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Given a large sphere with radius $R$, and given a set of smaller spheres with radius $r$, how many smaller spheres can fit in the larger sphere.

Ignoring the boundaries between spheres, the max number is $(R/r)^3$, but how much space do the gaps take? If the radius of the inner spheres continues to decrease and the quantity rise commensurately, will the surface area of all the inner spheres keep growing?

Some Context

After reading this article: https://www.edge.org/response-detail/27026, particularly this passage:

But physics has a surprise. How much information can you store in a volume of physical space? We learn from the pioneering work of physicists such as Gerard 't Hooft, Leonard Susskind, Jacob Bekenstein, and Stephen Hawking, that the answer depends not on volume but on area. For instance, the amount of information you can store in a sphere of physical space depends only on the area of the sphere. Physical space, like visual space, is holographic.

Consider one implication. Take a sphere that is, say, one meter across. Pack it with six identical spheres that just fit inside. Those six spheres, taken together, have about half the volume of the big sphere, but about 3 percent more area. This means that you can cram more information into six smaller spheres than into one larger sphere that has twice their volume. Now, repeat this procedure with each of the smaller spheres, packing it with six smaller spheres that just fit. And then do this, recursively, a few hundred times. The many tiny spheres that result have an infinitesimal volume, but can hold far more information than the original sphere.

I wanted to know how far this principle could go.

Imran Q
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  • One if $R/r<2$, then 2 if $R/r<5/3$, and for $R/r$ approaching infinity, the number will indeed approach $(R/r)^3$. But could you provide more context for this question? – Mefitico Oct 31 '19 at 20:56
  • Thanks! Could you provide some justification for those numbers? I added context. – Imran Q Oct 31 '19 at 21:09
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    If the smaller sphere is only slightly smaller, then only one will fit inside. As it shrinks, there will be room for a second once they both fit aligned. Now, if they shrink further, a triangle of spheres will fit inside, and the triangle formed by their centers, has its barycenter at the center of the larger sphere. Later, a tetrahedron of spheres will fit inside, and so on. But after this, it gets much more complicated, and I'm doubtful there will be an elementary approach to this problem. – Mefitico Oct 31 '19 at 21:22
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    Packing spheres in a cube: http://www.randomwalk.de/sphere/incube/ – Gerry Myerson Oct 31 '19 at 21:57
  • Thanks - I think this isn’t as simple as I thought – Imran Q Nov 01 '19 at 23:50

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