Firstly, that is legitimately unfortunate...
The number of moons that fit into earth can be determined by (radius earth/radius moon)^3
(6371/1737)^3 ends up to be 49.3... which would round to 1
Secondly, let's go Matt O'Dowd!Question: according to a interesting video I was watching (by PBS spacetime, it's probably worth checking out if your bored) the maximum amount of information (in bits) that can be stored in a system is = surface area (in plank units)/4. Given that the actual information stored in the universe is closer to around 10^90 bits how many times larger is the radius of the observable universe then the radius of the volume of sphere the total information in the observable universe can be stored in?
e90 = pi*r^2 (the four from the SA formula cancels out with the other four) 3e89ish = r^2 r = 45 (rounded up, because of the three) The radius of the observable universe is something like 40 billion light years, because of some reason that I don't remember. so you have 10 fermi light years (almost 11, but since we're rounding down later, we're fine) 16 meters per light year, so 26 meters. (but almost 27) one planck length is 2e-35m, so 5e34 planck lengths per meter. so, we get 26+35 (cuz of rounding up) = 61 planck lengths. 61-45 = [b]16[/b]
so, actually doing the calculation yields a Bekenstein radius of 5.6 e44 planck lengths (uh oh, my rounding will cause issues later...) observable universe is actually 46.6 billion light years, or 2.75e61 planck lengths dividing 2.75e61 by 5.6e44 gives 4.9e16, which just [i]barely[/i] makes my answer of [b]16[/b] correct.
If I made a "1-dimensional" cup tower, whose base stretched from Melbourne to Los Angeles, using 16 oz red Solo cups, how many cups would I need? Assume a flat earth (and confirm the Illuminati while you're at it...), though I'm not sure if it would make a difference.
Bonus question: how much would all these cups cost, in US dollars? (not necessary to answer; may be fun to add to your answer.)