The global positioning system (GPS) satellite that moves in a circular orbit with period of 11h 58 min has a speed of: 3872.9848 m/s
To solve this problem the formulas and the procedures we will use are:
Where:
Information about the problem:
By converting the time period of (h) and (min) from (s) we have:
T = (11h *3600 s/ 1 h) + (58 min * 60 s/1 min)
T = 39600 s +3480 s
T = 43080 s
Applying the orbital radius formula we get:
r =[(G* m *T²)/(4 * π²)]
r =[(6.67 × 10⁻¹¹ m³/kgs² * 5.972 × 10²⁴ kg *(43080 s)²)/(4 * (3.1416)²)]
r =[1.8725*10^22 m³]
r= 26.55465166*10^6 m
Applying the tangential velocity formula we get:
v = (2*π * r)/ T
v = (2 * 3.1416 * 26.55465166*10^6 m)/ 43080 s
v = 3872.9848 m/s
It is the maximum of the radial distribution curve of the outermost orbital.
Learn more about orbital radius at: brainly.com/question/14437688
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