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Habitability (Read 2930 times)
abyssoft
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Habitability
11/01/10 at 22:46:32
 
Given
L = 4π r2σT4
Lsol_w = 3.85 x 1026 W
σ = 5.67E-008 W m-2 K-4
 
and
 
Parameters of
L : in Sols
R : Distance from Star in AU
ε : Axial Tilt in degrees
α : Bond Albedo
 
Result
Teff ≈ ( ( ( L * Lsol_w ) / ( 4 * PI() * ( R * 149598000000 ) ^ 2 * σ ) ) * ( 1 - α ) * ( COS( RADIANS( ε ) ) ) ) ^ 0.25
 
Plugging in For Earth and Sol
Teff ≈  354.04 K (80.89 C)
 
Given that Teff [273.15..373.15] K : Hydro Habitable Zone
Looks to be that Earth is heading towards the inner edge of the HZ. Given that the Suns luminosity increases with time the Earth has less then 2 Gyr left in the HZ.
 
Plugging in For Mars and Sol
Teff ≈  289.77 K (16.62 C)
Looks like Mars is on the outer edge of the HZ. Good news is as the sun ages Mars could potentially become more habitable. although the lack of water is a problem, so delivery of a medium KBO as an impactor would probably be needed to render it more earth like.
 
****
Refinement still needed for axial tilt > 45 deg
range should always be 0..45 deg for the calculation
****
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« Last Edit: 11/02/10 at 08:43:47 by abyssoft »  
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abyssoft
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Re: Habitability
Reply #1 - 11/02/10 at 09:01:21
 
Extending on the previous
 
Lets look at Gliese 581 g (unconfirmed)
 
as both
ε : Axial Tilt in degrees  
α : Bond Albedo  
are as of yet unknown  
lets assume  
α = (Earth's) 0.29,
ε = [0 .. 45]
 

Lsol_w    3.85E+26    3.85E+26
L         0.013       0.013
R         0.14601     0.14601
σ         5.67E-08    5.67E-08
α         0.29        0.29
ε         0.00        45
           
Teff      319.75      293.21
C         46.60       20.06

 
This places the Gl581g either in the middle or the outer edge of the HZ. If they can confirm Gl581g look like we definitely have a potentially habitable on our hands 8)
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Re: Habitability
Reply #2 - 11/02/10 at 11:34:59
 
I don't think you need axial tilt in the equation there, I've never seen it in this equation at least (I guess that calculates the base temperature at the subsolar latitude on the planet?). Here's something I wrote on the SFRPG forums about this:
 
 
The equation below shows how to determine the surface temperature of a planet:  
 
Code:
Ts = 278.66 * { [(4th root of L) * (4th root of 1-A)] / (square root of D) } * GFX 


 
where L is the Luminosity of the star in Sols, D is the distance of the planet from the star in AU, A is the planet's Bond Albedo, and GFX is a Greenhouse Effect factor.
 
The 278.66 is a simplified constant that comes from the original (more complicated) equation. It's essentially a conversion factor since in the equation we use here we turn the distances and luminosities into AU and Sols instead of metres and watts. If anyone is curious it is (4th root of 1/16 * pi *sigma) * [(4th root of 3.846e26) / (SQRT(1.49598e11)] where sigma = stefan-boltzmann constant (5.87e-8), 3.846e26 = luminosity of the sun in watts, 1.49598e11 = distance between Earth and sun (astronomical unit) in metres.  
 
The Bond Albedo tells you the fraction of radiation that is reflected by the planet, so (1-Albedo) is the fraction of radiation that is absorbed (and therefore heating the planet and what we're interested in).  A Blackbody is an idealised perfect absorber of radiation (A=0). A perfect reflector (A=1) would have a temperature of Absolute Zero (0 K) since it's not absorbing anything at all, but nothing natural would perfectly reflect all radiation - the closest we know to that is Saturn's cryovolcanic moon Enceladus, which has a Bond albedo of 0.99 due to all that fresh ice on it (it absorbs only 1% of the incoming radiation). Earth's Bond Albedo is 0.309, Venus' is 0.9 (most incoming radiation is reflected by its clouds. But even the 10% that gets through is enough to raise its temperature significantly due to the atmosphere's massive greenhouse effect).
 
The GFX is a multiplier that increases the surface temperature of the planet. Essentiually, it's a fudge factor and is unfortunately almost impossible to actually calculate, and it just has to be guessed based on the atmosphere type and thickness. Its mininum value is 1.0, which corresponds to a planet with no atmosphere at all (with no atmosphere, the temperature doesn't increase). Earth's greenhouse factor is about 1.13, while Venus' is around 3.0.  
 
There are three types of "temperature" that we can get from this equation, however:
 
Tb (Blackbody Temperature): If we set A=0 and GFX=1, this gives us the Blackbody Temperature - the temperature of a perfect absorber at that distance from the sun. This effectively removes the A and GFX terms, so we get:
 
Code:
Tb = 278.66 * [ (4th root of L)  / (square root of D) ] 


 
 
Te (Effective Temperature): If we set GFX=1 and A=whatever the albedo of the planet is, we get the temperature of the planet's surface without the effect of an atmosphere, which is known as the Effective Temperature (this isn't really used much in practice):  
 
Code:
Te = 278.66 * { [(4th root of L) * (4th root of 1-A)] / (square root of D) } 


 
 
Ts (Surface Temperature): If we use the planet's actual values for A and GFX, we get the actual average Surface Temperature of the planet. For worlds with no atmosphere, Te=Ts.
 
Code:
Ts = 278.66 * { [(4th root of L) * (4th root of 1-A)] / (square root of D) } * GFX 

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abyssoft
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Re: Habitability
Reply #3 - 11/02/10 at 18:24:16
 
Axial Tilt is included as it acts as a moderator of global average temperature.  
 
0 deg = higher global average
45 deg = lower global average
 
this has to do with polar circle re-radiation, afterword this trend reverses again back to 90 deg.
 
inclination of orbit has no bearing on irradiance.
 
if Earth tilt was reduced to 0 the Teff for global not accounting for atmospheric attenuation or greenhouse effect.
 
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Re: Habitability
Reply #4 - 11/03/10 at 10:12:30
 
Quote from abyssoft on 11/02/10 at 18:24:16:
Axial Tilt is included as it acts as a moderator of global average temperature.

0 deg = higher global average
45 deg = lower global average

this has to do with polar circle re-radiation, afterword this trend reverses again back to 90 deg.

inclination of orbit has no bearing on irradiance.

if Earth tilt was reduced to 0 the Teff for global not accounting for atmospheric attenuation or greenhouse effect.


 
Got a link that explains this more? I'm not aware of this "polar circle re-radiation".
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Re: Habitability
Reply #5 - 11/03/10 at 10:26:54
 
If axial tilt may have an influence I also wonder if the rotation rate might have an influence ?
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Re: Habitability
Reply #6 - 11/03/10 at 18:02:09
 
Quote from frankuitaalst on 11/03/10 at 10:26:54:
If axial tilt may have an influence I also wonder if the rotation rate might have an influence ?

 
It does. Take a look at http://en.wikipedia.org/wiki/Effective_temperature , which shows the full equations.
 
I'm actually using:
 
SE = fσ(T^4)
 
S = Stellar flux = (L/4πRČ)
E = Energy Absorption = (1-A), where A is the albedo of the planet.  
F = "Flux Factor" (4 for "fast rotators", 2 for "slow rotators").  
 
 
(L/4πRČ)E = Fσ(T^4)
 
Make T the subject of the equation to get from that to:
 
T = (L(1-A)/4FσRČ)^(0.25)
 
So for "fast rotators" (e.g. Earth), 4F = 16. For "slow rotators" (which i think is where the rotational period is a significant fraction of the orbital period, e.g. Mercury), 4F = 8.  
 
So "slow rotators" should be hotter at a given distance than a "fast rotator", because the planet can absorb more heat over time.  
 
At least, that's what my notes say - unfortunately I don't have a reference for that :/.
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« Last Edit: 11/03/10 at 21:57:16 by EDG »  

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Re: Habitability
Reply #7 - 11/03/10 at 22:00:26
 
So how would the flux factor be modified for a fast rotator and tide locked?
 
and would you need 2 different flux factors to handle the continuously lit versus perpetually dark sides?
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Re: Habitability
Reply #8 - 11/04/10 at 00:26:29
 
That I'm not sure about Smiley. I can't even find anything that mentions this "flux factor" online... though I'm sure I got it from some online article or other.  
 
The dark side could potentially be almost at absolute zero (if there's no atmosphere to trap the heat). If you have a thick enough atmosphere though and sufficient circulation, then it could be considerably warmer due to winds bringing warm air from the dayside. IIRC you don't need that thick of an atmosphere (only about 0.1 atm pressure?) to keep a water ocean liquid on the darkside (and also, water will store more heat than rock due to its high heat capacity).
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Re: Habitability
Reply #9 - 11/06/10 at 23:58:14
 
After much research and examination of different papers on the matter I was able to come up with the following.
 
 
Finding Planetary Temperatures
 
ε※ := Manual Value for bond albedo
Cl := Non-fluid, non-Ice coverage (land) (%)
Cf := fluid coverage need not be water (%)
Ci := ice coverage need not be water ice (%)
Cc := cloud coverage need not be H2O (%)
L := Luminosity of star in L☉
R := Radius in R☉
d := distance between Star and body in AU
x := Greenhouse Effect, Attenuates effectiveness of Stefan-Boltzmann constant as caused by atmosphere thermal retention (%)
     Considered to contain adjustments cause by rotational influence, internal heating, and moderation due to axial tilt.  
     This number can be negative and must be less than 1.000.
 
x      for the Solar System Planets = { -180.00% , 99.61% , 37.1% , 0.00% , 80.3% , 86.6% , 65.7% , 82.4% }  
     for dwarf planets = { 1.8% , ? , ? , ? , ? } Ceres, Pluto, Haumea, Makemake, Eris
     can be expected to run from -300% to 99.99%, but should be assumed to be 0 until can be proved otherwise by measurement
     
 
For gaseous planets Tp is at 1 bar  
 
εl = 0.130
εf = 0.028
εi = 0.752
εc = 0.550
σ = 0.000000056704 W M-2 K-4
E☉ = 3.846 x 10^26 W
π = 3.14159265358979
 
 
ε = « ∃ { Cl , Cf , Ci , Cc } = {∅} ∨ Cl+Cf+Ci+Cc = 0  ⇒ ε※ ⇢ Cl*(1-Cc)*εl+Cf*(1-Cc)*εf+Ci*(1-Cc)*εi+Cc*εc »  
Es = L * E☉
S = Es / ( 4 * π * ( d * 149598000000m ) ^ 2 )
Ts = ( Es  / ( 4 * π * ( R * 695500000m ) ^ 2 * σ ) ) ^ 0.25
Tp = ( S * ( 1 - ε ) / ( 4 * σ * ( 1 - x ) ) ) ^ 0.25
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