Calculating the Temperature in Space

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I am trying to determine if I am on the right track by using the Stefan-Boltzmann Law to calculate the distance a point in space must be to have a given temperature. For example, the snow line (a.k.a. frost line, ice line) of a star is when the temperature reaches 160°K ± 10°K. Using the Stefan-Boltzmann Law :

313pt03.jpg

Where:
TE = The temperature of the black body object, in Kelvins (160°K in this case)
TS = The surface temperature of the star, in Kelvins
rS = The radius of the star, in meters
a0 = The distance the black body object is from the star, in meters​

Using Sol's data as an example, a black body object would have a temperature of 160°K at 3.04 AU. Yet, according to a paper published on May 9, 2003, the frost line for the Sol system should be ~5 AU.

Remote Infrared Observations of Parent Volatiles in Comets: A Window on the Early Solar System

Which leads me to believe that my approach is incorrect since it does not match our observations. If someone can set me right, it would be greatly appreciated.
 

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  • #2
Simon Bridge
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You are trying to work out the distance a test black-body is from a star for it to have a particular temperature?
 
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You are trying to work out the distance a test black-body is from a star for it to have a particular temperature?
Yes. I realize that, except for perhaps a black hole, there is no such thing as a perfect black body. That may be where my problem lies. What I am actually trying to determine is the point in space where a particular temperature can be found. The snow line is one example, but if you used 273.15°K and 373.15°K you could, in theory, determine the inner and outer range of "habitable zones" for any given star. If you know the star's effective surface temperature and radius.
 
  • #4
256bits
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The " frost line" refers to the distance for certain molecules to coalese into the solid phase during the formation of the solar system billions of years ago. Your PDF does actually refer to it as the nebular frost line can be easily missed while reading. With the different conditions at present, you will of course calculate a different distance.

Take in mind that a planet will reflect some of the radiation which will decrease the distance, and greenhouse effects with an atmosphere will increase the distance .
 
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The " frost line" refers to the distance for certain molecules to coalese into the solid phase during the formation of the solar system billions of years ago. Your PDF does actually refer to it as the nebular frost line can be easily missed while reading. With the different conditions at present, you will of course calculate a different distance.

Take in mind that a planet will reflect some of the radiation which will decrease the distance, and greenhouse effects with an atmosphere will increase the distance .
So you are saying that the frost line of ~5 AU was when the Sol system was still in its nebular state. Now, 4.5 billion years later, the frost line should be 3.04 AU, assuming 160°K is the temperature that must be achieved?

Yes, I understand that the albedo, radiative forcing, and atmospheric pressure of a planet will have an effect on whether or not it can support liquid water on its surface. However, absent any planetary information other than its radius and approximate orbit, it should provide a ballpark idea whether or not the planet is anywhere near that range.
 
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256bits
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The coma is generally made of ice and dust.[1] Water dominates up to 90% of the volatiles that outflow from the nucleus when the comet is within 3-4 AU of the Sun.[1] The H2O parent molecule is destroyed primarily through photodissociation and to a much smaller extent photoionization.[1] The solar wind plays a minor role in the destruction of water compared to photochemistry.[1] Larger dust particles are left along the comet's orbital path while smaller particles are pushed away from the Sun into the comet's tail by light pressure
http://en.wikipedia.org/wiki/Coma_(cometary [Broken])

If I interpret wiki correctly, the coma (atmosphere) for a comet begins to form roughly corresponding to your calculation.
 
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  • #7
Simon Bridge
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Yes. I realize that, except for perhaps a black hole, there is no such thing as a perfect black body.
It's all right, I was just making sure I understood you before replying.
The others have done the answering since so...
 
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