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FREE FACT: An oblate spheroid is a special case of an ellipsoid where two of the semi-principal axes are the same size.

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$Geosynchronous Orbits are WEIRD Go to https://givewell.org/minutephysics to have your first donation matched up to $100! This video is about the physics of geosynchronous and geostationary orbits, why they exist, when they dont, when theyre useful for communication/satellite TV, etc. REFERENCES Fraction of a sphere thats 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 Keplers third law https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law Keplers 3rd law (which can be derived from Newtons 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 isnt just a pile of stuff held together by gravity like most planets) would be able to spin faster. Calculate how much of a planets 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. Lets 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 earths surface. Support MinutePhysics on Patreon! http://www.patreon.com/minutephysics Link to Patreon Supporters: http://www.minutephysics.com/supporters/ MinutePhysics is on twitter - @minutephysics And facebook - http://facebook.com/minutephysics Minute Physics provides an energetic and entertaining view of old and new problems in physics all in a minute! Created by Henry Reich$

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What is Sea Level? @MinutePhysics