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Gravitation deep hole physics problem

  1. Nov 25, 2005 #1
    Hi I am stuck with this problem:
    A deep hole in Earth reaches a depth of one half of Earth's radius. How much work is done when a 1-kg mass is slowly lifted from the bottom of the hole to Earth's surface?
    I use the expression of the gravitation force, F=GMm/R3
    so the work would be:
    integral (F*(R/2)) from R=R/2 to R=R
    I am not sure I am right so I need your input please.
    Thank you
  2. jcsd
  3. Nov 25, 2005 #2


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    Make sure you include the dependence that M has on r (the integration variable).
  4. Nov 25, 2005 #3


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    Homework Helper

    Your Force function is incorrect.

    The Gravitational Force has an INVERSE SQUARE form [tex]F = m \frac{GM}{r^2}[/tex] .
    The Force by Gravity will only be due to the mass of Earth that is
    INTERIOR to the mass being lifted - presume uniform mass density, that is [tex]M = M_E \frac{V_{inside}}{V_E}[/tex] .
    Last edited: Nov 25, 2005
  5. Nov 25, 2005 #4
    Thank you.
    But How can I compute the Vinside? we only have the length nothing else?
  6. Nov 25, 2005 #5


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    Volume? LxWxH !
    Yeah, rxrxr = r^3 , the variable r ,
    compared with [tex]R_E^3[/tex] , the maximum R .
  7. Nov 25, 2005 #6
    OK!! I am sorry . sometimes I am a little bit way off!

    thank you so much
    I can finish this problem now.

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