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## Too cheap to meter

March 12th, 2009

In 1803, the fledgling United States purchased the Louisiana Territory from France, and thereby entered into what has wound up being one of history’s better real estate deals. Napoleon, as the principle on the sell side, remarked at the time, “This accession of territory affirms forever the power of the United States, and I have given England a maritime rival who sooner or later will humble her pride.” In somewhat typical fashion, the US House of Representatives was slower to grasp the stupendous advantage of the bargain, with Majority Leader John Randolph standing firmly against the purchase. Fortunately, a measure to axe the deal wound up failing by two votes, 59-57.

The Louisiana Purchase price was a (suspiciously spam-like) USD 15 million. For a payment of gold bullion and bonds, the United States obtained the entire western drainage of the Mississippi River. This constitutes ~2 million square miles, or roughly 1% of Earth’s ~200 million square mile total surface. Using the price of gold as a measure of inflation (Gold was USD 19.39 per oz. in 1803) the purchase in today’s currency was thus a mere USD 750 million.

Fast-forwarding two hundred years to the present, similarly good land deals are still to be had — not on Earth, but on potentially habitable terrestrial planets orbiting nearby stars! I think it’s fair to say that the successful launch of the Kepler Mission last weekend can be viewed as the first large-scale extraterrestrial land rush.

Oklo readers are doubtless familiar with the Kepler mission specs. The spacecraft will reside in an Earth-trailing orbit, and, during the 3.5-year mission will monitor ~100,000 main sequence stars with a photometric precision of 20ppm at 6.5h cadence. In all likelihood, it’ll detect of order 100 terrestrial planets. The total mission cost will be of order USD 600 million, remarkably close to the cost of the Louisiana purchase in 2009 dollars.

The advent of Kepler allows us to put meaningful prices on terrestrial extrasolar planets. I think the following valuation formula provides a reasonable start:

where $\tau_{\star}$ is the age of the planet-bearing star, and V is the apparent visual magnitude. Kepler’s best planets are likely going to come in with valuations of order 30 million dollars.

Applying the formula to an exact Earth-analog orbiting Alpha Cen B, the value is boosted to 6.4 billion dollars, which seems to be the right order of magnitude.

And applying the formula to Earth (using the Sun’s apparent visual magnitude) one arrives at a figure close to 5 quadrillion dollars, which is roughly the economic value of Earth (~100x the Earth’s current yearly GDP)…

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1. March 12th, 2009 at 21:06 | #1

In other words, a hypothetical Alf Cen Bb is worth roughly a dollar per person on this planet?

2. March 16th, 2009 at 11:21 | #2

Did you use the Earth’s actual or blackbody temperature for that valuation? The difference is about 50%…

3. March 17th, 2009 at 08:57 | #3

I repeated your Mars number, but I only get 1.4-2.6 quadrillion for Earth, depending on albedo. In fact, I calculate the maximum possible for our solar system to be 2.7 e15 dollars:
9.1 for the age (using CAI U/Pb value of 4.567 Ga)
5e7 for the visual magnitude (-26.7 from your blog)
And all the exponent components fixed to 1.

4. April 17th, 2009 at 21:05 | #4

I’m extremely interested in the derivation for this equation.

5. December 3rd, 2009 at 05:39 | #5

Hi Greg,

sorry for this out-of-date comment, but I it was impossible not to think about this post after reading today’s “dinosaur comic”. If you have 1 minute to spare, check this web-comic where three dinosaurs discuss “how much our planet (and life) is worth”:

http://www.qwantz.com/index.php?comic=1608

cheers,

Jose