Where is most "gravity", inside or out?

Discussion in 'Pseudoscience' started by nebel, Feb 29, 2016.

  1. nebel

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    well, others of course help me to learn, work over my efforts. here is the latest verified graph:

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    of course the redline level might be wrong if Gravitons have mass, and as a consequence the field strength g would increase with r. The level red line climb instead as a consequence. of that.
    Gravity strength compounded. with distance
     
    Last edited: Jan 28, 2020
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  3. DaveC426913 Valued Senior Member

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    No, you expect that others will teach you:
     
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  5. nebel

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    nebel, what learning material have you provided? here:
     

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  7. nebel

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    Dave, you did not actually say that, but provided clarity, below is news of a galaxy-size example of an inner empty space, or area. It would be interesting if we could observe velocities of objects near the center, (provided there is no lonely black hole there), at the inside of the "doughnut" wall, or the dime dilations at different locations.

    "ARC Centre of Excellence for All Sky Astrophysics in 3D (ASTRO 3D). "Cosmic Ring of Fire' 11 Billion Years Ago: How did structures in early universe form? Unusual galaxy set to prompt rethink on how structures in the universe form." ScienceDaily. ScienceDaily, 25 May 2020. <www.sciencedaily.com/releases/2020/05/200525115657.htm>."
     
    Last edited: May 25, 2020
  8. DaveC426913 Valued Senior Member

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    Here's the thing: if gravity behaved as you suppose, there wouldn't be any objects at the centre.
    According to your idea, gravity there is zero - which means any object that happens to be located there won't stay there - there's nothing to deflect its path back to the centre. Anything with motion will simply pass straight through.

    Your idea is wrong, and you don't know enough to form ideas of your own. Let it go and find something useful to learn.
     
  9. nebel

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    You are right about that. that is my idea too, if the cavity or the doughnut hole/area is truly empty. there is no gravity well in the center, as would be pictured by the depressing ball model
    Since in the inside of a perfect cage or ring like that, the opposing forces balance, per m/d^2, no force is felt when the object moves through, or away from the center, right up to the inner wall. Through inertia alone it will keep moving, so: Imagine
    A galaxy size inner space trampoline, acrobats freely floating, bouncing off the inner wall or this "ring of fire" . A class act for the Ringling Brothers.
    Different from a hole through the center of the Earth, or any other body, where the object would keep oscillating until it settles in the center of mass. (which is the only point [compared to the whole inner space] where the field forces balance).
     
    Last edited: May 26, 2020
  10. nebel

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    2,469
    you underestimate me, during my 9 decades of moving through time, I had several certified seminal ideas, one paying off in 6 figure rewards.
     
    Last edited: May 26, 2020
  11. river

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    What of the magnetic field though , its part in all this .
     
  12. DaveC426913 Valued Senior Member

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    No.

    Go do some reading. Learn some science.
    You have to know where the box is before you can try to think outside of it.
     
  13. nebel

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    The shape of the box is constantly changing. what was previously the perimeter of the box can easily become part of the confirmed inside.
    limiting ourself to "in the box thinking" has severely handicapped some before.
    re ' RING OF FIRE GALAXY: Rings are rare, because truly empty areas/ spaces, their stability, are hard to maintain.
     
  14. nebel

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    2,469
    Take the case of this Ring of fire galaxy, It could have a center of gravity with no central mass, no gravity in the center, formed as a rotating vortex, in an early universe that is now proposed to have had a rotation. Gravity not being the cause of the structure's permanent rotation velocities.
    Even in origin's solid body in post #2, endorsed by wiki, there is no gravity in the center of gravity. but here in the torus, you have not only a center of gravity without gravity, even a center of mass without mass.
    not only more gravity on the outside than inside but sometimes, mass too.
     
    Last edited: Jun 9, 2020
  15. DaveC426913 Valued Senior Member

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    There is gravity, it just happens that there'e no net gravitational potential.

    Consider the equivalent magnetic situation. You could have a hollow sphere that is positively charged. Drop an electron in the middle. There is not net pull in a particular direction, but that electron is definitely in a positive field.
     
  16. nebel

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    2,469
    so: with the effects of the opposing gravitational fields cancelled, undetectable, on the inside
    (you floating in space), what are the effects that have been proven to be un -effected by that overlapping cancellation? and:

    About the Gauss/Faraday cage analogy to gravity inside a torus, shell, ---- the cage has a real, specific shield, nothing shields against gravity. or?
     
    Last edited: Jun 9, 2020
  17. DaveC426913 Valued Senior Member

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    nebel: Here is a diagram of your idea of the potential outside and inside a hollow sphere.


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    R is a rocketship.
    A is a point at infinity (i.e entirely outside the gravity well of this system)
    B is a point anywhere inside the shell.
    Y is the energy potential at any given point.

    In order for R to get to A, it has to climb up out of the well - and requires a certain amount of energy. The amount of energy required is well-known - it is simply the escape velocity of the system. i.e. If the rocket can achieve escape velocity of the system, it bcan escape and continue forever, without ever falling back to the system, eventually reaching infinity (in theory). That would require Y energy.

    This is basic.



    But what happens in order for R to get to B?

    According to your idea, the centre of the hollow has the same potential (Y) as at infinity, and therefore requires the same amount of energy to reach.

    To repeat, in order for the rocket to go from R to B, it must expend the same amount of energy as is required to get it to escape to infinity.

    And that means, that, if the rocket were to try to enter a the hollow sphere, it would encounter an invisible wall - a wall so strong that it would need to expend an amount of fuel (Y) equivalent to escape velocity - in order to move just to move one more inch.

    That is what the vertical line in the graph means. It means an infinite gradient of gravity. An invisible wall. A rocket trying to enter the sphere would smash into ... empty space ... and stop dead.
     
    Last edited: Jun 10, 2020
  18. DaveC426913 Valued Senior Member

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    Now look at this (correct) diagram:

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    Rocket R loses a little potential energy (i.e. falls) to reach B (exactly as we'd expect), and once at B, requires no more fuel to drift through the centre of the sphere on inertia alone - (exactly as we'd expect in a hollow sphere).

    Note the fact that, once at the bottom of the well - the rocket requires energy equal to Y to escape the system - i.e. it needs to achieve its escape velocity (exactly as we see rockets do now).
     
  19. nebel

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    2,469
    DC426913, thank you for those renderings. wish I could reply in kind.
    1) Could we agree, that the first one, # 674 is correct in showing the strength of the gravitational field (or energy required to overcome it) in the empty inside of the shell? because it has to be at the same level as at infinity, O. and not at Y. Origin in post #2 showed that there is zero gravity at the center point, your diagram #674 correctly shows zero gravity at the center area (for a ring) or center volume for an empty shell. (slight possibility of confusion here because line of zero gravity is reversed top to bottom)
    2) Your vertical lines show that the walls of the shell although having no thickness, have all the mass equivalent to the solid interior of origin #2. Of course in that case there would have to be an instant jump from zero gravity to full surface strength. for gravity in the shell theorem is only a surface & outside phenomenon. ( Consider picture #2 interior a stack of ever bigger shells. )
    3) The departure of #R from your models# 674 is the same as from Origin's #2, because in the case of a shell full of matter, starting from zero at the center point , the ship has to fight it's way up against the linearly increasing gravity that peaks at the surface, the same as inside the much steeper g gradient in the thin walls, as you correctly stated.
    let it be resolved, in the end, there is more overall gravity field strength outside than the inside, empty or not. or?
     
    Last edited: Jun 11, 2020
  20. DaveC426913 Valued Senior Member

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    No, we cannot. Do you understand what the Y axis on this graph is?
    It is not net gravity. It is gravitational potential.
    At infinity, its value is zero and at the bottom of the well, its negative Y.

    Here - I've made it a little more intuitive for you.

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    I've also given the shell a non-zero thickness (your "gravity wall" is not vertical).

    So:

    To rise out of the well, rocket R must burn fuel, increasing its potential energy (climbing up the graph).

    If it turns its engine off, it will fall back into the well - which trades potential energy for kinetic energy as it falls (gaining speed).

    According to your (flawed) diagram, the rocket would need to burn fuel to go into the shell (rise out of the well), increasing its potential energy.

    If the rocket were inside the shell, coasting toward the wall, it would suddenly fall down a very steep slope, converting its potential energy to kinetic energy. If your diagram were correct, the rocket would come out of the hollow and be instantly accelerated to escape velocity (by converting all that potential energy to kinetic energy).

    Essentially, your graph turns any hollow shell into a rocket launcher that uses no energy.




    So let's correct a few of those misunderstandings before moving to anything more complex.
     
    Last edited: Jun 11, 2020
  21. Halc Registered Senior Member

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    What I think you mean by strength is the acceleration field at that point (a vector that you're graphing as a scalar), and (the magnitude of) it is indeed minimal inside a shell. Energy required to overcome it is the potential, and that is maximally negative and flat as Dave has correctly drawn it. You seem unable to separate the two concepts. This was pointed out in the very first reply to the OP, and 675 posts later you still haven't grasped this.

    No. It takes energy to get from inside the shell to infinity, but no fuel is needed to do it in the other direction.
    Doesn't work for rings, only shells.
     
    Last edited: Jun 11, 2020
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  22. nebel

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    2,469
    This is basically the same as my graph of post #664. and just as post #2, too, it shows the central local strength to be zero, at the level of the x axis. zero gravity whatever, at the center, not again as you reach infinite distance.
    The deep (or high in #2) well is not in the center, but outside of the wall surface, a circular trench really rather than a single negative y point/area.
    All along I have been talking about the strength of the gravitational field, zero in the center, zero at infinite distance. My image in# 664 should have been reversed to match that of origin in #2. so:
    i stand corrected, your graph at #675 for work required as you move off the center is correct, no work required until you hit the inner wall, and too infinity, unless you reach terminal velocity, then you float, like in the center cavity.
    thank you for your patience.
     
  23. nebel

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    2,469
    Is not a shell sometimes treated as a stack or rotation of many rings.?
    agreed the picture of "gravity" would be different along the plane of a ring/ torus than vertical to the center hole.
     

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