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11 August, 20:36

If HST has a tangential speed of 7,750 m/s, how long is HST's orbital period? The radius of Earth is 6.38 * 106 m.

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  1. 11 August, 23:55
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    Assuming that earth is a perfect circle. First solve the circumference of the earth,

    C = 2pi (r)

    Where are is the radius

    C = 2pi (6.38 x 10^6 m)

    C = 4.01 x 10^7 m

    O the orbital period

    T = 4.01 x 10^7 m / 7,750 m/s

    T = 5172.48 s
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