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Pressure inside a sphere

  1. Aug 9, 2008 #1
    I'm reading through Feynman's lectures, and just read his demonstration that:
    a) a sphere generates the same gravity force on a body outside of it as if all its mass was
    concentrated in its center;
    b) on the other hand, on a body placed inside the same sphere there is no gravity force at all.

    I think I understand the proof well, but how does b) agree with the structure of the Earth? I understand that as you go towards the center of the Earth, the density and the pressure increase greatly... but now I realize I don't quite understand why. If a rock inside the Earth at depth 100km and another one at depth 1000km experience the same force of gravity towards the center, where does the increased pressure come from? I suddenly feel like I'm missing something very basic here...

    Thanks in advance!
  2. jcsd
  3. Aug 9, 2008 #2
    A spherical shell, not a uniform sphere.
  4. Aug 9, 2008 #3


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    As you go down into the earth, the gravitational force on you decreases as r (distance from the center of the earth). However, the pressure on you, from the weight of the earth above you would be enormous. Those are two different things.
  5. Aug 10, 2008 #4
    This is correct - inside hollow shell there is no gravity from the shell
  6. Aug 10, 2008 #5
    It is the same reason as why the deeper you dive in water, the higher the pressure you experience, is. The reason is simply that you have a heckload of water being pulled down on you by gravity. The acceleration is approximately the same on all the mass, but the deeper you go into earth, the more of it is being pulled down on top of you. Even if you were just standing in a "hole in the ground" (imagine a sphere of air with just you in it) the rocks on top would push down on the air, forcing the air to push on you. Ergo, an increased pressure.
  7. Aug 10, 2008 #6
    Many thanks for your replies. I think I understand it much better now.
  8. Aug 10, 2008 #7


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    One thing that's counterintuitive is that even though the gravitational force drops off to zero, the cumulative weight continues to rise.

    Think about standing near the centre of the Earth, and pretend the Earth is made of 1 ton boulders all the way down.

    The boulder at zero depth (4000 miles above you) presses down on your with 1ton of weight. The one under that presses down on you with another ton of weight, etc. Now skip 3999.9 miles. The boulders immediately above you, since they are near the centre of the Earth weigh virtually nothing. They contribute little to the total weight ... but the total weight is still huge.

    Basically, you're adding up the weight of the boulders, like this (ultra-simplified):

    Code (Text):
    [FONT="Courier New"]Depth    weight of 1 boulder   total weight
      0 mile        1ton                  1ton
    .01 mile      .99tons              1.99tons
    .02 mile      .98tons              2.97tons
    .03 mile      .97tons              3.93tons
       .            .             .
       .            .             .
       .            .             .
    3999.98 miles .02tons        x zillion tons
    3999.99 miles .01tons        x zillion tons & a bit
    4000    miles  0 ton         x zillion tons  & a bit more
  9. Jan 17, 2010 #8
    If you are at the centre of the earth you also have a whole column on the other side bearing down upon you. Does that mean the confining pressure is doubled?
  10. Jan 17, 2010 #9


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    Wow. I've been waiting here for almost 18 months waiting for someone to come along and post... :smile:

    No. Pressure is pressure.

    If I lie on the ground and place a 5lb. book on my chest, the pressure I experience is 5lb., even though the Earth is pushing up on me with equal force.
  11. Jan 18, 2010 #10
    Dave, thanks, appreciate your reply. To me (being a geo) there are important implications - extremely high pressure being a key if we are going to postulate a solid core at the centre of the earth.
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