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March 29, 2017

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A spy satellite is in circular orbit around Earth. It makes one revolution in 6.04 hours.
(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?

  • Physics - ,

    YOu start with centripetal acceleration= gravitational attraction.

    w^2 * r= 9.8 (re^2/(r^2)) where re is the radius of the Earth, and r is the distance from the center of the Earth to the satellite.

    w= 2PIr/period work in m/s, m

  • Physics - ,

    so W = (2PI(6.37x10^6))/21,744sec
    6.04hr = 21,744 sec

    and then for r = 117,367
    I don't get what to do afterwards

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