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Derived equation initial speed

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Use the equation derived about to calculate the exact value of the initial speed that will produce a circular orbit of the satellite at an altitude of 1000 km above earth.


    2. Relevant equations
    -Derived equation provided:
    Finward = Fg
    m1 v^2 / r = Gm1m2/r
    v^2 = Gm2/r
    v= (square root of) Gm2/r


    3. The attempt at a solution
    Finward = Fg
    m1 v^2 / r = Gm1m2/r
    v^2 = Gm2/r
    v = (square root of) Gm2/r
    v = (square root of) (6.67x10^-11)(5.98x10^24)/1000
    v = (square root of) 3.98....x10^11
    I didn't round of 3.98....on my calculator; I saved the whole number and calculated the square. This gave me:
    v = 631558.39
    v = 6.3x10^5 m/s ????

    Is this initial velocity though? Or did I miss a step?
     
    Last edited: Dec 11, 2009
  2. jcsd
  3. Dec 11, 2009 #2

    cepheid

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    In Newton's law of gravitation, r is the distance from the *centre* of mass 1 to the *centre* of mass 2. This is true for the equation for centripetal force as well -- r is the distance from the object to the centre of its orbit. In other words, you have not used the correct distance for r.
     
  4. Dec 11, 2009 #3
    Draw a picture, and ask, what is r?
    [​IMG]
     
  5. Dec 11, 2009 #4
    Finward = Fg
    m1 v^2 / r = Gm1m2/r
    v^2 = Gm2/r
    v = (square root of) Gm2/r
    v = (square root of) (6.67x10^-11)(5.98x10^24)/(1000/2)
    v = (square root of) (6.67x10^-11)(5.98x10^24)/500
    v = (square root of) 7.977......
    I didn't round of 7.977....on my calculator; I saved the whole number and calculated the square. This gave me:
    v = 893158.4406
    v = 8.9x10^5 m/s ????

    Is this the initial velocity?
     
  6. Dec 11, 2009 #5

    cepheid

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    Where do you get 1000/2 from? This is not correct. I think you should read my previous post again. The distance 'r' has a very specific definition here.
     
  7. Dec 11, 2009 #6
    Finward = Fg
    m1 v^2 / r = Gm1m2/r
    v^2 = Gm2/r
    v = (square root of) Gm2/r

    G= gravitational constant = 6.67x10^-11
    m2= mass of earth = 5.98x10^24
    r= radius of earth + altitude of satellite above earth = 6.37x10^6 + 1000 ?????

    If everything else is right, would this equation give me initial velocity or average?
     
  8. Dec 11, 2009 #7

    cepheid

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    It would give you the orbital speed of the satellite (which is constant in magnitude).
     
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