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>> No.9891648 [View]
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9891648

>>9891428
It makes more sense than your buoyancy description, as I already stated earlier: >>9888683

But ignoring that, I've just calculated the distance required to get those two images, approximately speaking. The one on the right has the USA taking up about 3/5 the width of the globe, and since the USA is about 4500km wide, that would make this distance "L" equal to 5/3*4500 = 7500km. Using the equation [math] L = R*arccos(\frac{R}{r}) [/math] where r is the satellite's orbital radius and R is the radius of the Earth, 6371km. Or [math]r = R*cosec(\frac{L}{R})[/math]. By plugging in R and L, we get a radius of 7660km, which corresponds to an altitude of 1290km. This is well above the boundary of space, and if you were in the ISS, the entire USA would take up your entire vision of the Earth since L would be lesser than 4500km. Doing the same to the left-most image gives an L of perhaps 10000km, and so a radius of about 9000km, altitude of 2630km. Do the geometry, you should get the same result.

>>9891570
>dimensional analysis
Well for starters they'd have to start using measurements that don't have the same unit for force and mass, that's a hotbed of confusion just waiting for flatheads to sink their teeth into.

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