What would be the radius of a sphere made of many smaller identical spheres?

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Offline Maniax101

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Hey,

If i have a sphere of r=1, and take 1000 of those and arrange them into a larger sphere (for example if i let those go in an empty region of space and let gravity do the work) - what will the radius be of the larger sphere?

//Thanks

[MOD EDIT - PLEASE ENSURE THAT POST TITLES ARE FORMATTED AS QUESTIONS IN FUTURE. THANKS. CHRIS]
« Last Edit: 23/06/2010 08:07:21 by chris »

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Offline imatfaal

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The sphere would be roughly over 11 units radius.  This would be much more accurate with more spheres - if we presume that kepler's ratios holds  for such a small number then

1000 spheres unit radius gives total volume of 1000 * r^3 * 4/3 * pi

density of packed spheres is pi/root(18) - keplers ratio

r^3 of conglomerate sphere ≈ [1000 * r^3 * 4/3 * pi] / [pi/root(18)]

r^3 of conglomerate sphere ≈ 1350

r of conglomerate sphere ≈ 11

I cannot be bothered to look at the actual case of 1000 spheres to see if the ratio holds - my gut instinct is that it will not and that actual radius will be much larger as approximation of the larger sphere will be very inexact


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Offline imatfaal

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if you fancy working out the smallest arrangement of 1000 spheres, go ahead.  kepler's ratio is the lowest possible density - all others will be higher, possibly considerably higher.  the more individual components the closer the approximation to a large sphere.  the actual problem requires too much legwork
Thereís no sense in being precise when you donít even know what youíre talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n

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Offline Maniax101

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Thanks.
What got me started is that I have always disliked the notion of singularities.

And just as a personal mindgame, (if we presume that stringtheory is correct) and a string has a length of 10^-33m (and we presume that they are spheres and not one-dimensional)
And that there are about 1*10^57 H-atoms in an average star.
And since electrons are strings, and there are (just by playing roughly with numbers) three quarks in each proton, that would make (just approx.) 4*10^57 strings in a star.
Then if we crush the star to a neutron star, then to a quark star and finally to a string star (where strings are practically touching one eachother, and they according to theory can't be compressed more -
I don't have a good calculator - then how large would that ball o' strings be?

I guess it would be fairly equal to a black hole, but the singularity has been avoided...

Thanks for your time anyways... [:)]

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Offline saruz

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If we make a sphere from the identical small spheres, the volume of small spheres multiplied by number of sphere equals the volume of sphere.
Mathematically,

                      V=nv
                              where,
                                      V=Volume of Big Sphere
                                      v=Volume of Small Sphere
                                      n=number of small Sphere
                    Finally,
                               R^3=n(r^3)
                                 where,
                                      R=Radius of Big Sphere
                                      r=radius of Small Sphere
                                      n=number of small Sphere

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Offline imatfaal

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Saruz - that would work with cubes, but with spheres you are left with little gaps.  A thousand little unit sided cubes could be perfectly packed together to make one larger 10 unit sided cube.  Spheres, even ideally packed, leave gaps; the ratio of packing in an ideal case was as above. This wikipedia page has good graphics showing the two methods that give highest density
http://en.wikipedia.org/wiki/Sphere_packing

A more realistic answer would be the random packing ratio which is even lower
http://mathworld.wolfram.com/RandomClosePacking.html
Only about 60% of the space used is actually filled with small spheres - the rest is gaps.

Matthew
Thereís no sense in being precise when you donít even know what youíre talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n