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This video is about the physics of geosynchronous and geostationary orbits, why they exist, when they don’t, when they’re useful for communication/satellite TV, etc.

REFERENCES

Fraction of a sphere that’s visible from a given distance

https://math.stackexchange.com/questions/1329130/what-fraction-of-a-sphere-can-an-external-observer-see

Orbital period

https://en.wikipedia.org/wiki/Orbital_period

Kepler’s third law

https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law

Kepler’s 3rd law (which can be derived from Newton’s law of gravitation and the centripetal force necessary for orbit as mromega^2=Gfrac{mM}{r^2}, and using omega=frac{2pi}{T}) is

T = 2pi Sqrt(r^3/(GM)) where M is the mass of the central object, G is the gravitational constant. Alternatively, we can solve for r, r = (T^2/(4pi^2) GM)^(1/3) ~ T^(2/3)/M^(1/3) = (T^2/M)^(1/3).

There is a limit (kind of like the Roche limit but for rotations). A rotating solid steel ball or other chunk of metal that has tensile strength (ie that isn’t just a pile of stuff held together by gravity like most planets) would be able to spin faster.

Calculate how much of a planet’s surface you can see from a given geosynchronous orbit/radius? (Obviously for lower ones you can see less, etc) – d/(2(R+d)) where d is distance to surface, ie, R is sphere radius, R+d is object radius from sphere center.

Let’s plug that in with r being the geostationary orbit radius. That is, we have frac{1}{2} left(1- left(frac{4 pi^2 R^3}{T^2 G M }right)^{1/3}right)

Average density of a sphere rho is given by rho =M/(frac{4}{3}pi R^3), ie rho=frac{3M}{4 pi R^3} aka

frac{M}{R^3}=frac{4}{3}pi rho.

So we can convert the "fraction of planet surface seen" to

frac{1}{2} left(1- left(frac{3 pi}{G rho T^2}right)^{1/3}right)

So as either rho or Tto infty, the fraction goes to a maximum of frac{1}{2}. And the point of "singularity" where the orbit coincides with the surface is where Grho T^2=3pi, aka rho=frac{3pi}{GT^2}. For a rotation period of 3600s, that corresponds to a density rho approx 11000kg/m^3, which is roughly twice the density of the earth. For a rotation period of 5400s, we have rhoapprox 4800kg/m^3, which is basically the density of the earth.

Alternately, if we plug the density of the earth in to an orbit of period 5400s, we get as a fraction of the planet seen:

frac{1}{2} left(1- left(frac{3 pi}{G rho T^2}right)^{1/3}right) = 0.02

aka 2% of the earth’s surface.

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Created by Henry Reich