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Thread: Shell theorem and tide

  1. #1
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    Shell theorem and tide

    If you are inside a spherical shell of highly condensed matter, would you be affected by tidal forces at all?
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  2. #2
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    Not from the shell, but from any tide-raising mass outside the shell, yes.

    Grant Hutchison

  3. #3
    The gravity and pressure would affect you more.
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    Quote Originally Posted by The Backroad Astronomer View Post
    The gravity and pressure would affect you more.
    ??

    Could you expand on this statement?
    "I'm planning to live forever. So far, that's working perfectly." Steven Wright

  5. #5
    The more shells you have above more pressure and lesser the gravity. When you are at the exact center the gravity balances in all directions but you have all the mass around you presses inwards and squeezes. If you are near the surface the gravity all the mass between you and the center pulls you in and you don't have much mass creating the squeezing force.
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  6. #6
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    G, Mass, And Gas Problems.

    Quote Originally Posted by The Backroad Astronomer View Post
    The gravity and pressure would affect you more.
    What graviy? The idea of the shell theorem is that the summation of the gravitational attraction of the mass of the shell is zero for any object inside the shell.

    What pressure? The classic shell theorem contains a vacuum. If you fill it with a gas, the gas has mass, and the shell is no longer a shell.

    I get a headache jut thinking about the integrals.

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    I expect this was meant to be a thought experiment involving a hypothetical massive, dense but empty spherical shell. Is that right, Noclevername?

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    Quote Originally Posted by Van Rijn View Post
    I expect this was meant to be a thought experiment involving a hypothetical massive, dense but empty spherical shell. Is that right, Noclevername?
    Yes.
    "I'm planning to live forever. So far, that's working perfectly." Steven Wright

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    Quote Originally Posted by The Backroad Astronomer View Post
    The more shells you have above more pressure and lesser the gravity. When you are at the exact center the gravity balances in all directions but you have all the mass around you presses inwards and squeezes. If you are near the surface the gravity all the mass between you and the center pulls you in and you don't have much mass creating the squeezing force.
    There's nothing in the center, that's why it's a shell and not a solid.
    "I'm planning to live forever. So far, that's working perfectly." Steven Wright

  10. #10
    Sorry but in astronomy, stars and planets are modeled using layer of shells, to determine pressure, temperature, density and other properties, sorry got confused.
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    Quote Originally Posted by Noclevername View Post
    There's nothing in the center, that's why it's a shell and not a solid.
    Yeah, I think he misunderstood the question. I think I understand the question now. And just to clarify a bit further, when you say "tidal forces" you mean whether your feet would be pulled in a different direction than your head, and therefore whether you would be pulled apart?
    As above, so below

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    Quote Originally Posted by Jens View Post
    Yeah, I think he misunderstood the question. I think I understand the question now. And just to clarify a bit further, when you say "tidal forces" you mean whether your feet would be pulled in a different direction than your head, and therefore whether you would be pulled apart?
    Yes, whether there would be any tidal stress on a body anywhere inside the shell.
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    Quote Originally Posted by Noclevername View Post
    Yes, whether there would be any tidal stress on a body anywhere inside the shell.
    In that case, kind of echoing what Grant wrote in the second post, I think, though I don't really understand the mathematics, that Newton showed that an object anywhere within a spherical shell will not feel any gravity, so I think the answer is no, there are no tidal forces.
    As above, so below

  14. #14
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    There are no tidal forces caused by the shell, because the gravitational acceleration due to the shell cancels out everywhere within the shell. That is the shell theorem.

    Grant Hutchison

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