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Circular Orbit and Period

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data
    The Hubble Space Telescope orbits Earth 615 km above Earth's surface. What is the period of the telescope's orbit?


    2. Relevant equations
    T = 2(pi)r/v


    3. The attempt at a solution

    T = 2(pi)(6.37x10^6)/9.8
     
  2. jcsd
  3. Feb 6, 2008 #2

    mgb_phys

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    Remember the radius of the orbit is measured from the centre of the earth.
     
  4. Feb 6, 2008 #3
    Isn't 6.37 × 106 m the radius of the earth, or do I use another measurement.
     
  5. Feb 6, 2008 #4
    So, the radius of the orbit is radius of earth? That means the satellite is orbiting in surface, right?
     
  6. Feb 6, 2008 #5
    it orbit 615km above the Earths surface
     
  7. Feb 6, 2008 #6

    mgb_phys

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    The orbital radius is the distance from the centre of the Earth to the satelite.
    Hence the radius of the Earth PLUS the altitude = approx 7000km.
     
  8. Feb 6, 2008 #7
    Remember to express total distance (Earth's radium + orbital distance) in meters to mantain a coherence among unities.
     
  9. Feb 6, 2008 #8

    mgb_phys

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    And since you aren't given the speed you also need Kepler's 3rd law to work out the period. Not sure where you got the 9.8 from in your original equation - but it's wrong!
     
  10. Feb 6, 2008 #9
    how would you use Kepler's 3rd law in relation to this question though, you aren't given some of the information needed.
     
  11. Feb 6, 2008 #10
    Nope - you have everything you need. What information do you think you need for Kepler's 3rd Law that you aren't given?

    And by the way, mgb_phys - I'll be cstout was using the value for the approximate acceleration due to gravity at the Earth's surface, i.e. 9.8 m/s^2, which you'd just have to call a pretty bad guess.
     
  12. Feb 6, 2008 #11

    mgb_phys

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    Remember the simplified form of the law doesn't need the mass of the orbital body
    , you can assume that G(M+m) is pretty much equal to GM. Hubble only weighs about 10tons!
     
  13. Feb 6, 2008 #12
    Ok, I finally got it, the answer is 1.61 hrs. Thanks for the help
     
  14. Feb 6, 2008 #13

    mgb_phys

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    That sounds about right, 90 minutes is pretty typical for LEO
     
  15. Jan 29, 2009 #14
    Try using this formula: 4(pi)² r³/(M+m)G = P²

    Where:
    r = radius between the centers of mass of both bodies measured in meters
    M = the larger mass in kilograms (5.9742 X 10^24 for the earth)
    m = the smaller mass in kilograms (this mass is usually so much smaller than the
    larger mass that it can almost always be disregarded)
    G = 6.67259 X 10-¹¹
    P = the period of orbit in seconds

    This formula yields an answer of approx. 5,811 seconds, which is about 96.85 minutes.
     
    Last edited: Jan 29, 2009
  16. Feb 2, 2009 #15
    I don't think so..

    Since the radius,r of the telescope is just about 10% of the radius of the Earth.
    Then it can be ignored..

    Just use R for the radius of Earth.

    Hence, Period of revolution of telescope
    = 2Pi Square root R/g

    =84.43min
     
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