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Calculating planetary temperatures
Posted: 16.08.2003, 20:18
by ajtribick
How do you calculate the temperature of a planet given the luminosity and temperature of its parent star and the orbit radius?
I am planning creating an add-on for the Epsilon Eridani system as described in Alistair Reynolds' Revelation Space series.
Two planets and one moon are mentioned: the planet Yellowstone and its moon Marco's Eye, and a gas giant Tangerine Dream.
I am planning to make the planet currently designated "b" Tangerine Dream.
Now Yellowstone is supposed to be a Titan-like planet, oceans of ammonia and the like, and the books imply that it is closer to Epsilon Eridani than Tangerine Dream. Is there room for a stable orbit at the right temperature in the system?
Posted: 17.08.2003, 02:36
by granthutchison
Take a look at the latter part of this thread
http://www.celestiaproject.net/forum/viewtopic.php?t=2999
for probably as much as you care to know about calculating planetary temperatures.
Glad to see someone else out there is a Reynolds fan ...
Grant
Posted: 17.08.2003, 02:50
by Evil Dr Ganymede
How do you calculate the temperature of a planet given the luminosity and temperature of its parent star and the orbit radius?
It's close enough to:
Blackbody Temperature = (278.3 * (L^0.25))/[SQRT(R)]
Where Temperature is in Kelvin, L is luminosity in sols, and R is distance in AU. That's if it's a perfectly absorbing blackbody.
If you want to take Albedo into account, then multiply this temperature by
[(1-A)^0.25], where A is the albedo of the planet. (if you just want a blackbody temperature, then A = 0. A perfect reflector has A=1, but that won't happen in reality).
This is the same equation as in the thread that Grant linked to, just simplified for use with luminosity in solar units and distance in AU.
Posted: 17.08.2003, 06:02
by I am MackTuesday
Sorry if this is redundant. I started this response before anyone else posted.
The temperature of a planet is (1-A)^(1/4) * (r/2a)^(1/2) * T.
Here, A is the albedo of the planet, r is the radius of the star, a is the average distance from the planet to the star, and T is the temperature of the star. Let's plug in Earth's numbers and see what we get:
(1-0.3)^(1/4) * (700000/300000000)^(1/2) * 5840 = 260 K
This is below the freezing point of water. However, The above equation doesn't take Earth's minor greenhouse effect into account. Atmospheres complicate things quite a bit, unfortunately, but if we say that Earth's atmosphere makes things 325/260 = 1.25 times hotter, it gives us an idea of what a little water vapor and carbon dioxide can do.
How do we get the numbers for some other planet and star? Well, albedo is the fraction of radiation a body reflects. For airless, rocky bodies this number appears to be in the 0.05-0.15 range (Moon=0.12). Ices and gases can drive this number way up. A thick atmosphere will put a planet's albedo somewhere between 0.5 and 0.7 (Venus=0.65), and iceballs have a very wide range depending on the surface composition. They can go as low as about 0.2 (Callisto) or as high as 0.99 (Enceladus).
There are a number of very small moons around the gas giants in our system with albedos less than 0.1. I *think* this means they have very little ice in their composition, but I'm not sure what's going on there.
A fairly simple way to get the radius of a star is with this equation:
R/r = (T/t)^2 * (L/l)^(1/2)
Here, the Rs are radii, Ts are temperatures, and Ls are luminosities. Big R, T, and L are numbers for the star in question, and little r, t, and l are numbers for the sun. Given that the sun's radius is 695000 km, its temperature is 5840 K, and its absolute magnitude is 4.83, we can find the radius of say, Epsilon Eridani. To get the temperature and luminosity data, it's best just to look them up. Epsilon Eridani is a K2 star, which puts its temperature at about 4800 K. Its absolute magnitude is 6.18. (Note that magnitude is a logarithmic luminosity scale.) So its radius turns out to be
(5840/4800)^2 * (100^((4.83-6.18)/5))^(1/2) * 700000 = 560000 km
This is pretty rough, mind you, but it should serve. To have an earthlike atmosphere and earthlike temperature, a planet would then need to be
a = (1-A)^(1/2) * (T/260)^2 * R/2 = 80000000 km
from Epsilon Eridani. This is about 0.53 AU.
For ammonia oceans the temperature must be between 196 and 240 Kelvins if the pressure is 1 atm. Let's divide by 1.25 to get 158 to 192 Kelvins. The average is 175:
0.7^0.5 * (4800/175)^2 * 280000 = 176000000 km or 1.17 AU
BUT...any water on this planet is going to exist as ice unless the pressure is very very high, which it isn't. Water is plenty so this planet is certain to have it somewhere. This means that its albedo will be pretty high, which means it will need to orbit closer than we originally thought. If its albedo is 0.5, it will need to be 1.18 times closer, or 1.0 AU from Epsilon Eridani.
Interesting coincidence. Anyway, assuming Tangerine Dream is out at 3.3 AU, this leaves plenty of room for Yellowstone.
Posted: 17.08.2003, 23:53
by Evil Dr Ganymede
IAMT - where do those equations come from? I've never seen that method before... the ones I posted were derived directly from blackbody calculations.
Posted: 18.08.2003, 01:15
by I am MackTuesday
I think they're basically the same equation. If you consider that radius, luminosity, and temperature are all related, I'll bet you can easily go between the two equations by solving for luminosity and substituting. That constant in your equation probably has the Boltzmann constant and pi and stuff all wrapped up in it. I guess those things cancel out when you go to radius and temperature.
I first saw the equation in one of the appendices of an old book by Don Dixon called _Universe_, copyrighted 1981. A lot of great artwork in that book, by the way. I still have it in a box somewhere. You can also find the equation I used if you Google around.
I've read that you do or have done a lot of world building. Do you know of any algorithms for computing the evolution of a planet's atmospheric and hydrospheric compositions?
Posted: 18.08.2003, 04:08
by Evil Dr Ganymede
I am MackTuesday wrote:I think they're basically the same equation. If you consider that radius, luminosity, and temperature are all related, I'll bet you can easily go between the two equations by solving for luminosity and substituting. That constant in your equation probably has the Boltzmann constant and pi and stuff all wrapped up in it. I guess those things cancel out when you go to radius and temperature.
OK... I wasn't doubting their veracity, I was just curious... I suspected it was just a more convoluted way of saying the same thing
I've read that you do or have done a lot of world building. Do you know of any algorithms for computing the evolution of a planet's atmospheric and hydrospheric compositions?
Eee. That's far too fiddly... depends on all sorts of things - geochemistry, volcanic activity/outgassing, presence/absence of life... I don't know of any
algorithms for it, I think at best one would only be able to qualitatively guess at the evolution.
Posted: 18.08.2003, 14:46
by ajtribick
OK thanks everybody. I'll have to do some further calculations based on all this...
Annoyingly planet "b" (Tangerine Dream) has an eccentricity of 0.608, meaning it swings in to 1.29 AU at perihelion. I suspect this would constrict the location of the planet Yellowstone quite a lot.
Of course, if it has a thick atmosphere that means lots of cloud cover hence higher albedo, right? BTW What's the albedo of Titan, because Yellowstone is supposed to be fairly similar.
Thanks.
Posted: 18.08.2003, 15:51
by Cormoran
Titan's geometric albedo is 0.21, according to this site:
http://www.thehubbletelescope.com/titan.html
I guess this is the albedo averaged over the entire cloud layer. Most sites list the variances in albedo with latitude and in different wavelengths.
Hope this helps
Cormoran