Habitable zones for objects in orbit around brown dwarfs
Posted: Wed Apr 17, 2024 11:18 pm
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
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