# How much of an asteroid mass can be in platinum or rare metals?

There are many articles about the 2011 UW-158 asteroid which will pass earth tonight. The interesting thing about this asteroid is that it is said to contain $5.4 trillion worth of platinum. Many articles state that this asteroid is an X-type asteroid and is estimated to have 90-100 million tons of platinum in its core. The asteroid is said to be 300$\times$600m in dimensions, so this amount of platinum would constitute a large fraction of the asteroid mass. There is also something worth noting. The current platinum price is about 32 dollars/gram or 32 million per ton. So the estimated 100 million ton would be worth$3.2\times10^{15}$dollars which is almost 1000 times more than the estimated$5.4 trillion. So there must be something wrong, either the price or the mass, and that's why I ask this question.

Aside from that asteroid, there are obviously some types of asteroids with different metal content as percentage of the total mass. So what is the mass fraction of platinum/rare metals that we should expect in different types of asteroids ?

## 2 Answers

Bad journalism, bad economics, and bad engineering.

Bad journalism:
Someone played telephone with kilograms and metric tons. This kind of bad scientific reporting happens all the time. It's just sad.

Bad economics, part 1:
The current price of platinum is rather depressed. It's much more fun to use the \$1497/troy ounce value of platinum from a year ago than the current \$991/troy ounce. At that higher price, all that is needed is a mere 110 million kilograms to reach 5.4 trillion dollars. (It's about 170 million kilograms at the current price).

Bad economics, part 2:
Suppose there truly was that much platinum in 2011 UW-158. Suppose that it was mineable, and that it could be brought back to Earth. This would make platinum become a non-precious metal. Platinum costs so much precisely because it is so rare. Making it non-rare would drastically reduce it's value, by a much bigger factor than the 1497/991 reduction from a year ago.

Bad economics, part 3 (and bad engineering):
The current value of that rock is the same as the value of the immense amounts of gold and platinum dissolved in the oceans, and the value of the insanely immense amounts of gold and platinum at the center of the Earth: Negative. In all three cases, it costs more to obtain that material than that material is worth. Even if there is a lot of it and even if it's worth a lot, it's worthless if it costs a whole, whole lot to get it.

The simple answer is that the 90 to 100 million tons is wrong. It's kilograms.

Let's assume 300 x 300 x 600 meters, that's a volume of 54 x $10^6$ $m^3$. Note that an early estimate was 500 x 500 x 1000 meters, for a volume of 250 x $10^6$, and that is what should be used.

Assume it is an LL chondrite with a density of 3200 kg/$m^3$, and the mass is 800 x $10^9$ kg. Furthermore, if the abundance of Pt is 100 ppm, then the total platinum mass is 80 million kg, or 80,000 tons. At $32/g, that's a total value of 256 billion dollars. Actually, it gets worse. Here https://journals.uair.arizona.edu/index.php/maps/article/view/15569 is an analysis of nearly 300 chondrites, with a reported abundance in LL chondrites of 1000 ng/g, or 1 ppm. Here www.outofthecradle.net/WordPress/wp-content/uploads/srn_v3n12.pdf is an older paper which suggests 100 ppm both for LL chondrites and stoney meterites (at 99th richness percentile). How did this happen? Well, first keep in mind that all the estimates I've seen ran from 600 billion to 5 trillion, so of course the larger number made the headlines. And the error in confusing tons with kilograms is actually fairly common, although that's not much excuse. The same error, for instance, occurs in line 0087 of Rossi's notorious ecat cold fusion patent application http://appft1.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PG01&p=1&u=/netahtml/PTO/srchnum.html&r=1&f=G&l=50&s1=20110005506.PGNR.&OS=DN/20110005506&RS=DN/20110005506 Furthermore, X-class (a much sexier classification than the old M-class) asteroids have about 3 times the density of LL chondrites, so assuming the same abundance gives 75 billion as the value. Making the (unwarranted) assumption that 2011-UW158 is ten times richer than the "regular" M-class gets us up to 750 billion, and where the other order of magnitude comes from is beyond me. I'd explain it as a really unfortunate intersection of insanely optimistic promoters and lazy journalists. • Please check your stated density for LL chondrite, it's 3200 kg/$m^3\$ and so your calculations need to be corrected. And by the way, using the 100 ppm number, everything will make sense now. Also, the simple answer will be 90 to 100 million kilograms. Thanks. – Abanob Ebrahim Jul 19 '15 at 18:58
• @AbanobEbrahim - Ack! Dang, I do that too often. Thanks, and I've edited. – WhatRoughBeast Jul 19 '15 at 20:05
• You can't assume the same abundance of platinum in X-type asteroids and LL chondrites. M-type asteroids (a subset of X-type asteroids) are differentiated. – David Hammen Jul 19 '15 at 22:12