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Gravitational Potential

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the gain of potential energy of a mountaineer of mass 80kg who travels to the top of the mountain Everest 9km above sea level.


    2. Relevant equations

    ΔEp=mgΔh

    3. The attempt at a solution

    ΔEp=mgΔh=80x9.81x9x10^3=7.1x10^6J.

    This is the correct answer but surely it's only an estimate since the value of g 9km above the Earth's surface is 9.83830. Δg=9.8380-9.81= 2.8x10^(-3)?

    Is it possible to calculate the exact change is Ep as the mass moves through Δh, rather than using a fixed value of 9.81?
     
  2. jcsd
  3. Oct 13, 2011 #2

    Hootenanny

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    Yes. One would need to go back to Newton's law of gravitation and compute

    [tex]\Delta U = \int_{R}^{R+h} \frac{GM_E m}{r^2}\;\text{d}r,[/tex]

    where R is the radius of the Earth, ME is the mass of the Earth, m is the mass of the man and h is the height of Everest.
     
  4. Oct 13, 2011 #3
    Ah ok.

    How would you go about integrating GMm/r^2? My maths isn't great but this is what I get:

    Since G is a constant, factor it out.

    [tex]\Delta U = G\int_{R}^{R+h} \frac{M_E m}{r^2}\;\text{d}r,[/tex]

    [tex]\Delta U = G [-\frac{M_E m}{r}]_{R}^{R+h}[/tex]

    Sorry, my maths isn't at this level yet!
     
  5. Oct 13, 2011 #4

    Hootenanny

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    So we are left with
    [tex]\Delta U = GM_E m \left(\frac{1}{R} - \frac{1}{R+h}\right).[/tex]
    Simply plug the numbers in and you should get your result.
     
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