Derived equation initial speed

  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. cepheid

    cepheid 5,191
    Staff Emeritus
    Science Advisor
    Gold Member

    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. Draw a picture, and ask, what is r?
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  5. 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. cepheid

    cepheid 5,191
    Staff Emeritus
    Science Advisor
    Gold Member

    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. 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. cepheid

    cepheid 5,191
    Staff Emeritus
    Science Advisor
    Gold Member

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