a satellite is moving in a circular orbit at a certain height above the earth's surface.. it takes 5.26 x 10^3 s to complete one revolution with centripetal accelaration equal to 3.92 ./sec^2. the height of the satellite above the earth's surface is?

Question

a satellite is moving in a circular orbit at a certain height above the earth's surface.. it takes 5.26  x 10^3 s to complete one revolution with centripetal accelaration equal to 3.92 ./sec^2. the height of the satellite above the earth's surface is?

anonymous · Answer

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For the satellite to be in circular orbit, the centripetal force would have to be provided by the gravitational force. You know F = ma, and that the gravitational force equals$$F _{G} = (m_{S}M _{E}G)/L ^{2}$$. Setting the two equations equal to each other, the mass of the satellite cancels, and you can solve for "L", the distance between the satellite and Earth. (Look up the earth's mass.)