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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