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Keppler's law

  1. May 27, 2008 #1
    Use Kepler’s laws to determine the period of the Hubble Space Telescope which orbits the earth at an altitude of 610 km.

    I'm using the eqation i know to find the period of an object revolving around another object

    T^2=Kr^3
    T^2=(4pi^2/GM)*r^3
    G=6.67 x 10^-11
    M=5.98× 10^24 (Mass of Earth)
    r=610,000

    T^2=(39.48/3.99 x 10^14)*2.27 x 10^17
    T^2=9.898 x 10^-14*2.27 x 10^17
    T^2=22465.81736
    T=149.88

    I don't know what this means...
    but i have another equation for the period of a satelite given by
    T=2pi*r/v

    mv^2/R+h = GMm/(R+h)^2
    v^2/R+h = GM/(R+h)^2
    v^2/6.99 x 10^6 = 3.989 x 10^14/4.886 x 10^13
    v^2= 8.16 * 6.99 x 10^6
    v=7554.3

    back to original eqation:
    T=2pi*r/v
    T=5813.84 seconds
    T=96.9 minutes

    wikipedia agrees with that number
     
  2. jcsd
  3. May 27, 2008 #2

    alphysicist

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    Homework Helper

    Hi warmfire540,

    I believe this number is your error; this r is supposed to be the radius of the orbit, so r is the altitude of 610 km plus the radius of the earth.
     
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