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Author Topic: What would be the radius of a sphere made of many smaller identical spheres?  (Read 2963 times)

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 »


 

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


 

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
 

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... :)
 

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
 

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
 

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