Let’s say I’m a billionaire and wanted to have an amazing conversation piece in my mansion’s main reception area.
I want to have a meter wide perfect sphere of polished, pure 24 carat gold displayed on a marble slab. Will the weight of the sphere of metal deform it if it is placed on a flat marble surface destroying the perfection of the sphere or will it be OK? If I grind a 30 cm deep matching “cup” into the marble to match the sphere’s shape will it be OK or will it still deform due to the weight?
How much would it cost me to obtain this much gold at current market prices?
Depends on what your definition of “perfect sphere” is. Certainly, at some level, there will be some distortion from perfect sphericity just being in a gravitational field, due to tidal forces.
We’re talking a solid gold sphere? The formula for a sphere’s volume is 4/3 pi r cubed. With a 500 cm radius that comes to 523,583,333 cubic centimeters. Gold weighs 19.3 grams per cubic centimeter so the total weight would be 10,105,158,327 grams. The price for gold today was US$29.89 per gram (EUR 18.92). So the cost of the sphere would be 302,043,182,391 dollars or 191,189,595,545 euros.
See Hertzian contact stress. [url=http://tx.technion.ac.il/~merei02/public/kogut1.pdf]Here is a paper developing a finite element validated model of plastic-elastic contact stress between a sphere and a rigid flat surface. In short, the sphere will deform, and with a material with a low yield point and small elastic modulus like gold, it’ll deform pretty readily. A 1 meter sphere will pretty much squash itself into a flattened ellipsoid pretty quickly. The 30cm cup would help to distribute the load considerably, but I’m pretty sure that the material would still measurably sag out of round under its own weight, although that’s kind of a complicated problem. I’m sure it is possible of analytical solution, but not at this point in the evening.
Hajario has it right. I goofed and used 500 cm as the radius instead of the correct figure of 50 cm. This threw my subsequent calculations off by a factor of a thousand. The correct cost would be a bargain at 302,043,182 dollars or 191,189,596 euros.
OK. Actually, I was wrong at first but only by a factor of 10 which I quickly changed. I thought that we matched after that but I misread what you typed.
Just look for the house that has collapsed because the foundation was smashed by a massive weight.
Me? I used to keep a lump of neutronium in my kitchen, but it was just too much to cart around every time I moved, plus having to recalculate the Earth’s rotational momentum was a real bitch.
would it deform if it were constantly kept spinning at random directions by say… water jets and water being pumped up from beneath to lower the friction?
(I can’t believe I’m entertaining the thought of this)…
And… if so… how long until your water bill (or at least your electric bill if you’re recirculating it in a closed system so that it doesn’t evaporate) out paces the value of the sphere? (what? It would… eventually)
Other than some super conducting magnetic levitation effect (and gold is only very weakly magnetic) I don’t think there is any medium that’s going to “float” a dense ball of gold.
Buy a space station and live on it, therefore making it your house. In the airlock/main reception put lots of padding and let it float around. If you’ve got enough money to buy the ball of gold you’ve got enough to play around with rockets.
I have no cite but I remember reading that all of the gold in the world would fill an Olympic-sized swimming pool and that all of the platinum would fill the pool up to the top of one’s ankles. I can’t confirm the accuracy of that statement…
