In march 2009, after the Kepler Space Telescope was put into orbit, an astrophysicist posted an interesting equation.
It's initial purpose was to calculate how much the discovery of exoplanets worth, but in the end, you can basically price any planet with it.
Other than just label planets for sake of "fun", it is more of a way to measure the perceived potential of a world relative to our needs.
At the time, the planet Gliese 581c, thought to be the most Earth-like planet known outside the solar system, was worth merely 160 USD according to the calculation result.
So, how is this made? There are a lot variables here, but nothing we should be worried about, as I'll make sure to provide a spreadsheet for that at the end of his demonstration.
Let's use the Vol system and planet Paart as our inputs, and fill in as we advance.
V = (6*10^6) * (τ/0,5Gyr) * (Mplanet / Mstar)^(1/3)
The Value is the output in dollars, having the Kepler mission cost as baseline (about 600 million USD), divided by the 100 Earth-like planets they were initially expecting to find, ie, they initially evaluated each of those as ~6 million USD.
This is then multiplied by the star's age - divided by half a billion years, this way, older stars get the preference, as they systems and planets are more mature.
Then, multiplied by the ratio of the masses of the planet and the star (both in Msol), to the power of 1/3rd, this favor low mass stars, as the more massive ones live less.
exp - ((log10(Mplanet / MEarth)) / 0,2)^2
This section picks where in a Gaussian distribution this planets are located relative to Earth, the top of the bell-curve represents Earth's mass, this way, it favors planets that physically resemble Earth.
exp - ((Teff - 273) / 30)^2
Another Gaussian distribution, this time, evaluating how much lighting does the planet receive from it's star, this way, we discard terrestrials that are too far away like Mars, or too close to their stars, like Venus.
The 273 is the triple point of water in kelvins (~0ºC) at STP, the division by 30 means the equation will largely devaluate planets too hotter or colder than Earth.
exp - ((Tyr - 2009) / 4)
The next section evaluates how closer to Kepler's mission start was the planet discovered.
The value decreases every 4 years away from the mission start, as over time, is more likely the telescope finds more and more planets alike the one we're calculating.
(2,5^(12 - v))^(1/2)
This part kind of measures how close the star is to Earth - actually it's related to it's apparent magnitude of the star that harbors the planet.
Being V ≃ -26,7 the Sun at Noon as seen from Earth, and V≃ +30 the dimmest far away galaxies HST can see.
This section exists in case for future missions. A star may be particularly dim from Earth, but if you go there and establish an outpost, and later find another habitable planet in this system, the mere fact you're already there already raises the interest over the planet, so it's linked to how bright the star is in the sky.
The only problem with this equation so far is that it can over-evaluate inhospitable / hellish planets like Venus, or under-evaluate Earth-like planets, if we don't have enough data about them.
Filling the required values for Paart, we get about 416,6 million USD in 2020 values, or ~22,2 million if back in 2009.
I included a list of star luminosities and their respective magnitudes, in case you don't know where to convert your star's brightness.
- M.O. Valent, 31/05/2020
