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If the standard gravitational parameter of earth, μ is 398,600 km3/s2 and the radius of earth is 6378 km, what is the period of a satellite revolving around earth in a circular orbit 500 kms above the surface of the earth?

Correct Answer: 5,801.06 s
The Time Period of a circular orbit can be derived as: T = (2πR3/2/√(μ) R = radius of earth + distance from the surface of earth to satellite R = 6378 + 500 R = 6978 km Therefore, T = (2π(6978)3/2/√(398600) T = 5,801.06 s

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3. The orbital velocity of a satellite increases with the radius of the orbit.
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