Question
Can someone please help me understand why my answer is wrong?
In Section 12-2, there is a cakculation showing that the acceleration due to gravity at the Earth's surface is g = 9.819"9 81mw. Furthermore. Example 12-2 displays a calaleion of the aceleraton due to gravity on top of Mt Everest Using these as examples, solve the following: I the Tracking and Data Relay Satelite System (TDRSS) is in a geostationary orbt at A typical satellite in an ahtude of 22.236 miles atove mean-sea-level (MSL) Ignoring effects of the Moon, Sun, etc, what is the acceleration due to the Earth's gravity at that altitude? 0005 m/s 981 m/s 9.81 ms 0.223 mis 1.033 m/s
Explanation / Answer
R =6371 km , h =22236 mile= 35785373 m
g = GM/r^2
g = (6.67*10^-11*5.98*10^24)/(6371000+35785373)^2
g = 0.223 m/s^2
Correct option is (c)