## Essential University Physics: Volume 1 (3rd Edition)

$h = 1.69\times 10^6 \ m$
We simplify the equation for the period of a satellite for r to find: $r =\sqrt[3] {\frac{GMT^2}{4\pi^2}}$ Plugging in the value of G, a constant, the mass of the earth, and 7200 seconds for the period (T), we find: $r=8.06\times 10^6$ Subtracting the radius of the earth gives: $h = 1.69\times 10^6 \ m$