6. More and more countries are exploring space. The number of states that launch their satellites using their own launch vehicles is already 12. There are also countries that use the services of the main space powers to launch their satellites for economic purposes. Due to the increasing number of participants in space activities and the intensification of these activities, issues of ensuring the safety of space operations are arising. For example, NASA has approached the corporation "Energia" with a request to reduce the altitude of the ISS orbit.

a) Considering the Earth as a sphere with radius $R$, determine the maximum number of satellites that can simultaneously be in orbits around the Earth at the same height $H$ from its surface, such that the distance between the satellites is greater than $\sqrt{2}(R+H)$.

b) For the maximum number of satellites found, specify the coordinates of their possible locations in a coordinate system with the origin at the center of the Earth and the x-axis directed along the vector connecting the center of the Earth with one of the satellites. (20 points)
Solution. According to the condition, at the current moment, all satellites should be located on a sphere with radius $R+H$. Let $O$ be the center of the sphere, and its radius $R_{H}=R+H$. Denote the satellites by points $C_{i}, i=1, \ldots, n$. We need to determine the maximum value of $n$.

Let's find the distance $C_{i} C_{j}$, using the cosine theorem for triangle $O C_{i} C_{j}$.

$C_{i} C_{j}^{2}=O C_{i}^{2}+O C_{j}^{2}-2 O C_{i} \cdot O C_{j} \cdot \cos \left(\angle O C_{i}, O C_{j}\right)=2 R_{H}^{2}-2 R_{H}^{2} \cdot \cos \alpha=2 R_{H}^{2}(1-\cos \alpha)>2 R_{H}^{2}$,

Thus, $1-\cos \alpha>1 \Rightarrow \cos \alpha<0$, and the third axis is directed towards the nearest to $C_{3}\left(x_{3}, y_{3}, z_{3}\right)$, so that the z-coordinate is positive, i.e., $z_{3}>0$. Consider the scalar products of the vectors, noting that all of them are negative.

$\overrightarrow{O C_{1}} \cdot \overrightarrow{O C_{i}}=x_{i}, \quad i=\overline{2, n}, \quad \overrightarrow{O C_{2}} \cdot \overrightarrow{O C_{i}}=x_{2} x_{i}+y_{2} y_{i}, i=\overline{3, n}, \quad \overrightarrow{O C_{3}} \cdot \overrightarrow{O C_{i}}=x_{3} x_{i}+y_{3} y_{i}+z_{3} z_{i}, i=\overline{4, n}$.

Since $\cos \left(\angle O C_{i}, O C_{j}\right)<0$, the scalar products of the vectors are negative, and thus, $\quad x_{i}<0, i=\overline{2, n}, \quad$ meaning all $y_{i}<0, i=\overline{3, n}, \quad x_{3} x_{i}+y_{3} y_{i}+z_{3} z_{i}<0, i=\overline{4, n}$. Therefore, all $z_{i}<0, i=\overline{4, n}$. Thus, the coordinates of all vectors starting from $\overrightarrow{O C_{4}}$ will be negative. Consequently, the scalar product of vectors $\overrightarrow{O C_{4}}, \overrightarrow{O C_{5}}$ will be positive. We have reached a contradiction. The assumption of the existence of more than 4 satellites satisfying the required conditions is incorrect. Answer: 4 satellites.

b) 4 satellites are possible, for example, with coordinates that can be written as follows:
if the $o x$ axis is directed from the center of the sphere in the direction of one of the satellites, then the coordinates of this satellite will be $\left(R_{H}, 0,0\right)$, the other three satellites will be located on the plane $x=-R_{H} / 3$ and their coordinates will be respectively

$$
\left(-R_{H} / 3,0, r\right), \quad\left(-R_{H} / 3, \sqrt{3} r / 2,-r / 2\right), \quad\left(-R_{H} / 3,-\sqrt{3} r / 2,-r / 2\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-16.jpg?height=291&width=354&top_left_y=217&top_left_x=1528)

where $r=\sqrt{R_{H}^{2}-R_{H}^{2} / 9}=R_{H} \sqrt{8} / 3$.

These relationships are derived from considering the relationships between the elements of a regular

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-16.jpg?height=568&width=694&top_left_y=698&top_left_x=264)
tetrahedron with side $a$. The height $H=\sqrt{a^{2}-\left(a \frac{\sqrt{3}}{2} \cdot \frac{2}{3}\right)^{2}}=a \sqrt{\frac{2}{3}} . \quad$ The radius $\quad$ of the sphere circumscribed around a regular tetrahedron is found from the similarity of triangles

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-16.jpg?height=323&width=323&top_left_y=1049&top_left_x=1506)

$\frac{H}{a}=\frac{a}{2 R} \Rightarrow R=\frac{a^{2}}{2 \sqrt{\frac{2}{3}} a}=\frac{a \sqrt{3}}{2 \sqrt{2}} \Rightarrow \frac{H}{R}=\frac{\sqrt{\frac{2}{3}} a}{\frac{a \sqrt{3}}{2 \sqrt{2}}}=\frac{4}{3}$.