Asked by Renee
A spy satellite is in circular orbit around Earth. It makes one revolution in 5.95 hours. (Radius of the Earth=6.371 106 m)
(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?
(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?
Answers
Answered by
bobpursley
velocity=2PI*(re+h)/period
but velocity can be found..
9.8(re/(re+h))^2=v^2/(re+h)
or v=sqrt (9.8 (re^2/(re+h))
then put that v into the first equation, and solve for h.
This is a form of Keplers' law.
but velocity can be found..
9.8(re/(re+h))^2=v^2/(re+h)
or v=sqrt (9.8 (re^2/(re+h))
then put that v into the first equation, and solve for h.
This is a form of Keplers' law.
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