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Gravitational problem

  1. Apr 3, 2005 #1
    In the attachment the answer is that there would be higher pressure and lower acceleration due to gravity. Again, please explain this concept to me. Thanks in advance.

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  2. jcsd
  3. Apr 3, 2005 #2


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    Acceleration due to gravity can be simplified by only considering the mass below you. The mass above you pulls up on you. There's a bit more to it that than that, but I hope that's a good start.
  4. Apr 3, 2005 #3


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    The reason for the acceleration part has taken Newton 20 years to explain. So there is no simple explanation. It would have been cruel of your teacher to ask you this if you haven't seen the theory behind it.

    In a nutshell, the theory is this...

    Consider the gravitationnal field

    [tex]\vec{g}(r) = \frac{\vec{F}_{grav}}{m}[/tex]

    The statement of Gauss' law for gravitation is

    [tex]\int_S \vec{g} \cdot d\vec{A} = -4\pi G M_{enclosed}[/tex]

    Consider S to be a sphere centered on the earth that has said mine on its surface. Due to symetry consideration, the integral becomes

    [tex]\int_S \vec{g} \cdot d\vec{A} = g(4\pi r^2)[/tex]


    [tex]g(4\pi r^2) = -4\pi G M_{enclosed} \Rightarrow g = \frac{-GM_{enclosed}}{r^2}[/tex]

    That is to say, the grav. potential (and hense the force) at any location INSIDE the earth, is the same as the potential that would be created atthis point by a point-mass situated at the core of the earth whose mass is that of the portion of the earth enclosed by a centered sphere that has said location on its surface. (Excuse me, I lost me :grumpy:)

    In other word, the gravitational force inside the earth depends only on the mass "under" you. So the deeper the mine, the lesser the attraction.
  5. Apr 3, 2005 #4
    hmm... i think i understand it a little more now. So if the person was in the bottom of a hole that was at half the radius of the earth from the centre of the Earth, then the acceleration will only be GM(underneath)/(r/2)^2?
  6. Apr 3, 2005 #5


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