Posted by Jess on Wednesday, March 23, 2011 at 11:14pm.
The velocity required to keep a satellite in a circular orbit derives from
Vc = sqrt(µ/r) where r = the radius of the orbit in feet and µ = the Earth's gravitational constant = 1.407974x10^16 yielding r = 57,064 miles.
Ignore the previous response as it did not answer your question.
The required altitude to maintain a period of 48 hours derives from
T = 2(Pi)sqrt[r^3/µ] where T = the period in seconds, r = the orbital radius in feet and µ = the Earth's gravitational constant = 1.407974x10^16.
From this, r = 41,668 miles making the altitude 37,705 miles.
The velocity at this altitude is 8000 fps.
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