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Keplers laws of motion - quick question on units

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data
    find the radius
    R3=((T2GM/(4∏2))1/3


    3. The attempt at a solution
    simple enough, my only question is when I plug in T (units: sec) G (units: m/s^2) and M (units: kg) there is nothing to cancel out the kg so my R is not in units of meters...the equation is correct because we were given it in class but what's going on with the units?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    G in this problem isn't an acceleration. It's the gravitational constant G. The units are m^3/(kg*s^2).
     
  4. Jan 30, 2013 #3

    gneill

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    Staff: Mentor

    That's G, not g :wink:

    EDIT: Beaten to the punch!
     
  5. Jan 30, 2013 #4
    Ah hah! Thank you both!
     
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