any ongoing theory on anti-gravity cores?

Question

Just as an amateur I wonder what happens at a center of a planet gravity. The standard picture where space curves like a \/ due to gravity or the curve beeing the gravity puzzles me about what do happen at the \/ center to were all the "force" converge. Can it be possible it reflects up to a small distance as anti-gravity? instead of just be confined to a point, just move on? as long it move pass that point it would be pushing not pulling, just because of beeing past the center of gravity. Is there anyone this also makes sense?

Answer 1

You are asking a question which shows the following: that you have thought about an apparent paradox that at first sight follows from the functional form for gravitational attraction that "everybody knows." But it's not so simple when the interiors of planets, stars, or other material bodies are concerned.

Here is how gravity behaves INSIDE a material body. It will necessarily involve a fairly long discussion.

Since you specifically asked about a PLANET'S INTERIOR, it is sufficient to discuss only the "weak-field," or NEWTONIAN, approximation. Even for stars like the Sun, "post-Newtonian gravitational effects" only begin to be of any importance inside them once they reach the white dwarf stage --- and even then, it's relatively small, up to a certain point. (For the Sun the relevant parameter, 2 G M / (c^2 R_*), or R_Schwarzschild/R_* is ~ 10^(-5). For a white dwarf it's ~ 10^(-3) to ~ 0.4x10^(-2); however, for reasons I won't go into, an idealized cold white dwarf will go dynamically unstable at that still apparently quite small parameter value.) In the much denser pre-collapse cores of supernovae, and particularly in neutron stars (parameter value up to ~ 0.2, 0.3) one approaches some quite strong-field effects.

[[I must give you one warning. It is often not appreciated that ALL --- and I mean ALL --- of Newtonian gravitational effects follow SOLELY from the local behaviour of TIME. This is unfortunately obscured by discussions in the popular press, and even in books by professional popularizers, about the "bending of spacetime." However, the separate contributions of "space bending" and "time bending," if I can use that latter term, are as ~ 1 to 10^(7 or 8) in solar system dynamics. The trouble is, it's easy to make a table-top demonstration that really only illustrates the space part; but to illustrate the far larger, dominant time distorting part would mean that we'd have to alter the local rate at which time itself runs!]]

Enough of that sober aside. To help understand what happens back on the terra firma of Newtonian gravitational attraction, it's useful to also think in terms of the "gravitational potential." This is generally denoted by the Greek letter phi, but we can employ another symbol often used in physics for potential in mechanical problems, namely 'V'. I'll break up the discussion into THREE regions:

1. Completely outside a spherically symmetric material body or distributed material region (that is, in empty space), the gravitational potential is given by:

V(r) = - G M / r.

Here G is the Gravitational Constant, M is the mass of the body, and r is the radial distance from the centre.

The local gravitational acceleration, g(r) = - dV/dr = - G M / r^2.

(You can see that this is the traditional formula; the " - " sign indicates that vector g(r) points towards the origin.)

I think that the question you're asking relates to how this functional form, containing 1/r^2, clearly diverges to (- infinity) as r ---> 0.

2. Inside a material body things are different. If there's a diffuse outer region, so that most of M(r) is still inside the position you're considering, both the potential and local g(r) still behave mainly as before. However, once you've passed a fair amount of matter, the potential function more and more assumes the shape of a symmetrical bowl, and the strength of local (g(r) starts getting smaller.

3. Near the very centre, the following happens. Isaac Newton showed that spherically symmetric matter OUTSIDE any position 'r' has NO GRAVITATIONAL EFFECTS ! Only the stuff whose radial distance from the centre is LESS THAN 'r' MATTERS.

So, VERY CLOSE to the centre, M(r) ~ 4 (pi/3) (rho_c) r^3, where 'rho_c' means "central density." Then CENTRALLY:

g(r) = - G M / r^2 ~ - 4 (pi/3) G (rho_c) r + ... (higher order terms),

and V(r) ~ Const. + 2 (pi/3) G (rho_c) r^2 + ... (higher order terms).

What these show is that near the centre, V(r) looks like a shallow bowl with a flat centre. [The purpose of the "Const." term(s) in region(s) of different V(r) dependences is this: they enable ALL forms for V(r) to be smoothly stitched together.] Correspondingly, g(r) goes to zero at the centre, as indeed it must BY SPHERICAL SYMMETRY. (If it didn't, in which direction would it point, at r = 0 ?!)

If you like, you can think of "going past the centre" as proceeding to "negative r." The gravitational acceleration would then change from being "negative" to being "positive," when graphed against 'r' [if the latter variable is now allowed (unconventionally!) to have negative as well as positive values], but it would STILL BE POINTiNG AT THE CENTRE --- which is exactly what it should do, of course.

There aren't "anti-gravity cores" --- this whole, desirable behaviour of the formulae for gravitational acceleration, with its magnitude tending to zero at the centre, follows from considering only the properties of entirely attractive matter.

You thought of quite a sophisticated question. I hope that this rather long, but detailed and complete answer has helped you to understand this.

Live long and prosper.

Answer 2

You have a misunderstanding about how gravitation is related to spacetime curvature. We see in popular print a curved surface with a cone "dip" in it, and this is supposed to be what happens when we have gravitational mass. What people frequently do not understand is that it doesn't matter if it dips down or up, what matters is that local spacetime curvature everywhere in the gravitational field is negative. Non-Euclidean geometries of negative curvature are termed Lobachevskian after the mathematician who studied them, and those of positive curvature are termed Riemannian after another mathematician. In order to have "anti-gravity", one must have POSITIVE local spacetime curvature, and as a matter of fact, that is the reason why the universe is expanding. The avarage large scale curvature of spacetime is positive, like a ball as a 2D analogy. A region of negative curvature would be more like a saddle, not a ball. If you were able to alter the geometry of a local negatively curved spacetime, and make it positive, yes, that would be like anti-gravity, but the equations of General Relativity doesn't make it easy for this to ever happen locally. However, you might want to check the following article, see link:

Answer 3

For every object there is an exact center of mass. Theoretically this center would be located at the precise center of the proton in the single atom at the center of mass. The total gravity of this object would converge FROM ALL DIRECTIONS to meet at the center of the proton. At that single point gravity would be near zero, discounting the gravitational energy from any other masses in the vicinity, such as Earth's moon.