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I Problem with gravitation field perpendicular vector.

  1. Nov 22, 2018 #1
    since it is known that ##\vec{A_\perp} = -{mG \over R^2}## why did the professor write it as ##\vec{A_\perp} = {- R G \rho \over 3}## for perfect sphere with perfect mass distribution ? Shouldn't it be ##\vec{A_\perp} = -{4 \over 3} \pi R G \rho##? I need help thanks.
     
    Last edited: Nov 22, 2018
  2. jcsd
  3. Nov 22, 2018 #2

    Doc Al

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    Staff: Mentor

    If I understand your question, shouldn't that be a 4/3 instead of a 4? (But I agree that your professor's version seems off.)
     
  4. Nov 22, 2018 #3
    Sorry got it off by a bit there. fixed
     
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