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?
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.
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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