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Proof of no net gravitational force on a particle inside uniform shell

  1. Dec 7, 2013 #1
    In my book it is written that Newton's shell theorem can be used to show that a uniform shell of matter exerts no net gravitational force on a particle located inside it. How?
     
  2. jcsd
  3. Dec 7, 2013 #2
  4. Dec 7, 2013 #3
    proof?
     
  5. Dec 7, 2013 #4

    jtbell

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    Staff: Mentor

    Last edited: Dec 7, 2013
  6. Dec 7, 2013 #5
    Thanks.
     
  7. Dec 8, 2013 #6
    Its written on wikia :If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass. I don't understand its reason.

    Maybe I don't understand how the spherical shell is divided into infinite shells. Are they shaped like infinitesimally thin rings? If yes then I don't understand the reason of the given statement in bold.
    Help!
     
  8. Dec 8, 2013 #7
    Yes. You have to consider the sphere as infinitesimally number of spherical shell.
     
  9. Dec 8, 2013 #8
    okay.so?
     
  10. Dec 8, 2013 #9
    You have to consider the sphere as infinitesimally number of spherical shell. And if you take one spherical shell then the mass of spherical shell consider to be concentrate at the center of shell.. Like that the mass of all the spherical shell to be concentrate at the center of the shell. And finally you will get the total mass of sphere concentrated at the center of the sphere.
     
  11. Dec 8, 2013 #10
    How?
     
  12. Dec 8, 2013 #11
    because we have considered the sphere as innumerable number of spherical shell, and the center of all shell is at the center of earth
     
  13. Dec 8, 2013 #12
    how? if we take the centers of all the shells they would lie on the diameter.
     
  14. Dec 8, 2013 #13

    Bandersnatch

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    I think this is the source of your confusion. You're thinking of dividing the sphere as if you were slicing a tomato - into very thin rings or discs of increasing(and after half-way through, decreasing) diametre, each ring having its centre lying along the radius of the sphere.

    What is being done, is dividing the sphere as if you were peeling an onion - into very thin layers, each laying on top of the other. Each layer's centre(the point equidistant from all of its constituent points) lies in the centre of the sphere.
     
  15. Dec 8, 2013 #14
    Oh ! Thank You! I finally understand.
     
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