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Gravitational Force Units Question

  1. Mar 12, 2013 #1
    I just have a question concerning a previous thread about gravitation.

    The formula is from post 19 at https://www.physicsforums.com/showthread.php?t=635188&page=2

    v= dr/dt

    a = dv/dt

    multiply by dv/dt by dr/dr:

    a = (dr dv)/(dt dr) = v dv/dr

    This gets you to the first step:

    v dv/dr = -G (m1 + m2) / r2

    My question is: Why is it possible to set a = v dv/dr = -G (m1 + m2) / r2 ?

    On the left side there are units of acceleration (m/s^2) and on the right side is the gravitational force (kg*m/s^2). Shouldn't the units at all time be the same on both sides?

    Thanks very much for an answer.

    Regards, Philipp
  2. jcsd
  3. Mar 12, 2013 #2


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    hey philipp, please use [itex]\LaTeX[/itex]. it ain't hard.
  4. Mar 12, 2013 #3


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    Science Advisor

    Are you sure? Look up the units of G and check that again.
    Last edited: Mar 12, 2013
  5. Mar 12, 2013 #4
    Thanks for the reply.

    Well usually -G (m1 + m2) / r2 = F = m * a

    This is not equal to just acceleration as in a = (dr dv)/(dt dr) = v dv/dr = -G (m1 + m2)
  6. Mar 12, 2013 #5

    Doc Al

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    Staff: Mentor

    No. Gravitational force requires two masses multiplied, not added: Gm1m2/r2

    The left side of your equation has units of acceleration, not force.
  7. Mar 12, 2013 #6
    aaa, thanks. so silly from me :)
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