So I'm not a mathematician but a writer. I'd like to know if there's a formula that could solve the following problem for a sci-fi book I'm writing:
How many satellites would it take to surround a sphere of diameter $x$, where the satellites are $y$ miles apart from each other, and $z$ miles above the sphere?
e.g. $x = 8,000$ miles, $y = 1$ mile, $z = 100$ miles.
Your satellites occupy a sphere of diameter $x+2z$. That sphere has a surface area of $4\pi(x/2+z)^2$. If each satellite is a distance about $y$ from its neighbors, we can assume that for each satellite, there is a circle of diameter $y$ centered on that satellite with no other satellites in it. That circle has area approximately equal to $\pi (y/2)^2$ (this is approximate since we are talking about a circle on the larger sphere of diameter $x+2z$, and areas on a sphere are different from areas in the plane, but since $x+2z$ is much bigger than $y$, the difference is negligible). It is not possible to perfectly cover a sphere in circles, but it is possible to cover about $$\frac{\pi\sqrt{3}}{6}\approx 91\%$$ of a sphere in circles. Therefore, to get the number of satellites, we divide the area of the sphere they occupy (adjusted to $91\%$) by the area of the circle for each individual satellite: $$\text{number of satellites}=\frac{\pi\sqrt{3}}{6}\frac{4\pi (x/2+z)^2}{\pi (y/2)^2}=\frac{\pi}{\sqrt{3}}\frac{2(x+2z)^2}{y^2}.$$ Plugging in $x=8000$, $y=1$, and $z=100$, we get $243,919,738$ satellites.
To estimate the humongous number $N$ of satellites to be seeded in a cloud on area basis.. we can assume hexagonal packing of nodes in the geodesic dome of Buckminster Fuller's designation $np,$ the number of subdivisions of a spherical triangle side of the basis icosahedron.
$y= 1$ is the blue satellite separation distance between the centers of equilateral triangles in a hexagonal close-packed triangular array of each side red length $a=\sqrt3 y $. With sphere radius $R= x/2+z = 8000/2+100 =4100 $ miles, and assuming uniform distribution valid,
$$ N= \frac{4\pi R^2}{\sqrt{3}a^2/4}=\frac{16 \pi}{3\sqrt3}(R/y)^2 \approx 9.674 (R/y)^2 \approx 1.626\; 10^8 $$
Square root of this is around $12,750$ satellites spread over so many miles occupying almost a hemisphere cloud shell.
But they won't stay put at one place even if injected at same escape velocity in different directions, rather they may swarm around a bright street lamp ( the Earth) like so many insects in low earth orbits. They may need to have omni-directional antenna receiver-transmitters to avoid collisions in a sort of Brownanian motion.. better to address a Physics website about what might possibly happen.
