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Acceleration given distance from earth (r)

  1. Sep 20, 2012 #1
    Question: Satellites are placed in a circular orbit that is 3.59 X 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

    So far I've been using g=(M/r2) and get .31m/s2. There must be something wrong with either using this equation or that g isn't the acceleration. Can you explain how to reach acceleration from this question?
     
  2. jcsd
  3. Sep 20, 2012 #2
    In the equation you posted, you're missing the gravitational constant "G." However, I suspect you used it in your calculation because I got the same answer when I included G. You seem to be assuming that your answer of .31 m/s2 is wrong, but it seems fine to me.

    I wouldn't use small g to denote the acceleration in orbit though. Small g usually refers to the acceleration due to gravity on earth's surface. Instead, you could call it "a" for acceleration.
     
  4. Sep 20, 2012 #3
    Normally I'd use the traditional gravitational formula, but since I don't have a mass for the second object (satellite) I'm not sure if I can use it. F=(GM1M2)/r2. Since F=ma, I wonder if once I solve the equation I should divide it by the mass of earth again to isolate a, that would lead to a very small number though and I don't think that would be accurate, any suggestions about doing it that way?
     
  5. Sep 20, 2012 #4
    Sure, you can use Newton's law of gravitation no problem. That's what the equation you used is derived from. Here's how:

    [itex]\displaystyle F=G\frac{mM_E}{r^2}[/itex]

    [itex]\displaystyle a=F/m[/itex]

    Substitute F:

    [itex]\displaystyle a=G\frac{mM_E}{r^2}/m[/itex]

    and m cancels to give:

    [itex]\displaystyle a=G\frac{M_E}{r^2}[/itex]

    Then, plugging in the radius given by the problem, and putting in the numerical values for G and ME (mass of earth):

    [itex]\displaystyle a=(6.673×10^{-11})\frac{5.972×10^{24} kg}{(3.59×10^7 m)^2} = 0.31\frac{m}{s^2}[/itex]

    Are you convinced that you got the right answer yet? :)
     
    Last edited: Sep 20, 2012
  6. Sep 20, 2012 #5
    I'm convinced but it's this online system that won't mark the answer as correct, I'll check with my professor. Thanks for you help.
     
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