Expansion ? Pulling !

[Naeel]

Hell-o !

 

First of all, sorry for my English, I hope you’ll understand the proposed idea.

 

By Hubble’s law, the universe is expanding, the farther away the observed object, the faster.

Like as some anti-gravitational forces exist.

 

Anti-gravitation means objects with mass are PUSHED away.

 

But – if some masses exist outside of our vision, that PULL ? By “regular” gravitation.

 

Nice to name it as Universe Halo, but same computer game exists, so let’s call this far-far away area Hell-o (you will understand later why).

 

Let's consider a sphere with a density lower than that of its surrounding infinity.

It will behave as massive ball, but with anti-gravitation !

 

Same (with opposite force) as Earth gravitation – maximum on the surface, decreasing down to zero to the centre of the sphere.

 

Here is calculation of anti-gravitational acceleration, r – is distance from “centre” of our universe

 

​

 

Calculation of acceleration by Hubble’s law

 

 

Equation …

 

 

At first I planned to get the value of this density to know is there any sense on this calculations …

 

 

But – it appeared 2 times bigger than critical density !

 

And – it is not Hell-o’s density, it is difference with universe density …

So, dark hell is 3 times more populated, than our lighted heavens.

 

Why “hell” ?

If university’s density is more than critical, it will collapse.

 

How is such future for all of us, what to you think ?

[Janus]

Your Hell-o's mass would not have any net gravitational effect on anything within the visible universe. This is something that Newton proved with the Shell theorem. If we consider the galaxy in your diagram, it will feel no net pull in any direction due to the mass surrounding the visible universe. Look at it this way, If you draw a vertical line through it extending up and down through the "Hell-0", you divide it into two halves, one to the left of the galaxy and one to the right. Now while the galaxy is closer on average to the right half, there is more mass to the left. The result is that the gravity pulling to the left will be exactly equal to the gravity pulling to the right and the galaxy has no tendency to accelerate to the right.

[AbstractDreamer]

Your Hell-o's mass would not have any net gravitational effect on anything within the visible universe. This is something that Newton proved with the Shell theorem. If we consider the galaxy in your diagram, it will feel no net pull in any direction due to the mass surrounding the visible universe. Look at it this way, If you draw a vertical line through it extending up and down through the "Hell-0", you divide it into two halves, one to the left of the galaxy and one to the right. Now while the galaxy is closer on average to the right half, there is more mass to the left. The result is that the gravity pulling to the left will be exactly equal to the gravity pulling to the right and the galaxy has no tendency to accelerate to the right.

 

Whilst there are plenty of things I disagree with OP's model on, I think Shell theorem assumes a lot of things; its a Theorem; it doesn't invalidate OP's model. Straw man fallacy.

 

For instance it's a theory of a perfect symmetrical sphere. Any tiniest deviation from perfection will lead to tiny variance in gravitation potential, which over time will distort the shape of the sphere, leading to greater variance in potential and so on. OP's model is not dependent on the sphere being perfectly spherical.

 

It also assumes symmetrical mass distribution. Whilst the cosmos is isotropic, i don't believe it is perfectly gravitionally isotropic, even within the limits of our event horizon, let alone just our particle horizon.

 

However if the sphere is not perfectly spherical, and assuming the outside forces are acting uniformly "pulling" the universe outwards with an even force in all directions, then expansion "inside" would not be even - indeed there would probably be some areas of contraction and other areas of expansion, although it might conceivably expand eventually into a perfect sphere, once the outside forces reach equilibrium with the "other side", and then Shell theorem would be valid lol.

 

As I've contradicted myself, I'm definitely wrong somewhere.

Edited November 29, 2016 by AbstractDreamer

[Naeel]

Thank you for answering.

 

Shell produces zero gravity inside it's hollow when absolutely symmetrical and same thickness.

 

Let's consider the point on the edge of the cavity.

For it's gravity balance it is necessary to have the same cavity symmetrically located.

If the second cavity is filled, it will have gravity.

Which is equivalent to the fact that the hollow has the property of anti-gravity

 

 

 

 

The same situation if the point is located inside the cavity (not in the centre, of course).

But the cavity will not be whole ball shape, but bitten apple or the crescent

Edited November 29, 2016 by Naeel

[Phi for All]

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Moderator Note

I've hidden a post with another speculative addition. Please don't hijack someone else's thread with speculative guesswork, even if it's a speculative thread.