Find the period of a low earth orbiting satellite using Kepler Laws
earth radius = 6.38E6 meters
T^2/R^3 for earth = 2.97E-19 (sec^2/m^3)
2.97E-19 = T^2/(6.38E6)^3
T^2 = (2.97E-19)(6.38^6)^3 = 77.1
T = 8.72 sec
The Attempt at a Solution
This is not correct since I know the period of a low earther is about 90 minutes or about 5600 seconds.