A brown dwarf is an object made mostly of hydrogen and helium that is more massive than a gas giant planet, but less massive than a red dwarf star. The range of masses for brown dwarf runs from 13 Jupiter masses to about 75 Jupiter masses. One Jupiter mass is equal to 1.8986e+27 kilograms. One solar mass is equal to 1047.3486 Jupiter masses, or 1.9885e+30 kilograms.
Let n = the mass of a brown dwarf in Jupiter masses.
Then the mass of the brown dwarf in solar masses is
m = n/1047.3486
And the mass of the brown dwarf in kilograms is
M = 1.9885e+30 (m)
Let's use, as an example, a brown dwarf of 70 Jupiter masses.
n = 70 Jupiter masses
m = 0.0668354 solar masses
M = 1.32902e29 kilograms
The radius of a brown dwarf, aged 3 billion years or more, in Jupiter radii can be found from these polynomial curve-fits.
if 0.013≤m≤0.070 then
r = 1.066309 − 2.090367 m − 226.4943 m² + 4694.703 m³ − 27097.9 m⁴
if 0.070≤m≤0.100 then
r = −556.0091 + 34499.87 m − 848679.3 m² + 10356770 m³ − 62693940 m⁴ + 150666700 m⁵
With m=0.0668354,
r = 0.7757601
One Jupiter radius is equal to 71492000 meters. So the brown dwarf radius in meters is
R = 71492000 (r)
R = 55460642 meters
Brown dwarfs are puffy for about their first three billion years, after which they settle down to their permanent size. So let us set the age, t, of a brown dwarf equal to 3.5 billion years. (For t, we will use units of billions of years.)
t = 3.5
There's a variable called the Rosseland mean opacity, κ, which has to do with how well electromagnetic radiation can penetrate the material near the brown dwarf's photosphere. We will assume that
κ=0.02 cm²/gram
Then the effective temperature of the brown dwarf is found from [Reference: Burrows & Liebert (1993)]
T = (58.74 K) t⁻⁰·³²⁴ n⁰·⁸²⁷ (κ/0.01)⁰·⁰⁸⁸
Where
t = 3.5
n = 70
κ/0.01 = 2
In which case,
T = 1396.5 K
And the brown dwarf's bolometric luminosity becomes
L = 4πσR²T⁴
where the Stefan-Boltzmann constant, σ = 5.670374419e-8 W m⁻² K⁻⁴.
With R=55460642 meters and T=1396.5 K,
L = 8.3360794e+21 watts
We pick an equilibrium temperature, T₀, and a Bond albedo, A, suitable for a habitable planet.
T₀ = 250 K
A = 0.3
Then the orbital radius, d, for the planet is found from
d = √[ L(1−A) / (πσ) ] / (4T₀²)
Given the parameters in our example,
d = 723950925 meters
The sidereal orbital period of the planet, in hours, would be
P = (π/1800) √[ (d³ / (GM) ]
Where
G = 6.6743e-11 m³ kg⁻¹ sec⁻²
With M=1.32902e29 kilograms and d=723950925 meters,
P = 11.4149 hours
The brown dwarf’s average density in kilograms per cubic meter is
ρ = 3M/(4πR³)
With M=1.32902e29 kilograms and R=55460642 meters,
ρ = 185990 kilograms per cubic meter
The Roche limit, D, of the brown dwarf with regard to a planet having an average density of 5000 kilograms per cubic meter is
D = 2.46 R ∛(ρ/5000)
With R=55460642 meters and ρ=185990 kilograms per cubic meter,
D = 455434975 meters
So our planet orbits the brown dwarf at a distance that is 1.5895813 times greater than the Roche limit.
So, yes. There is a dynamically stable habitable zone for a planet orbiting a brown dwarf. Brown dwarfs cool as they age, but the heavier brown dwarfs (n>50) do so slowly enough that a planet might remain, or be kept, habitable for several billion years.
Of course, a planet here would be tidally locked to the brown dwarf. However, this wouldn't present any problem for a space station. Space stations in such orbits can increase the amount of time they can enjoy habitable conditions by adjusting their albedos. White acrylic paint has an albedo of 0.8, while black acrylic paint has an albedo of 0.05, providing a simple and low-tech method for extending their thermal range by a factor of 1.4763.
https://jenab6.livejournal.com/photo/al ... &id=177223
Habitable zones for objects in orbit around brown dwarfs
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David_Sims
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Will Williams
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Re: Habitable zones for objects in orbit around brown dwarfs
Hmmm? The brown one sounds like it would be a nice place to visit but I don't think I'd like to live there. 
If Whites insist on participating in "social media," do so on ours, not (((theirs))). Like us on WhiteBiocentrism.com; follow us on NationalVanguard.org. ᛉ
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Jim Mathias
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Re: Habitable zones for objects in orbit around brown dwarfs
The potential to reach such places to get a closer look at the conditions for habitability can be developed, so long as we Whites make this world safe for White habitation first. It's well that Mr. Sims has thoughts well into the future, though.
Activism materials available! ===> Contact me via PM to obtain quantities of the "Send Them Back", "NA Health Warning #1 +#2+#3" stickers, and any fliers listed in the Alliance website's flier webpage.
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David_Sims
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Re: Habitable zones for objects in orbit around brown dwarfs
A question from Quora:
"Do brown dwarfs have a habitable zone?"
For a while they do.
Brown dwarfs have a Rosseland mean opacity of about 0.02 cm²/g at their photospheres.
Brown dwarfs range in mass from 13 to 74 Jupiters’ worth of mass.
The more massive brown dwarfs work best for habitable zones. So let’s examine a brown dwarf of 70 Jupiters’ mass. That’s the same as 0.0668354 solar masses or 1.32902e+29 kilograms.
A 70 Jupiter mass brown dwarf has a radius of 0.775760 Jupiter radii, or 55461 kilometers. Its average density is 185990 kg/m³. Its Roche limit (versus a satellite having a density of 5000 kg/m³) is 455435 kilometers.
At an age of 3.5 billion years, a 70 Jupiter mass brown dwarf has an effective temperature of 1396.5 K and a bolometric luminosity of 8.33608e+21 watts.
A space station having a Bond albedo of 0.5 (matte surfaced metal) would have an equilibrium temperature of 255K (which is the same as Earth’s equilibrium temperature) at an orbital distance of 588091 kilometers, which is 1.2913 times the Roche limit. Its orbital period would be 8.3575 hours.
A space station having a Bond albedo of 0.5 would have an equilibrium temperature of 210K (which is the same as Mars’ equilibrium temperature) at an orbital distance of 867135 kilometers, which is 1.9040 times the Roche limit. Its orbital period would be 14.9637 hours.
… time passes …
At an age of 5.19 billion years, a 70 Jupiter mass brown dwarf has an effective temperature of 1229.2 K and a bolometric luminosity of 5.00285e+21 watts.
A space station having a Bond albedo of 0.5 would have an equilibrium temperature of 255K at an orbital distance of 455588 kilometers, which is 1.0003 times the Roche limit. The space station’s orbital period would be 5.6986 hours.
A space station having a Bond albedo of 0.5 would have an equilibrium temperature of 210K at an orbital distance of 671760 kilometers, which is 1.4750 times the Roche limit. Its orbital period would be 10.2030 hours.
The Roche limit will soon begin eating away at the inner edge of the brown dwarf’s habitable zone! Something must be done!
The space station dwellers grab some black paint and go out to paint the outside surface of their space station with a coating that absorbs infrared light very well, so that its Bond albedo becomes 0.05. Then they go back inside. The space station warms up because it is absorbing more of the brown dwarf’s (mostly infrared) radiation.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 255K at an orbital distance of 627984 kilometers, which is 1.3789 times the Roche limit. The space station’s orbital period would be 9.2221 hours.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 210K at an orbital distance of 925957 kilometers, which is 2.0331 times the Roche limit. Its orbital period would be 16.5118 hours.
… time passes …
At an age of 8.52 billion years, a 70 Jupiter mass brown dwarf has an effective temperature of 1046.8 K and a bolometric luminosity of 2.63162e+21 watts.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 255K at an orbital distance of 455463 kilometers, which is 1.0001 times the Roche limit. The space station’s orbital period would be 5.69623 hours.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 210K at an orbital distance of 671575 kilometers, which is 1.4746 times the Roche limit. Its orbital period would be 10.1988 hours.
Again, the brown dwarf’s Roche limit has begun to gobble up the inner (warmer) end of the habitable zone. But the space station can’t reduce its albedo any further, and it wouldn’t matter much if it could. From here on, the part of the habitable zone that remains outside the Roche limit will become smaller and, on average, colder.
But, notice that the space station people were able to get 5.02 billion years’ worth of use out of the brown dwarf, with one timely modification of their home, before any of its habitable zone went below the Roche limit. And they probably could drag things out for another five billion years, if they could get used to the cold. Warm clothes, maybe?
… time passes …
Probably, sometime around when the brown dwarf is aged 15 billion years, the orbital distance of a space station having a Bond albedo of 0.05 and an equilibrium temperature of 210K will slip below the brown dwarf’s Roche limit, and everybody freezes to death. The end.
"Do brown dwarfs have a habitable zone?"
For a while they do.
Brown dwarfs have a Rosseland mean opacity of about 0.02 cm²/g at their photospheres.
Brown dwarfs range in mass from 13 to 74 Jupiters’ worth of mass.
The more massive brown dwarfs work best for habitable zones. So let’s examine a brown dwarf of 70 Jupiters’ mass. That’s the same as 0.0668354 solar masses or 1.32902e+29 kilograms.
A 70 Jupiter mass brown dwarf has a radius of 0.775760 Jupiter radii, or 55461 kilometers. Its average density is 185990 kg/m³. Its Roche limit (versus a satellite having a density of 5000 kg/m³) is 455435 kilometers.
At an age of 3.5 billion years, a 70 Jupiter mass brown dwarf has an effective temperature of 1396.5 K and a bolometric luminosity of 8.33608e+21 watts.
A space station having a Bond albedo of 0.5 (matte surfaced metal) would have an equilibrium temperature of 255K (which is the same as Earth’s equilibrium temperature) at an orbital distance of 588091 kilometers, which is 1.2913 times the Roche limit. Its orbital period would be 8.3575 hours.
A space station having a Bond albedo of 0.5 would have an equilibrium temperature of 210K (which is the same as Mars’ equilibrium temperature) at an orbital distance of 867135 kilometers, which is 1.9040 times the Roche limit. Its orbital period would be 14.9637 hours.
… time passes …
At an age of 5.19 billion years, a 70 Jupiter mass brown dwarf has an effective temperature of 1229.2 K and a bolometric luminosity of 5.00285e+21 watts.
A space station having a Bond albedo of 0.5 would have an equilibrium temperature of 255K at an orbital distance of 455588 kilometers, which is 1.0003 times the Roche limit. The space station’s orbital period would be 5.6986 hours.
A space station having a Bond albedo of 0.5 would have an equilibrium temperature of 210K at an orbital distance of 671760 kilometers, which is 1.4750 times the Roche limit. Its orbital period would be 10.2030 hours.
The Roche limit will soon begin eating away at the inner edge of the brown dwarf’s habitable zone! Something must be done!
The space station dwellers grab some black paint and go out to paint the outside surface of their space station with a coating that absorbs infrared light very well, so that its Bond albedo becomes 0.05. Then they go back inside. The space station warms up because it is absorbing more of the brown dwarf’s (mostly infrared) radiation.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 255K at an orbital distance of 627984 kilometers, which is 1.3789 times the Roche limit. The space station’s orbital period would be 9.2221 hours.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 210K at an orbital distance of 925957 kilometers, which is 2.0331 times the Roche limit. Its orbital period would be 16.5118 hours.
… time passes …
At an age of 8.52 billion years, a 70 Jupiter mass brown dwarf has an effective temperature of 1046.8 K and a bolometric luminosity of 2.63162e+21 watts.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 255K at an orbital distance of 455463 kilometers, which is 1.0001 times the Roche limit. The space station’s orbital period would be 5.69623 hours.
A space station having a Bond albedo of 0.05 would have an equilibrium temperature of 210K at an orbital distance of 671575 kilometers, which is 1.4746 times the Roche limit. Its orbital period would be 10.1988 hours.
Again, the brown dwarf’s Roche limit has begun to gobble up the inner (warmer) end of the habitable zone. But the space station can’t reduce its albedo any further, and it wouldn’t matter much if it could. From here on, the part of the habitable zone that remains outside the Roche limit will become smaller and, on average, colder.
But, notice that the space station people were able to get 5.02 billion years’ worth of use out of the brown dwarf, with one timely modification of their home, before any of its habitable zone went below the Roche limit. And they probably could drag things out for another five billion years, if they could get used to the cold. Warm clothes, maybe?
… time passes …
Probably, sometime around when the brown dwarf is aged 15 billion years, the orbital distance of a space station having a Bond albedo of 0.05 and an equilibrium temperature of 210K will slip below the brown dwarf’s Roche limit, and everybody freezes to death. The end.
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David_Sims
- Posts: 9
- Joined: Sun Apr 07, 2024 9:39 pm
Re: Habitable zones for objects in orbit around brown dwarfs
White dwarfs can have habitable zones, too!
A question from Quora:
"Could a white dwarf have a habitable zone?"
For a while, it does.
Let’s consider a white dwarf having a mass of 0.54 solar masses. That’s the mass that the sun is expected to have, after it becomes a white dwarf.
A 0.54 solar mass white dwarf has a radius of 0.013245 solar radii, an average density of 3.2751e+8 kg/m³, and a Roche limit (versus a satellite having an average density of 5000 kg/m³) of 913879 kilometers.
When the white dwarf is aged 2.282 billion years, it will have an effective temperature of 6000K, meaning that its spectrum of light will have become comfortable to humans. It will have a bolometric luminosity of 7.8433e+22 watts.
A space station having a Bond albedo of 0.05 (painted black on the outside) will have an equilibrium temperature of 255K (the same as Earth’s equilibrium temperature) at an orbital distance of 2486515 km, which is 2.7208 times the Roche limit. Its orbital period will be 25.562 hours.
A space station having a Bond albedo of 0.05 will have an equilibrium temperature of 210K (the same as Mars’ equilibrium temperature) at an orbital distance of 3666341 km, which is 4.0118 times the Roche limit. Its orbital period will be 45.768 hours.
… time passes …
At a WD age of 11.059 billion years, the white dwarf will have cooled to 3000K and will have a bolometric luminosity of 4.9021e+21 watts.
A space station having a Bond albedo of 0.05 will have an equilibrium temperature of 255K at an orbital distance of 621628.807328 km, which is only 0.68021 times the Roche limit. If the station could safely orbit the white dwarf at this distance, then it would have a sidereal period of 3.1953 hours.
A space station having a Bond albedo of 0.05 will have an equilibrium temperature of 210K at an orbital distance of 916585 km, which is about equal to the Roche limit. Its orbital period would be 5.7210 hours.
Notice that the amount of time elapsed between the onset of a comfortable spectrum of light from the 0.54 solar mass white dwarf, until the habitable zone around the white dwarf fell below the Roche limit, is 8.777 billion years. That’s more time than Earth will have been a habitable planet while the sun was a main sequence star.
Now, generally the higher mass white dwarfs offer less time for the existence of a dynamically stable orbit within their habitable zones. A white dwarf of 1 solar mass, for example, would have such orbits for only about two billion years.
See also my answer on habitable zones around brown dwarfs.
A question from Quora:
"Could a white dwarf have a habitable zone?"
For a while, it does.
Let’s consider a white dwarf having a mass of 0.54 solar masses. That’s the mass that the sun is expected to have, after it becomes a white dwarf.
A 0.54 solar mass white dwarf has a radius of 0.013245 solar radii, an average density of 3.2751e+8 kg/m³, and a Roche limit (versus a satellite having an average density of 5000 kg/m³) of 913879 kilometers.
When the white dwarf is aged 2.282 billion years, it will have an effective temperature of 6000K, meaning that its spectrum of light will have become comfortable to humans. It will have a bolometric luminosity of 7.8433e+22 watts.
A space station having a Bond albedo of 0.05 (painted black on the outside) will have an equilibrium temperature of 255K (the same as Earth’s equilibrium temperature) at an orbital distance of 2486515 km, which is 2.7208 times the Roche limit. Its orbital period will be 25.562 hours.
A space station having a Bond albedo of 0.05 will have an equilibrium temperature of 210K (the same as Mars’ equilibrium temperature) at an orbital distance of 3666341 km, which is 4.0118 times the Roche limit. Its orbital period will be 45.768 hours.
… time passes …
At a WD age of 11.059 billion years, the white dwarf will have cooled to 3000K and will have a bolometric luminosity of 4.9021e+21 watts.
A space station having a Bond albedo of 0.05 will have an equilibrium temperature of 255K at an orbital distance of 621628.807328 km, which is only 0.68021 times the Roche limit. If the station could safely orbit the white dwarf at this distance, then it would have a sidereal period of 3.1953 hours.
A space station having a Bond albedo of 0.05 will have an equilibrium temperature of 210K at an orbital distance of 916585 km, which is about equal to the Roche limit. Its orbital period would be 5.7210 hours.
Notice that the amount of time elapsed between the onset of a comfortable spectrum of light from the 0.54 solar mass white dwarf, until the habitable zone around the white dwarf fell below the Roche limit, is 8.777 billion years. That’s more time than Earth will have been a habitable planet while the sun was a main sequence star.
Now, generally the higher mass white dwarfs offer less time for the existence of a dynamically stable orbit within their habitable zones. A white dwarf of 1 solar mass, for example, would have such orbits for only about two billion years.
See also my answer on habitable zones around brown dwarfs.
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David_Sims
- Posts: 9
- Joined: Sun Apr 07, 2024 9:39 pm
Re: Habitable zones for objects in orbit around brown dwarfs
White dwarf mass: 0.54 solar masses
WD age (Gyrs), Effective Temperature (K)
0.10, 15857.7
0.17, 14133.2
0.28, 12596.2
0.42, 11226.4
0.58, 10005.5
0.79, 8917.4
1.05, 7947.6
1.39, 7083.3
1.85, 6313.0
2.66, 5626.5
4.09, 5014.6
5.94, 4469.3
7.98, 3983.2
8.88, 3760.4
9.69, 3550.1
10.43, 3351.5
Five parameter logistical curvefits
t = WD age (Gyr), T = effective temperature (K)
T = 1431.045 + 17094.985/(1 + (t/0.1772761)^1.196982)^0.4263873
t = 0.1403198 + 11.3169002/(1 + (T/3728.029)^10.17064)^0.3433481
I'm rather disappointed in the quality of these curvefits, but they are the best single-function equations that I could find. When I code the age-temperature relation for white dwarfs, I plan on using an average from two 2nd degree Lagrange interpolating polynomials — one including the point prior to the interval being interpolated on, and the other including the next point after the interval being interpolated on. The result should be more accurate than the above 5PL equations.
https://jenab6.livejournal.com/photo/al ... &id=179692
WD age (Gyrs), Effective Temperature (K)
0.10, 15857.7
0.17, 14133.2
0.28, 12596.2
0.42, 11226.4
0.58, 10005.5
0.79, 8917.4
1.05, 7947.6
1.39, 7083.3
1.85, 6313.0
2.66, 5626.5
4.09, 5014.6
5.94, 4469.3
7.98, 3983.2
8.88, 3760.4
9.69, 3550.1
10.43, 3351.5
Five parameter logistical curvefits
t = WD age (Gyr), T = effective temperature (K)
T = 1431.045 + 17094.985/(1 + (t/0.1772761)^1.196982)^0.4263873
t = 0.1403198 + 11.3169002/(1 + (T/3728.029)^10.17064)^0.3433481
I'm rather disappointed in the quality of these curvefits, but they are the best single-function equations that I could find. When I code the age-temperature relation for white dwarfs, I plan on using an average from two 2nd degree Lagrange interpolating polynomials — one including the point prior to the interval being interpolated on, and the other including the next point after the interval being interpolated on. The result should be more accurate than the above 5PL equations.
https://jenab6.livejournal.com/photo/al ... &id=179692