## 4 Starlink as Case Study
### Stage 3: Parametric Estimation of Propellant Mass
\[V_{\rm{txA}}=\mu[(2/r_{\rm{A}}-1/a_{\rm{tx}})]^{(1/2)} \tag{7}\]  
where inserting values of \(\mu\), \(r_{\rm{A}}\) as 300 km, and \(a_{\rm{tx}}\) as 6796 km returns \(V_{\rm{txA}}\) = 7.8 km/s. \[V_{\rm{txB}}=\mu[(2/r_{\rm{B}}-1/a_{\rm{tx}})]^{(1/2)} \tag{8}\]  
where inserting values of \(\mu\), \(r_{\rm{B}}\) as 550 km, and \(a_{\rm{tx}}\) as 6796 km returns \(V_{\rm{txB}}\) = 7.51 km/s. \[\Delta V_{\rm{A}}=|V_{\rm{txA}}-V_{\rm{iA}}|=0.08\ \rm{km/s} \tag{9}\]  
\[\Delta V_{\rm{B}}=|V_{\rm{txB}}-V_{\rm{fB}}|=0.07\ \rm{km/s} \tag{10}\]  
\[\Delta V_{\rm{total}}=\Delta V_{\rm{A}}+\Delta V_{\rm{B}}=0.15\ \rm{km/s}\ \rm{or}\ 150\ \rm{m/s} \tag{11}\]  
Hence, 150 m/s \(\Delta\)V is required for orbit rising and for 5 years of station keeping in LEO. For this purpose, the 6-26 equation from [24] is used. The coefficient of drag (CD) is assumed to be 1.01 (for a cube). The best possible match for the Starlink design (rectangular cross-section) when compared to available CD data for different shapes is a cube. The cross-sectional area \(A\) value for Starlink is taken from [65]. The dimensions are given as length 3.2 m, width 1.6 m, and height 0.2 m. The cross-sectional area exposed to the flight path can be either 3.2 m \(\times\) 0.2 m or 1.6 m \(\times\) 0.2 m. For the worst case scenario, 3.2 m \(\times\) 0.2 m is taken as \(A\). The maximum density value of 9.25 \(\times\) 10\({}^{-13}\) kg/m\({}^{3}\) at 550 km altitude at solar maximum is taken for this calculation, a value that corresponds to the semi-major axis, which is 6921 km. The change in velocity experienced by Starlink for one revolution by inserting all these values is represented by the following:  
\[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]0  
\(\Delta V_{\rm{rev}}\) is the change in velocity experienced by Starlink for just one revolution. For calculating the \(\Delta\)V for the entire mission duration, the total revolution Starlink takes in its operational lifetime at 550 km altitude should be determined. In order to determine total revolution K, we need to find the orbital period value. Equation (13) is used to find the mean motion, and Equation (14) is used to find the orbital period P.  
\[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]1  
\[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]2  
P from Equation (14) is 96.07 min, with one sidereal day equal to 1436.068 min. Hence, the total minutes for 5 years of mission duration will be 1825 \(\times\) 1436.068 = 27,298.95 min. Dividing this value with 96.07 min for one revolution, we obtain the total revolutions K Starlink will undergo in 5 years. K = 27,280 revolution  
When we multiply \(\Delta\text{V}_{\text{rev}}\) with total revolution K, we obtain the total change in velocity \(\Delta\text{V}\) experienced by Starlink for 5 years. \[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]3  
The \(\Delta\text{V}\) value excludes solar panel drag, and the next step is to calculate the \(\Delta\text{V}\) during de-orbit from 550 km operational orbit to 80 km. Starlink satellites start to demise around 80 km altitude. So, 80 km is assumed here from the demise report of Starlink by SpaceX to FCC [66]. The deorbit \(\Delta\text{V}\) equation is taken from [24], where H\({}_{\text{i}}\) is the operational altitude (550 km), and H\({}_{\text{e}}\) is the end orbit altitude, which is 80 km. By equating these values, we obtain the following:  
\[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]4  
Therefore, the total \(\Delta\text{v}\) can represented as given below:  
\[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]5  
Equation (1) can be rewritten, and mass of propellant can be represented as follows:  
\[a_{\rm{tx}}=(r_{\rm{A}}+r_{\rm{B}})/2 \tag{3}\]6  
Now, the total \(\Delta\text{v}\) value from Equation (16) and \(\text{v}_{\text{o}}\) value are equated in Equation (17) and can be expressed as \(\text{I}_{\text{sp}}\times\text{g}\), where \(\text{I}_{\text{sp}}\) is the specific impulse, which is 1500 s, as discussed before, and g is acceleration due to gravity.