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m1omg
2009-Jan-10, 10:43 AM
If there are these (hypothetical) (proto)planets around this; http://en.wikipedia.org/wiki/Cha_110913-773444 ; brown dwarf, assuming its current size, temperature, and bolometric (total) luminosity, how much radiation (compared to Sun from Earth, bolometric not depending on the wavelenghts just total sum) would they recieve?Also. where would the radiation belts be?

1st planet, 0.007 AU from the primary
2nd planet, 0.013 AU f. t. p.
3rd planet 0.04 AU
4th planet 0.15 AU
5th planet 0.4 AU

Also, can such a system form if the masses of the planets are in the Moon-Earth mass range and there are no other companions (it is a solitary brown dwarf) other than smaller icy and rocky bodies such as asteroids and disk of dust and stones because the system is still in formation?

And what if these hypothetical planets orbited this second smallest brown dwarf instead http://en.wikipedia.org/wiki/OTS_44 ?

EDIT - I was not able to use any online temperature calculator as I have not found any that would calculate temperature for bodies orbiting brown dwarfs, just main sequence stars.

antoniseb
2009-Jan-10, 01:41 PM
Brown Dwarfs do not radiate a fixed amount of energy, but rather cool off as they age.

Concerning the size of rocky planets that could potentially form around a BD, that would depend on the concentration of heavy elements in the original cloud. I am not aware of anything preventing an Earth sized planet from forming.

m1omg
2009-Jan-10, 02:51 PM
Brown Dwarfs do not radiate a fixed amount of energy, but rather cool off as they age.

Concerning the size of rocky planets that could potentially form around a BD, that would depend on the concentration of heavy elements in the original cloud. I am not aware of anything preventing an Earth sized planet from forming.

Yes, I know that of course but I stated that I want to know the temperatures as for now, and the cooling occurs very slowly on the scale of billions of years + more massive BDs have a relatively steady source of energy at least for some time in form of lithium and deuterium fusion, through not in the case of the first brown dwarf I mentioned which is too small to burn lithium or deuterium, the second is massive enough to burn deuterium.

So let's forgot the long term cooling consequences for now and focus on current luminosity.

Also, I found something about the cooling process http://www.astro.princeton.edu/cgi-bin/browndwarf3.cgi , but it refuses to run in my browsers, I tried both Firefox and Safari and it still doesn't work, just shows a blank black page.

StupendousMan
2009-Jan-10, 04:26 PM
If there are these (hypothetical) (proto)planets around this; http://en.wikipedia.org/wiki/Cha_110913-773444 ; brown dwarf, assuming its current size, temperature, and bolometric (total) luminosity, how much radiation (compared to Sun from Earth, bolometric not depending on the wavelenghts just total sum) would they recieve?Also. where would the radiation belts be?

1st planet, 0.007 AU from the primary
2nd planet, 0.013 AU f. t. p.
3rd planet 0.04 AU
4th planet 0.15 AU
5th planet 0.4 AU

Also, can such a system form if the masses of the planets are in the Moon-Earth mass range and there are no other companions (it is a solitary brown dwarf) other than smaller icy and rocky bodies such as asteroids and disk of dust and stones because the system is still in formation?

And what if these hypothetical planets orbited this second smallest brown dwarf instead http://en.wikipedia.org/wiki/OTS_44 ?

EDIT - I was not able to use any online temperature calculator as I have not found any that would calculate temperature for bodies orbiting brown dwarfs, just main sequence stars.

Why don't you simply read this earlier thread -- which you started! (http://www.bautforum.com/questions-answers/83163-temperature-object-deep-space.html) and use the information therein? Plug in the luminosity of the brown dwarf and the distance of the planets from the brown dwarf, and off you go.

m1omg
2009-Jan-10, 05:29 PM
Why don't you simply read this earlier thread -- which you started! (http://www.bautforum.com/questions-answers/83163-temperature-object-deep-space.html) and use the information therein? Plug in the luminosity of the brown dwarf and the distance of the planets from the brown dwarf, and off you go.

Sorry, but when I plugged that example formula from Wikipedia to my calculator program on my computer, it took me 15 minutes and according to my calculations the Earth would have been some 520 K hot:lol:

EDIT - Really, my calculator program screws it up, any advice?

m1omg
2009-Jan-10, 08:14 PM
Well, this isn't the entire answer to your first question, but it will get you started. If an isolated spherical "quickly rotating" body of radius r and emissivity e and albedo a is placed at a distance R from a blackbody-like star of luminosity L, the equilibrium temperature of the body will be



a L 1/4
T = ( ------------------------------ )
16 pi R^2 e sigma


where "sigma" is the Stefan-Boltzmann constant.

For rough purposes, set a = e = 1.

You could check to make sure that this works by plugging in values for the Sun and Earth and verifying that the temperature ends up around 250 K.

Thanks, but in what units I am supposed to insert the numbers?
Seriosly, just please, calculate it for me and spare me the math.
I am about as good at math as dyslexiac is good at reading.
Math is my Achilles heel.

George
2009-Jan-10, 10:43 PM
Thanks, but in what units I am supposed to insert the numbers?
Seriosly, just please, calculate it for me and spare me the math.
I am about as good at math as dyslexiac is good at reading.
Math is my Achilles heel. :) It isn't as hard as you think, I think.

Perhaps you might try this approach, though it is essentially the same..

You need to know how much energy is coming off your star. You know the energy from the Sun: 1366 Watts/sq. cm at 1 AU. So, how much weaker will this be with a lower temperature and smaller star?

The energy radiated is easy to determine if you know their temperatures. The Sun's Planck temperature is about 5850K. Let's say your stars temperature is 2500K. The energy flux per unit area will be simply their ratio to the fourth power. (2500/5850)^4 or (0.4275)^4 or 0.033 or 3.3% that of the Sun.

But, the smaller star does not have as much surface area to cast off this energy, so that needs to be determined.
The area of a sphere is 4 pi r^2, right? So the difference in their total areas will be their ratio of their radii squared, almost like above. Let's say you get a 10% result, then you multiply both these two values and then multiply this with the 1336 w/m^2 at 1 AU. This would give you about 4.4 watts/m^2.

Moving your planet closee will increae this unit energy value at that distance as the square of the increase: (D1/D2)^2, where D1 is the new, closer distance to the star. Multiply this times the 4.4 value (or whatever) and you have the amount of total radiant energy acting on that planet at that distance.

It won't be exact since brown dwarfs are probably not great blackbody radiatiors, but it should be close.

Once

m1omg
2009-Jan-11, 12:18 AM
:) It isn't as hard as you think, I think.

Perhaps you might try this approach, though it is essentially the same..

You need to know how much energy is coming off your star. You know the energy from the Sun: 1366 Watts/sq. cm at 1 AU. So, how much weaker will this be with a lower temperature and smaller star?

The energy radiated is easy to determine if you know their temperatures. The Sun's Planck temperature is about 5850K. Let's say your stars temperature is 2500K. The energy flux per unit area will be simply their ratio to the fourth power. (2500/5850)^4 or (0.4275)^4 or 0.033 or 3.3% that of the Sun.

But, the smaller star does not have as much surface area to cast off this energy, so that needs to be determined.
The area of a sphere is 4 pi r^2, right? So the difference in their total areas will be their ratio of their radii squared, almost like above. Let's say you get a 10% result, then you multiply both these two values and then multiply this with the 1336 w/m^2 at 1 AU. This would give you about 4.4 watts/m^2.

Moving your planet closee will increae this unit energy value at that distance as the square of the increase: (D1/D2)^2, where D1 is the new, closer distance to the star. Multiply this times the 4.4 value (or whatever) and you have the amount of total radiant energy acting on that planet at that distance.

It won't be exact since brown dwarfs are probably not great blackbody radiatiors, but it should be close.

Once

Thanks, but why I am getting much smaller results?When I multiply the ratio of their radii squared I must multiply it by 1000 to get approx. 4.4 and then the final result is even smaller, what I am doing wrong?

George
2009-Jan-11, 06:04 AM
Ok, let's use a more realistic radius for a brown dwarf rather than my example. They don't get much larger than Jupiter, so lets just use Jupiter, which is a nice 1/10th the radius of the Sun. Squaring this value shows that our b.d. will have 1/100th the surface area of the Sun, or 1%.

Since the energy output per unit area is only 3.3% of the Sun, then we would have ~ 44 watts per cm^2 at 1 AU if the b.d. were the same size of the Sun. Since it is 1/100th the surface area (size) then we will only see 1% of the 44 watts at 1 AU, or 0.44 watts/m^2.

Now let's move the planet to a closer orbit. Simply square the ratio of the orbital distances. For instance, let's take your 1st planet at 0.007AU. Thus, (1AU / 0.007 AU) = 143 and its square is = 20,400x. At this distance, the energy flux per unit area is the 1 AU value of 0.44 watts/m^2 times the 20,400, which equals about 9000 w/m^2, or about 6.5x more intense that we experience at 1 AU.

m1omg
2009-Jan-11, 11:37 AM
Ok, let's use a more realistic radius for a brown dwarf rather than my example. They don't get much larger than Jupiter, so lets just use Jupiter, which is a nice 1/10th the radius of the Sun. Squaring this value shows that our b.d. will have 1/100th the surface area of the Sun, or 1%.

Since the energy output per unit area is only 3.3% of the Sun, then we would have ~ 44 watts per cm^2 at 1 AU if the b.d. were the same size of the Sun. Since it is 1/100th the surface area (size) then we will only see 1% of the 44 watts at 1 AU, or 0.44 watts/m^2.

Now let's move the planet to a closer orbit. Simply square the ratio of the orbital distances. For instance, let's take your 1st planet at 0.007AU. Thus, (1AU / 0.007 AU) = 143 and its square is = 20,400x. At this distance, the energy flux per unit area is the 1 AU value of 0.44 watts/m^2 times the 20,400, which equals about 9000 w/m^2, or about 6.5x more intense that we experience at 1 AU.

Thank you very much, I've finally gotten the calculations right, with the values for the first BD http://en.wikipedia.org/wiki/Cha_110913-773444 ;

1st planet 0.007 AU from BD will recieve 2449 w/m^2, so about the insolation at Venus.

2nd planet at 0.013 AU = 710 w/m^2, like the sunshine on Mars in perihelion.

3rd planet at 0.04 AU = 75 w/m^2, a bit warmer and sunnier than on Jupiter.

4th planet at 0.15 AU = 5.33 w/m^2, a bit warmer and sunnier than on Uranus.

5th planet at 0.4 AU = 0.75 w/m^2, lower than half of the sunshine on Neptune.

And for the second BD I choose, this time hotter and bigger http://en.wikipedia.org/wiki/OTS_44 ;

1st planet at 0.007 AU = 43673 w/m^2, around 4-6x that of Mercury, HOT
2nd planet at 0.013 AU = 12663 w/m^2, this time around as hot as Mercury
3rd planet at 0.04 AU = 1338 w/m^2, HEUREKA, we have a Brown Dwarf orbiting Earth twin

4th planet at 0.15 AU = 95 w/m^2, around 2x as sunny as on Jupiter
5th planet at 0.4 AU = 13 w/m^2, like the coldest days on Saturn

I decided to do more calculations around other interesting brown dwarfs;
Let's do 2MASS 0415-0935, massive at 63 Jupiter masses, but one of the coldest brown dwarfs, T8 spectral class, at just 800 K, and just 0.095x the radius of our Sun;

1st planet at 0.007 AU = 86 w/m^2, around 2x as sunny as Jupiter
2nd planet at 0.013 AU = 25 w/m^2, 2x as sunny as on Saturn
3rd planet at 0.04 AU = 2.63 w/m^2, around the half on sunshine on Uranus, or cca 4/3 as bright as on Neptune

4th planet at 0.15 AU = 0.19 w/m^2, a bit more than tenth of the sunlight on Neptune

5th planet at 0.4 AU = 0.026 w/m^2, almost a hundred times fainter than Sun from Neptune, frigid cold and darkness, like in the Kuiper Belt


Now let's do Proxima Centauri to check it (radius 0.15x that of Sun, temperature 3000 K);

1st planet at 0.007 AU = 42388 w/m^2, again as the OTS 44, around 4-6x that of Mercury, HOT, I don't even know if such planet can exist around Proxima Centauri as it is much massive that all the previous objects so maybe it would get shredded if it is within it's Roche limit.

2nd planet at 0.013 AU = 12290 w/m^2, Mercury
3rd planet at 0.04 AU = 1298 w/m^2, again, Earth twin
4th planet at 0.15 AU = 92 w/m^2, again, Jupiter*2
5th planet at 0.4 AU = 13 w/m^2, again, coldest days on Saturn

And let's add 1 AU distant planet;

6th planet at 1 AU = 2.077 w/m^2, so yeah, this is about right, the sunshine on a planet 1 AU from Proxima is about the sunshine on Neptune, or more precisely about 4/3 the sunshine on Neptune.

I rounded values a bit.
And it seems that young hot brown dwarfs like OTS 44 can even outshine late M stars such as Proxima Centauri, even if they have very low mass as they still have enough of their primordeal heat and are bloated a bit (OTS 44 has 0.258x the Sun's radius while Proxima only around 0.15x the Sun's radius).

I got an idea, brown dwarfs may be the best havens for life, why?Because they can supply heat by radiating AND tidal heating AND they can have Earth mass and composition planets AND these planets can be outside the radiation belt, so they can have normal atmospheres and be free of horrible radiation, unlike for example Europa around Jupiter, so a habitable planet around a brown dwarf would be like young Earth first - Earthlike or a bit higher but not enough to cause runaway greenhouse effect insolation AND tidal heating = wild and active planet where life can thrive, and continue to thrive even when the brown dwarf has cooled down a bit and thus it normally would be frozen, because the continuous emission of greenhouse gases from tide supported volcanism would heat the planet + most of the brown dwarf's radiation is in infrared so the greenhouse gases will be able to retain heat more efficiently + in this stage abundant photosynthesis will be possible, on other wavelenghts than our plants use, in infrared.Then, after a few billion years, when the brown dwarf cools to temperature when the planet crosses the snow line and surface is frozen, life can still develop around hydrotermal vents and this does not need to be the end of all open air surface life - as the planet has Earthlike composition and geography there should be many open air hot springs and lakes, much more abundant than on Earth as the planet is strongly tidally heat + its natural geothermal heat.

The advantages are that life can thrive and develop basically until the heat death of the universe, brown dwarfs will not fry their planets when they die like other stars do, even the lightest red dwarfs will brighten 100-1000x eventually after trillion years and fry their planets, then shrink to white dwarfs + red dwarfs have strong stellar winds that would blow the atmospheres away after a long time, so brown dwarfs seems to me like the best long-term abodes for life.

The best combo will be a brown dwarf massive enough to mantain enough light and heat for the longest time without surface being frozen but not so massive that the life zone would be in the radiation belts, with a planet either Earthlike or Super-Earth, of conventional water-rock composition,not a dry rockball or iceball.

Also, how do I convert these data to get the temperature in Kelvins?

m1omg
2009-Jan-11, 02:19 PM
Next batch;

Gliese 229 B, the first discovered T dwarf, class T6, radius 0.12x Solar, temperature of 1020 K

1st planet, 0.007 AU = 367 w/m^2, a bit weaker than Mars at aphelion
2nd planet, 0.013 AU = 107 w/m^2, like 2/3 of sunshine that Ceres gets at aphelion or about 2x as much as on Jupiter

3rd planet, 0.04 AU = 11.25 w/m^2, a bit weaker than on Saturn
4th planet, 0.15 AU = 0.8 w/m^2, like on Pluto
5th planet, 0.4 AU = 0.11 w/m^2, like on Eris when it's furthest away from the Sun

I neglected the effect of the primary red dwarf star as it is so far away and too faint to provide any significant light or heat at that distance.

m1omg
2009-Jan-11, 03:35 PM
Now how do I convert these insolation values to surface temperatures in Kelvins?

m1omg
2009-Jan-11, 04:28 PM
I decided to calculate the insolations for planets orbiting white dwarf Sun (radius 0.01x, temperature 20 000 K, orbits of planets widened cca 1.85x due to Sun's mass loss during the RGB stages);

Earth, 1.85 AU = 5.33 w/m^2, a bit stronger than on Uranus today
Mars, 2.81 AU = 2.26 w/m^2, a bit weaker than on Uranus today
Jupiter, 9.62 AU = 0.197 w/m^2, a bit more than the intensity of sunshine 100 AU from Sun in the present, like Eris aphelion

Saturn, 17.72 AU = 0.058 w/m^2, like 150 AU from the Sun
Uranus, 35.58 AU = 0.014 w/m^2
Neptune, 55.69 AU = 0.006 w/m^2, a little less than sunshine 1000 AU from the Sun-aprox. distance of Sedna's aphelion

Pluto, 73.04 AU = 0.003 w/m^2

George
2009-Jan-11, 09:23 PM
There ya go! :)

m1omg
2009-Jan-12, 03:18 PM
There ya go! :)

:) , but how do I convert these values to temperature in Kelvins?

George
2009-Jan-12, 07:24 PM
:) , but how do I convert these values to temperature in Kelvins?
Are you wanting to know surface temperature? This is an area I haven't looked into, but planetary atmospheres will complicate such things in a big way. Consider how Venus has a outrages surface temperature compared with Earth.

m1omg
2009-Jan-13, 01:21 PM
Are you wanting to know surface temperature? This is an area I haven't looked into, but planetary atmospheres will complicate such things in a big way. Consider how Venus has a outrages surface temperature compared with Earth.

I just want to know simple blackbody/albedo temperature.

StupendousMan
2009-Jan-13, 02:13 PM
I just want to know simple blackbody/albedo temperature.

Post number 4 in this thread takes you to another recent thread which contains the formula you need to do this.

There are only three items you need to know -- L, the luminosity of the star (you could use Watts), R, the distance from star to planet (you could use meters), and sigma, the Stefan-Boltzmann constant (you should use a value which has units containing MKS units - that is, meters and kilograms, not centimeters and grams).