## 3 Results
### Selection of the Optimum Nanosatellite Dedicated to Maximise the Specific Operational Task
The drop in the altitude of the orbit results from the loss of orbital energy caused by drag. It was calculated based on the relation between energy and orbit altitude:  
\[\Delta h=-\frac{\Delta E}{m\cdot g(h)}=-643\;\mathrm{m}, \tag{30}\]  
where \(\Delta E\) = 27307 J, m is the weight of the satellite. This value is necessary to calculate the proportion of energy loss per unit of weight. In the analysed case, the satellite has a weight of m = 5 kg. g(h) is the acceleration of gravity. The altitude of the orbit influences the value of g(h), which decreases with altitude as the force of gravity diminishes on higher orbits. g(h) = 8.49 \(\frac{\mathrm{m}}{\mathrm{s}^{2}}\). Then, the total decrease in altitude during half a year \(\Delta h_{\mathrm{total}}\) was calculated:  
\[\Delta h_{total}=\Delta h\cdot 6=-3860\;\mathrm{m}, \tag{31}\]  
This means that the satellite's orbit will drop by 3860 m over a period of six months. Then, the required number of manoeuvres was determined. The authors considered the efficiency of orbital manoeuvres, where increasing the orbit by 643 m will be a reasonable compromise between the number of manoeuvres and the amount of fuel necessary. The lower the increase in altitude per a single manoeuvre, the more manoeuvres have to be performed, which increases the complexity of the mission. However, increasing the orbit by higher values (e.g., by 2 km at once) would require more fuel per a single manoeuvre, which might exceed the capacity of the satellite. In the analyses for low Earth orbits, changing the altitude by approx. 1 km causes a relatively small change in velocity (\(\Delta\)v), which facilitates fuel management and orbit control, as the manoeuvres are easier to plan and perform. The strategy that involves more frequent manoeuvres prevents the degradation of the orbit, as the drops are corrected more often. Therefore, \(\Delta\)h\({}_{\text{manoeuvre}}\)\(\approx\) 643 m.  
\[number\ of\ manoeuvres=\frac{|\Delta h_{total}|}{\Delta h_{manoeuvre}}=\frac{|-3860 |}{643}=6, \tag{32}\]  
Then, the increase in velocity (\(\Delta\)v) was calculated for one manoeuvre:  
\[\Delta v_{1}=\sqrt{\frac{GM_{\text{Z}}}{r_{start}}}\bigg{(}\sqrt{\frac{2r_{ target}}{r_{start}+r_{target}}}-1\bigg{)}=5.66\frac{\text{m}}{\text{s}}, \tag{33}\]  
\[\Delta v_{2}=\sqrt{\frac{GM_{\text{Z}}}{r_{target}}}\bigg{(}\sqrt{\frac{2r_{ start}}{r_{start}+r_{target}}}-1\bigg{)}=5.66\frac{\text{m}}{\text{s}}, \tag{34}\]  
\[\Delta v_{total}=|\Delta v_{1}|+|\Delta v_{2}|=11\ \frac{\text{m}}{\text{s}}, \tag{35}\]  
where G is the gravitational constant 6.6743 \(\times\) 10\({}^{-11}\), M\({}_{\text{Z}}\) is the weight of the Earth 5.972 \(\times\) 10\({}^{24}\), h\({}_{\text{start}}\) is the initial altitude before each manoeuvre (482 \(-\) 0.643 = 481.357 km), r\({}_{\text{start}}\) is the radius of the initial orbit 6852.357, and r\({}_{\text{target}}\) is the radius of the target orbit 6853. The next stage involved calculating the amount of fuel necessary to perform the manoeuvre:  
\[\Delta m=m\cdot\bigg{(}1-e^{\frac{-\Delta v}{S\cdot Up}}\bigg{)}=0.026\ \text{kg}, \tag{36}\]  
where \(\Delta\)v = 11 \(\frac{\text{m}}{\text{s}}\), m = 5 kg, g = 8.49 \(\frac{\text{m}}{\text{s}^{2}}\), Is\({}_{\text{p}}\) = 220 s.  
The final stage consisted of calculating the total weight of fuel required to perform six manoeuvres over a period of six months:  
\[F_{D}=\frac{1}{2}\rho\cdot v^{2}\cdot C_{D}\cdot A=0.00000135\;\mathrm{N}, \tag{28}\]0  
where \(\Delta\)m = 0.026 kg. Table 12 presents the results of the analysis aimed at selecting the optimum manoastellite configuration based on parameters such as initial and fuel weight, specific impulse, drag force, orbital energy loss, and altitude drop over time.