Asked by faaash

A spy satellite is in circular orbit around Earth. It makes one revolution in 6.01 hours.
how high above earth?
acceleration?

Answers

Answered by bobpursley
change time to seconds for the period.

v=2PI h/period
v^2=(2PI)^2/period^2

v^2/r= 9.8*(re^2/r^2) where r is the orbital radius.

solve for r, then altitude is r-re.
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