrm{min}}\) is the user's minimum elevation angle. Then, \(\mathrm{Y}(t_{0})\) can be expressed as  
\[\mathrm{Y}(t_{0})=\arccos(\frac{R_{e}}{R_{e}+h}\cos\theta_{\mathrm{min}})- \theta_{\mathrm{min}}, \tag{4}\]  
Figure 3: The geometric relationship of \(\mathrm{Y}(t_{0})\). Figure 2: Satellite coverage at different times. As show in Figure 4, \(\rm Y_{min}\)[20] can be obtain as follows:  
\[\rm Y_{min}=arccos\sqrt{1-\frac{d_{min}^{2}}{R_{e}^{2}}}, \tag{5}\]  
where \(d_{\rm min}\) is the shortest distance of user terminal to the satellite footprint trajectory plane. In the actual system, the latitude and longitude of the satellite footprint point \((\phi_{t}^{s},\varphi_{t}^{s},0)\) and user \((\phi_{t}^{u},\varphi_{t}^{u},\zeta_{t}^{u})\) for time slot t can be obtained according to the GPS positioning system. Converting latitude, longitude and altitude coordinates \((\phi,\varphi,\zeta)\) to Earth-Centered Earth-Fixed (ECEF) coordinates \((x,y,z)\). The conversion formula is:  
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