SOLUTION: When a satellite is h kilometres above Earth, the time, t in minutes, to complete one orbit is given by the formula:
t= (√(6370+h) power of 3)/6024
A satellite is placed
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Question 882942: When a satellite is h kilometres above Earth, the time, t in minutes, to complete one orbit is given by the formula:
t= (√(6370+h) power of 3)/6024
A satellite is placed in geosynchronous orbit about Earth. What must its altitude be?
*Geosynchronous orbit: orbit around Earth matches the rotation of Earth
The answer at the back of the textbook says that it's 35 800km
I get the basic of the question but I just can't get the calculation right please help! Thank you~
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
When a satellite is h kilometres above Earth, the time, t in minutes, to complete one orbit is given by the formula:
t= (√(6370+h) power of 3)/6024
A satellite is placed in geosynchronous orbit about Earth. What must its altitude be?
*Geosynchronous orbit: orbit around Earth matches the rotation of Earth
The answer at the back of the textbook says that it's 35 800km
======
The satillite would have to complete its orbit in 24 hr = 24*60 = 1440 minutes
----
Using your formula, let t = 1440 and solve for "h":
1440 = (sqrt(6370+h)^3)/6024
8674560 = (6370+h)^(3/2)
Raise both side to the (2/3) power to get
----
6370+h = 42218
h = 35848
When rounded you get h = 35800 km
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Cheers,
Stan H.
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