Using Keplers law of planetary motion, we can drive the following expression for a circular orbit: T ^ 2 = (4 ^ 2 a ^ 3 / ) where T = orbital period; a = orbital radius in km = distance from the center of the earth to the orbit; = Keplers constant = 3.986004418 * 10 ^ 5 km ^ 3 / s ^ 2. The earth rotates once per sidereal day of 23 h 56 m 4.09 s.

a. Determine the orbital radius of a GEO satellite.

b. Assuming an earth radius of 6370 km, what is the orbit height h (Figure 16.1) of a GEO satellite? Note: your answer should differ slightly from the figure used in the chapter. Different sources in the literature give slightly different values.