A%20satellite%20moves%20in%20a%20circular%20orbit%20around%0Athe%20Earth%20at%20

Question

A%20satellite%20moves%20in%20a%20circular%20orbit%20around%0Athe%20Earth%20at%20a%20speed%20of%206.1%20km%2Fs.%0ADetermine%20the%20satellite%E2%80%99s%20altitude%20above%0Athe%20surface%20of%20the%20Earth.%20Assume%20the%0AEarth%20is%20a%20homogeneous%20sphere%20of%20radius%0A6370%20km%20and%20mass%205.98%20%C3%97%201024%20kg.%20The%0Avalue%20of%20the%20universal%20gravitational%20constant%0Ais%206.67259%20%C3%97%2010%E2%88%9211%20N%20%C2%B7%20m%0A2%0A%2Fkg2%0A.%0AAnswer%20in%20units%20of%20km%20Your%20answer%20must%0Abe%20within%20%C2%B1%201.0%25

Explanation / Answer

Use Newton's Law of Gravitation combined with the centripetal force to solve this

GMm/(R + r)^2 = m*a = m*v^2/(R + r) where R is the radius of Earth and r is the height above the surface (Note a = v^2/(R + r) centripetal a)

So GM/(R + r) = v^2 So GM/v^2 = (R + r)
Therefore r = GM/v^2 - R = 6.67259x10^-11*5.98x10^24/(6100^2) - 6370x10^3 = 4.35x10^6m = 4353km

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