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Derivation of mgh

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  1. Dec 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Derive E=mgh from Newton's law of Gravitation where h is very small. (Use binomial expansion)

    2. The attempt at a solution
    [tex]E = \frac{GMm}{(r+h)^2}-\frac{GMm}{r^2} [/tex]
    [tex]E = \frac{GMm}{r^2}(\frac{1}{(1+\frac{h}{r})^2}-1) [/tex]
    [tex]E = \frac{GMm}{r^2}((1+\frac{h}{r})^{-2}-1) [/tex]
    [tex]E = \frac{GMm}{r^2}(1+\frac{-2h}{r}-1) [/tex] Other powers of h/r becomes negligible for h<<r
    [tex]E = \frac{GMm}{r^2}(\frac{-2h}{r}) [/tex]

    Not sure where I went wrong or what to do next
    Note:- I know this can be solved in a million other methods but I want the answer specifically from this method. I had a book which used this method and I cant remember how this method works out.
     
  2. jcsd
  3. Dec 14, 2016 #2

    Orodruin

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    The potential in Newtonian gravity is negative. You are taking potential(reference level) - potential(height h) instead of the other way around.
     
  4. Dec 14, 2016 #3
    Okay so I will have 2h/r instead of -2h/r. Doesn't solve the problem though
     
  5. Dec 14, 2016 #4

    Orodruin

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    Then what is your problem? It looks fine to me.
     
  6. Dec 14, 2016 #5
    Instead of "2h/r", I should get "h" so I can rewrite the answer as mgh
     
  7. Dec 14, 2016 #6

    Orodruin

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    Move 2/r from the h/r term to the term with G and M ...
     
  8. Dec 14, 2016 #7
    so 1/r^2 will become 1/r^3.
     
  9. Dec 14, 2016 #8

    Doc Al

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    Start with gravitational PE, not force. (What you are calling "E" is force not energy.)
     
  10. Dec 14, 2016 #9
    ohhhhhhhhhh
     
  11. Dec 14, 2016 #10
    Okay now I am getting GMm/r *h/r
     
  12. Dec 14, 2016 #11
    Oh okay done thanks
     
  13. Dec 14, 2016 #12

    Orodruin

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    True that. The same basic principle applies though. The entire point is expressing g in terms of G, M and r.
     
  14. Dec 14, 2016 #13

    Doc Al

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    Yep.
     
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