Problem 9-76 Geosynchronous orbit - Part 8 - C
Most telecommunications satellites are in geosynchronous orbits above the earth, that is, they have periods of \(24 \;h.\) As a result, since the earth turns on its axis once in \(24\; h\) and each satellite goes around the earth once in \(24\; h,\) any individual satellite stays positioned above a particular point on the earth. (a) How far above the earth's surface must a geosynchronous satellite be? The earth's mass and average radius are \(5.98 \times 10^{24}\; kg\) and \(6368 \;km.\) (b) What is the satellite's speed?
Accumulated Solution
\(F = GmM/r^2 \\ F = mv^2/r \\ v^2 = GM/r\)
No. That is not the relation between distance, speed and time. Again dimensional analysis can come to your aid.
\(v = dt = 2\pi rT | \text{Dimension of} \; v = L·T^{-1} \\ \text{Dimension of} \; rT = L·T\)
They are not the same so the equation is incorrect.