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Newtonian gravity theory and energy

  1. Jul 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1.0kg object is released form rest 500km above the earth. What is its impact speed as it hits the ground? Ignore air resistance.

    2. Relevant equations
    ##U_g = \frac{GmM_e}{r}##
    ##K = \frac{1}{2}mv^2##
    ## \Delta U = - \Delta K##

    3. The attempt at a solution
    Using energy conservation I arrived at
    ##\frac{GmM_e}{r} = \frac{1}{2}mv^2##, then ##v = sqrt{\frac{2GM_e}{r}}##

    Plugging in values, ##v = \sqrt{\frac{2(6.67\times10^{-11})(5.98\times10^{24})}{(6.37\times10^6 +5.00\times10^5)}} = 10776 m/s##, never mind the significant figures.
    This is rather far from the correct answer though, which is 3.02km/s.
     
  2. jcsd
  3. Jul 12, 2015 #2

    Nathanael

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    Homework Helper

    (According to your equation for the gravitational potential energy) is the GPE zero at the surface of Earth?

    P.S.
    There should be a negative sign on Ug
     
  4. Jul 12, 2015 #3
    Forgot to put the negative sign in there. By the look of it the GPE is zero at the centre of the earth, which accounts for the excess energy, correct?
     
  5. Jul 12, 2015 #4
    Solve for v
    upload_2015-7-12_14-43-4.png
     
  6. Jul 12, 2015 #5

    Nathanael

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    It has nothing to do with the center of the Earth.

    You have two positions, 500km above the surface, and at the surface. The change in the GPE between these two positions is equal to the change in kinetic energy. Your calculation neglected the GPE at the surface of Earth.
    (Your equation for the GPE takes the zero-potential to be at infinity, not at the surface.)
     
  7. Jul 13, 2015 #6
    True. I got that mixed up. Problem solved and thanks for the input. I feel I need to have another closer look at this section though.
     
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