Posted by Kelsey on Wednesday, April 7, 2010 at 1:26pm.
A satellite moves in a circular orbit around Earth at a speed of 4390 m/s.
(a) Determine the satellite's altitude above the surface of Earth.
m
(b) Determine the period of the satellite's orbit.
h
am I supposed to use rotational equations for this? I don't know what to start out with.

physics  Kelsey, Wednesday, April 7, 2010 at 1:31pm
I tried to use V^2/r = GM(earth)/r^2 but I am not getting the correct answer.

physics  tchrwill, Thursday, April 8, 2010 at 10:04am
V = sqrt(GM/r)
GM = 3.986x10^14
V = 4390
r = 3.986x10^14/4390^2 = 20,682,748.6m = 20,682.7km
Earth radius = 6378km
Orbital altitude h = 20,682.7  6378 = 14,394.7km.
The time it takes a satellite to orbit the earth in a circular orbit, its orbital period, can be calculated from
T = 2(Pi)sqrt[r^3/GM]
where T is the orbital period in seconds, Pi = 3.1416, r = the radius of the circular orbit and GM = the Earth's gravitational constant = 6378x10^42.
I'll let you carry out the computation of the period.
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