Gravity inside hollow sphere?

Discussion in 'Physics & Math' started by Dinosaur, Mar 18, 2018.

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  1. Dinosaur Rational Skeptic Valued Senior Member

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    My not infallible memory tells me that there is no gravitational force due the shell inside a hollow spherical shell with constant density.

    Outside the shell, I think the force acts as if the mass is concentrated at the center.

    An advanced technological culture could build an interesting amusement park environment, allowing folks to experience a large weightless space.
     
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  3. sculptor Valued Senior Member

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    Perhaps?
    However, it seems that the closer you are to an object the stronger would be the gravitational force/pull.
    If so, then you would feel more gravity while close to the shell(and could stroll along it quite comfortably), and only experience weightlessness when at the center of the sphere?
     
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  5. Dinosaur Rational Skeptic Valued Senior Member

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  7. DaveC426913 Valued Senior Member

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    No. Gravity is zero at every point inside the shell.

    Newton's Shell Theorem.
     
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  8. mathman Valued Senior Member

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    Amusement park idea would work only if the sphere itself was in a gravity free environment.
     
  9. exchemist Valued Senior Member

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    ..or free-fall, as in orbit for instance.
     
  10. Q-reeus Banned Valued Senior Member

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    That article follows Newton's original approach working directly with elemental forces. A much quicker and easier approach is via Gauss's law for gravity:
    https://en.wikipedia.org/wiki/Gauss's_law_for_gravity#Spherically_symmetric_mass_distribution
    http://www.sparknotes.com/physics/gravitation/potential/section3/
     
  11. nebel

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    on the alternate theory forum, nebel has a model of the expanding universe through time, where it is represented as a hollow sphere, the interior being the past, and no activity, gravity or electromagnetically induced in the inside.
    The interior of a sphere is also a cage described by Gauss and Faraday, not just keeping gravity at bay.
     
  12. TheFrogger Banned Valued Senior Member

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    A weightless ride would itself weigh nothing. As many people as you like could enter the ride and it would still weigh nothing. A whole society would breed on the ride and it would be akin to Noah's Ark.
     
    Last edited: Mar 30, 2018
  13. nebel

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    Frogs are great jumpers, they still have the same rest mass though, even without gravity*, so: care has to be taken in any movement. Procreation in Noah's Ark? don't give fundamentalists new ideas. Enough hollow spheres there holding bizarre stories like that.
    * during any less than terminal velocity you are in zero gravity, simulating the inside of a big sphere, or time in orbit. If I counted up all the seconds spend in "air" 3m boards, 5-10 meter towers--
     
  14. nebel

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    Dave, you are right- , so: look at the 3 body problem above, and
    Frogger, keep that ark moving fast to counter the aging effect of zero gravity.
     
  15. James R Just this guy, you know? Staff Member

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    If you put that shell on the ground, of course, it wouldn't magically negate the gravity from the Earth below. There'd be no extra gravity from the matter in the shell, but then you'd hardly notice that anyway...
     
  16. loverbal Registered Member

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    Could there be opposing force inside of the shell that stops the shell contracting? Meaning the inner ''walls'' of the shell push against each other to retain spherical form in accordance with Newtons third law by pushing back.


    e.g A football remains inflated because of air pressure

    ƒ: d = ←F→

    Where d is diameter and F is force?
     
  17. Q-reeus Banned Valued Senior Member

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    Of course. Standard expression for biaxial stress in thin shells under pressure here: http://www.wermac.org/pdf/calculation_pressure_vessels.pdf
    [Apologies - just checked and above article link now seems defunct. The same article is currently available here:]
    https://ocw.mit.edu/courses/materia...materials-fall-1999/modules/MIT3_11F99_pv.pdf
    For thin spherical shell, see eqn(1) top of p3.
    Or if you prefer heavier going with more exact maths: http://solidmechanics.org/text/Chapter4_2/Chapter4_2.htm

    Instead of internal gas pressure, you can substitute for gravitational inward 'pressure', once the surface mass density and shell mean radius is specified. There will be a Newtonian 'g' acceleration acting on that aerial mass density - defining the inward 'pressure', thence the shell tangential stress. I'm sure you can work out the specifics from there.
     
    Last edited: Apr 13, 2018
  18. loverbal Registered Member

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    Could we also apply ''solid'' mechanics to fields ?

    Would a spherical ''energy'' shell have likewise polarity, interior forces acting?

    →←
    F. F
     
  19. Q-reeus Banned Valued Senior Member

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    Not sure what 'polarity' means there but in principle a 'shell of field energy' will self gravitate since in GR the source of gravity is stress-energy momentum. The fields will have stress as well as energy density. Big problem - how do you propose to make a stable shell out of 'pure fields'. What kind of fields?
    As far as quantum physics bods are concerned, everything even 'solid matter' is at rock bottom just configurations of quantum fields. But you obviously have something different to ordinary solid matter in mind. Again - how would you seriously propose to obtain mechanical stability?
    [PS: Have just edited #14 re defunct link - updated as shown there]
     
    Last edited: Apr 13, 2018
  20. loverbal Registered Member

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    Maybe I should of said opposite poles.

    I do not wish to propose anything in this forum section.

    I only wish to ask questions and gain knowledge by answers.


    How does an atom have mechanical stability ?

    I assume an energy field would have components like an atom?


    A mono-pole field could not retain stability if all the points of the field were likewise pole points?

    Any given point of a mono-pole field would be repulsive points to other points of the same field?
     
  21. Q-reeus Banned Valued Senior Member

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    Vaguely suggests say a radial E field emanating from uniformly charged conducting shell? You need to be far more specific.
    OK but just keeping asking questions without acknowledging worth of previous answers is not the best way to go. Consolidation bit-by-bit is important.
    Of itself is has quantum mechanical stability. If you wish to go from an isolated atom to real solids, familiarity with some basic solid-state physics is advised.
    A bit word salady. Most physical fields have directionality of some sort. All depends on the specifics.
    Mono-pole field? Again - seems like word-salad. In GR, a static E field for instance has an energy density and triaxial stress state, all of which acts as a source of gravity. But typically the gravitational effects are infinitesimal compared to the EM interactions i.e. Maxwell stresses involved. And you need a source i.e. electric charge which should always be included in the overall interactions picture. Imagination is good but must be accompanied by a grasp of the mathematical & physical basics.
    Sorry must go.
     
  22. exchemist Valued Senior Member

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    I'm baffled by your question about "how" an atom has "mechanical" stability.

    By "how" do you mean "why"?

    And why "mechanical". Do you mean "quantum mechanical"?

    And do you have particular reasons for thinking an atom might be unstable?
     
  23. loverbal Registered Member

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    I mean a binding force , an atom has the binding force of two opposite poles attract then the atom has a strong nuclear force that binds the two opposites together. This gives the atom stability rather than having no stability.

    So for a field to maintain stability , it must be composed of two opposite poles that bind it together ?
     
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