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
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?
2 answers
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
1 answer