# Where inside a large uniformly dense, symmetrical sphere would its gravity toward the center be the strongest?

Imagine a sphere of uniform density with similar volume and average density of our Earth. There is a bore leading to the center of the sphere from the surface with a scale at regular intervals. At what part of your journey towards the center would the scale read your greatest weight?

There are some caveats to consider. The sphere is in a very isolated region of space and has no angular motion.

I wrote this question down last night while I was having trouble sleeping. My intuition tells me that the surface of the sphere will have the greatest force of gravity. Any distance beyond the surface and the mass of the sphere above you will be reducing your weight. Is this correct?

If however the core is much more massive similar to Earth, one would feel a greater strain to stand upright as they descend.

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The question title does not reflect your question. Are you allowing nontrivial radial density profiles, as your last remark suggests? If so then gravity can be stronger below the surface - just don't call the sphere uniform. – Emilio Pisanty Jan 21 at 20:11
My question is the first paragraph plus the caveats. After the line breaks are my thoughts. – Leonardo Jan 21 at 20:14
@Leonardo My personal suggestion is renaming to reflect that this is about a spherically symmetric distribution of mass, and not a uniform sphere. The strange reality of matter and gravity distribution in rocky planets is a topic that has tangentially come up in several questions and I don't think it adds much to the site to have another untitled discussion of it. – AlanSE Jan 21 at 20:41
Suggestion to the question(v3): Replace the word sphere with e.g. planet, because in mathematics a sphere by definition is just the $2$-dimensional surface of a massive $3$-dimensional ball or planet. – Qmechanic♦ Jan 21 at 21:08

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Sorry to those who feel the question is worded so incorrectly. Apparently I do not understand how to write it correctly but those who answered it understood it just fine. If it is unclear then feel free to edit it yourselves. – Leonardo Jan 22 at 0:10

## 2 Answers

Your intuition is correct, the gravity force will be the largest at the surface. The contribution of a spherical shell will cancel as soon as you are inside this shell (you can prove this by integrating the gravitational force along this shell). This can intuitively be explained by the fact that the mass of this shell will have contributions in all directions (which appear to exactly cancel).

Therefor, as you are descending down the bore, the effective mass attracting you, is only the mass closer to the center than you. This reduces the gravitational pull. The mass depends on the cube of the distance to the center

In addition, you are coming closer to the center of that part of the earth that is attracting you. This increases the force in inverse proportion to the square of your distance from the center. Combining these, the force of gravity is proportional to your distance from the center,

In the edit you describe a much denser core. In that case, it depends on the density distribution what happens. Simple example: Suppose you have a core with certain density and a thick outer shell with zero density, than the largest gravitational pull will occur at the surface of the core. As long as you have spherical symmetry, you can in principle, given the density as a function of radial position, calculate the maximum.

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– AlanSE Jan 21 at 20:28

Correct to the first part, it is strongest at the surface. This follows from the Shell theorem.

It is true that if the core were much denser then gravity would increase as you go down. The critical threshold is if the density of the material you are passing through is 2/3 as dense as the average density of everything under you. So for the Earth you would need an estimate of the densities of various layers.

Wikipedia has a graph showing the gravity of the earth as a function of depth based on a certain density model.

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