Gold (and related topics)

The below is why I seldom discuss politics. This sort of argument doesn't win any battles... but it is fun, so here you go.

As I was driving home last night, I noticed that the moon was up. I wondered why the moon was white. I mean, the Earth is mostly brown. I know a lot of that is organic matter, and if you dried it out and killed all the organics, it would be a sorta dingy grey. Since the moon came from the Earth (possibly from some sort of wayward asteroid), it should be grey, right?

Then I thought that maybe it is. After all, to get good photos of the moon, there's a white balance process... and moon rocks are grey. Maybe it just looks white because our eyes adjust (as they do in such circumstances) and our brain tells us it's white and not just a dingy grey.

That means that the moon isn't nearly as bright as it could be.

You know, like if we gold-plated it.

I've heard that gold can be hammered out incredibly thin. In fact, it cam be hammered such that a nugget 5mm can be spread out to cover half a square meter. If we assume that "a nugget" is a sphere, that means that the nugget has a volume of 65.42 cubic millimeters. Given that the surface area of the moon is 3.793 × 107 km^2, it would only take 75,860,000 nuggets to gold plate it. Of course, that's ridiculous, as we'd only have to plate the side of the moon that faces us. Converting half of 75,860,000 nuggets to cubic meters, we get 2.48 cubic meters.

Yep. Only 2.48 cubic meters of gold is needed to make our side of the moon all shiny! (Awesome!)

Given a density of gold of 19,300 kg/m^2, that's 47,888.21 kg of gold. Unfortunately, at today's rate of $35,028 for a kilogram of gold (in USD), that means it would cost $1,677,428,059.34 in USD.

However, I bet that if we bought the gold in that sort of quantity, we'd get a discount. Let's say we get a fairly standard 15% discount. That makes the cost of gold $1,425,813,850.43. Or 1.4 trillion dollars.

Does that number look a bit familiar?

It should.

It's the US deficit.

For the amount of money we lost under George W. Bush, we could have gold plated the moon!

I'm astonished that we're arguing about health care.

ETA I screwed up the math. Sorry. Geekery that is hopefully more correct is here

OH NOES! I was wrong on the Internet!

Oh dear. fivemack pointed out a small error in my math for my previous post. I was only off by a factor of a thousand or so. (I do find it amusing that I was wrong on the Internet for less than an hour before I was corrected. :) Ah well, I've been wrong before and I'm sure it will happen again... perhaps in this very post. :)

Here's the correct math:

If we convert the surface area of the moon (3.793 × 10^7 km^2) to m^2 and halve it (to just get the bit that faces us), we have a total area of 1.8965 x 10^13 m^2 that needs to be covered. Using the same gold nugget from Wikipedia that's 5mm in diameter, we get 65.42 mm^3 nuggets that can be expanded to half a meter. Thus, we would need 3.79300 × 10^13 nuggets or 2,481,380.6 m^3 of gold. With the density of Gold being 19.30 g/cm^3 or 19,300 kg/m^3, we would need 47,890,645,580 kg of gold. Using the price of $35,028 / kg, we are left with a total cost of $1,667 trillion (USD) for the project.

Oops.

That won't work. Clearly goldplating the moon isn't a workable project. Sadly, gold is one of the most malleable of elements, so perhaps my dreams of a bright shiny moon are lost for ever...

Good thing we know about electroplating. :)

Lots of things can be electroplated. Many of them result in a minimal thickness of 0.1 microns. So we would need a layer that takes up 1,896,500 m^3 (3.793 × 10^7 km^2 times 0.1 microns).

Let's take a quick look at silver.

Silver has a density of 10.49 g/cm^3 or 10,490 kg / m^3. At a cost of $523.8675 / kg, the total cost to cover 1,896,500 m^3 would be $10.42 trillion, which is close, but won't quite get us there.

Another common electroplating element is rhodium. It has very high reflectivity, so would be ideal for this project. Sadly, it's both more dense than silver (12.41 g/cm^3 or 12,410 kg/m^3) and more expensive ($2630/oz or $92,770/kg), so the price to maximize the reflectivity would be $2,183 trillion, which is ridiculous.

The next obvious option is chrome. Sadly, I learned that chrome itself isn't highly reflective. Most of that comes from the nickel layer that's usually underneath it. Even then, the numbers don't quite add up, coming in at $2.717 trillion, which is close, but not quite there... and who wants to half-plate the moon?

So I thought why not look at nickel itself. More specifically, at the high-phosphorous electroless nickel alloy recommended by Anthony J. Covey. (I love the Internet :) This type of alloy ranges all the way up to 13% phosphorous. So we get to break it down. Also, alloys tend to be a bit more brittle, so we're going quadruple the thickness of the coating to 0.4 microns... just to be safe. This raises the total volume of the material needed to 7,586,000 m^3.

The density of nickel is 8.908 g/cm^3 or 8,908 kg/m^3. Thus, the total amount of Nickel needed is 8,908 kg/m^3 * 7,586,000 m^3 * 0.87 (it's 13% phosphorous, remember) or 58,791,196,560 kg. The density of phosphorous is 1.823 g/cm^3 or 1,823 kg/m^3 (I assumed white phosphorous, other forms are slightly denser). Thus, the total mass needed would be 1,823 kg/m^3 * 7,586,000 m^3 * 0.13 or 1,797,806,140 kg.

Cost is hard to determine, as there are lots of different chemical compositions that are used to make this alloy, and it's after 10pm so none of the companies that sell them will give me a quote for 15 billion kg of the stuff. So, we'll just look at the main raw components. Nickel costs $8.2327 / lb or $18.1499966/kg, so 58 billion kg would cost us $1 trillion dollars. Phosphorous comes in at 43.90 € / 500g or $122/kg, so in the quantities we'd need, would cost us $219 billion. Now if we assume that we can find a company that would combine the two for us for a profit of $200 billion (seems likely), we have the magic number of $1.4 trillion.

It may not be as classy as gold plating, and "let's electroplate the moon with a high-phosphorous electroless nickel alloy" doesn't have quite the ring to it that I'd like... but let's face it, it would be a classic American-style solution. :)

It also manages to avoid some of the commodity pricing issues raised by markgritter, as there's a lot more nickel and phosphorous around than gold. ;)

(And before anyone asks, no, I don't have too much time on my hands. I just wanted something irrelevant to temporarily distract my mind from more important concerns... and doing math is more fun than watching TV.)