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A spy satellite is in circular orbit around Earth. It makes one revolution in 5.99 hours. (a) How high above Earth's surface is...Asked by Max
A spy satellite is in circular orbit around Earth. It makes one revolution in 6.04 hours.
(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
YOu start with centripetal acceleration= gravitational attraction.
w^2 * r= 9.8 (re^2/(r^2)) where re is the radius of the Earth, and r is the distance from the center of the Earth to the satellite.
w= 2PIr/period work in m/s, m
w^2 * r= 9.8 (re^2/(r^2)) where re is the radius of the Earth, and r is the distance from the center of the Earth to the satellite.
w= 2PIr/period work in m/s, m
Answered by
Max
so W = (2PI(6.37x10^6))/21,744sec
6.04hr = 21,744 sec
and then for r = 117,367
I don't get what to do afterwards
6.04hr = 21,744 sec
and then for r = 117,367
I don't get what to do afterwards
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