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When gravity spacetime is visualized, it's a sag with the lowest point in the middle of the planet/star as you can see here:

For all planets and stars, there is zero G in the core due to that the surrounding mass cancels out each other. This means that the gravity pull is strongest at the surface or close to it? Shouldn't this mean that the gravity spacetime should be visualized as a doughnut/ring instead with lots of gravitational pull at the edge, but not in the middle/core?

HDE 226868
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Unibyte
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    Probably related: http://physics.stackexchange.com/q/149936/, http://physics.stackexchange.com/q/155328/, http://physics.stackexchange.com/q/51198/, http://physics.stackexchange.com/q/185220/, and probably many more. – Kyle Kanos Jan 01 '16 at 20:39
  • @KyleKanos Those are certainly related, but the question is about something specific not covered therein. – garyp Jan 01 '16 at 22:03
  • I'm with @EdYablecki What exactly do you mean by core? The point at the center of the sphere, or all the mass beneath the surface? Or something else? – garyp Jan 01 '16 at 22:05
  • @garyp: I didn't use "possible duplicate" here, in case you were thinking that was the intent. I wrote "probably" to signify that it's not guaranteed and "related" to signify that readers might be interested in those other questions. Nothing more – Kyle Kanos Jan 01 '16 at 22:06
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    That is an awful visualization of the gravitational potential. It looks completely wrong. So would be your doughnut idea, of course. – CuriousOne Jan 01 '16 at 23:06
  • The related post that was marked as a duplicate is not really a duplicate. I was asking about the visualization of spacetime/rubber sheet that is commonly used, and @Amphibio asked if there's zero G in the middle of black holes as well as for planets/stars. – Unibyte Jan 04 '16 at 20:40
  • The visualization feels misleading as @CuriousOne pointed out. Are there better examples? – Unibyte Jan 04 '16 at 21:25

2 Answers2

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  1. On one hand, the graph in OP's picture seems to represent the gravitational potential $U$ of a spherically symmetric planet, which indeed attains its minimum at $r=0$, cf. a potential well.

  2. On the other hand, OP is presumably referring to the magnitude $|\vec{g}|$ of the gravitational field strength $\vec{g}= -\nabla U$ as being donut shaped, because $|\vec{g}|$ vanishes for $r\to 0$ and $r\to \infty$, cf. e.g. this and this Phys.SE posts. This recent Phys.SE question uses the word donut in the same sense. Topologically, the relevant shape is a hollow ball or a thick sphere $S^2\times I$; while a donut usually refers to either a solid torus $S^1\times D^2$, or its surface: the torus $S^1\times S^1$.

Community
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Qmechanic
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  • I would give this a bump if I had the reputation for it, eventough I don't understand the mathematics. And Yes, if you would depict this in not just one plane, as with the spacetime/rubber sheet, it would be a thick sphere. Are there any visualizations of this @Qmechanic ? – Unibyte Jan 04 '16 at 21:46
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As I read your question, I think you are confusing the idea of the "core". The core is considered a point inside the sphere. I am assuming you are looking at the picture and possibly confusing the core as the central mass above the larger depression in the spacetime graphic representation. I don't know if that helps, because my assumption of your assumption can be wrong.

Ed Yablecki
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  • The "duplicate" post mention the same thing, that the footprint on the spacetime/rubber sheet maybe should be a ring/donut, and not that the place with most gravitational pull is in the middle? Because that's how I interpret the image in my question: that the G's increase the closer you get to the center of the object, not the core itself @garyp and Ed. – Unibyte Jan 04 '16 at 21:24