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March 26, 2017

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(a) Find the speed of a satellite moving around the earth in a circular orbit that has a radius equal to six times the earth's radius of 6.38 x 10^6 m.

(b) Find the satellite's orbital period.

  • Physics - ,

    centripetalforce=gravity force

    v^2/r=9.8/(6)^2

    V= sqrt (6*re*9.8/36)

    checkthat.

    Period? Well, T=distance/velocity=2PI*6Re/above V

    That will give the period in seconds, you probably want to convert it.

  • Physics - ,

    (a) centrieptal acceleration rate =
    V^2/r = GM/r^2

    where M is the mass of the earth and r = 6 Re.
    Re is the radius of the earth that you were given
    G is the universal constant of gravity.

    Solve for V

    V^2/(6 Re) = GM/(6Re)^2

    You can save yourself the trouble of looking up M and G by using the relationship
    g = GM/Re^2. Then
    6 V^2 /Re = GM/Re^2 = g
    V^2 = (Re*g/6)
    V = 3230 m/s

    (b) V*(period) = 2 pi R = 12 pi Re

    Solve for the period

  • Physics - ,

    Vc = sqrt(µ/r) where µ = 3.9863x10^14 and r = 6378km.
    At 38,268km,
    Vc = sqrt[3.9863x10^14/6x6,378,000] =

    The period derives from
    T = 2(Pi)sqrt[r^3/µ]

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