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Change in gravitational Potential energy

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok, I am deriving the change in gravitational potential energy at the surface of the earth and have met with some difficulties.


    2. Relevant equations
    [tex]g=GM/r^2[/tex]
    [tex] P.E.=-GMm/r[/tex]


    3. The attempt at a solution

    [tex] \delta P.E. = -GMm/r_f+GMm/r_i[/tex]

    Where [tex] r_f > r_i[/tex]
    Since

    gravitational field strength is approaximately a constant for small changes of r,

    [tex] \delta P.E. = -mgr_f + mgr_i[/tex]

    But this gives me a negative answer! How can my change in gravitational potential energy be negative when I bring my object from a lower position to a higher poistion in the gravitational field. I have probed the mathematics of the above proof, but it all makes sense to me since g is a constant.
     
  2. jcsd
  3. Sep 8, 2007 #2
    [​IMG]
    for rf>ri, this quantity is positive, correct?

    When using this form:
    [​IMG]

    You must remember that change in PE is final - initial, and both are positive quantities. You have written it as initial minus final. The potential at a point is positive, a change however, can be negative.

    I'm not sure that helps, sorry.
     
  4. Sep 8, 2007 #3
    thats all right i appreciate your effort!

    The problem is I derived this equation and i can't think of anything wrong with my derivation!!
     
  5. Sep 8, 2007 #4
    How did you derive the equation? I am not absolutely sure, but I think you expand
    [​IMG]
    with the binomial theorem (expand each term). I am not sure, I will check when I get home.
    Anyway, the point is to not blindly follow equations to tell you about the physics. You need to understand the physics principles underlying a problem, and then apply the relevant formulae, not the other way around. This usually helps when dealing with signs.
     
  6. Sep 8, 2007 #5
    Oh, i just substituited the gravitational field strength equation into the change in potential energy equation. the signs couldnt be wrong.
     
  7. Sep 8, 2007 #6
    You'd need to know where the gravitational field strength eqautions come from and how they are to be applied. I can't really help you until I get home and work this out myself, which may be only this night. But I am sure someone else on here will be happy to help. Sorry, I know I wasn't much help :(
     
  8. Sep 8, 2007 #7
    Ok let's follow strictly mathematics precedures.

    [tex]\delta P.E. = -GMm/r_f + GMm/r_i [/tex]

    we can write [tex]r_f = r_0+\delta r_f ~ where~ of~ course~ r_0 >> \delta r_f [/tex]

    Similar we can write for [tex]r_i[/tex]. Now we write [tex]\delta P.E. = -GMm/(r_0(1+\delta r_f / r_0)) + GMm/(r_0(1+\delta r_1 / r_0)) [/tex].

    Now write [tex] g = GMm/r_0 [/tex] and use series: [tex] (1+x)^{-1} =1-x [/tex]
    What you get is exactly :

    [tex] \delta P.E. = mgr_f - mgr_i [/tex]

    Hope I was clear.
     
  9. Sep 8, 2007 #8
    ooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhh, i get it my mistake was that i took the gravitational field strength approaximation for granted. It is taken only for the surface of the earth, i included the respective heights for both positions into the gravitational field strength equation. Just to check, is this undergraduate stuff? or just high school physics
     
  10. Sep 8, 2007 #9
    Here in Europe we had that kind of things in introductionary courses of physics in the first year of undergraduate. Actually it is more question of math than physics. You can easily solve it if you are familiar with use of power series.
     
  11. Sep 8, 2007 #10
    huh, i am actually aiming to get through the first year exemption test before enrolling into a university.
     
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