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Newton universal gravitation formula, how f is dimensionally derived?

  1. Sep 2, 2014 #1
    If f dimensions are ml/t^2, where does t^2 come from in the equation of
    F = G*m1*m2/r^2 where I believe G to be a constant, m1 and m2 to be masses and r to be the distance between two masses - so length. To dimensionally analyse this then, where would the dimension time come from if I were to check if the equation is dimensionally correct and I cant see where it comes from? I do think that it may come from G but i have seen various answers on what the dimensions of G are so I cant cancel the dimensions out to check if it is the same as F? Help would be appreciated, many thanks.
     
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  3. Sep 2, 2014 #2

    A.T.

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  4. Sep 2, 2014 #3

    Integral

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    According to Newton

    F=ma since acceleration has units l t-2 the units of force must be m Lt-2.

    Using that you can figure out the units G needs to make the given equation dimensionally correct. Time is factored into the constant.
     
  5. Sep 2, 2014 #4
  6. Sep 2, 2014 #5
    Hi thanks for the help however I focusing on F instead of G giving that G = N m^2 kg^-2 or what G is?
     
  7. Sep 2, 2014 #6

    nrqed

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    Isolating, ## G= F ## ##
    r^2 / (m_1 m_2 )##. You can tell right away the units of G.
     
  8. Sep 2, 2014 #7
    ah I think I understand so by working out the dimensions of G you can show that F must equal to Gxm1xm2 because it will balance the equation to = F?
     
  9. Sep 2, 2014 #8

    A.T.

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    No. Observation shows that F equals Gxm1xm2/r^2. The dimension of G are defined based on this. So there is no point in checking if the equation is dimensionally correct, because it was defined to be correct.
     
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