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m1omg
2009-Jan-05, 01:17 PM
1.What would be the temperature of an object in interstellar space given that the nearest Sunlike star is about 5 light years away and normal intensity of cosmic rays and the CMB?

2.What would be the temperature of an object in intergalactic space, with the nearest dwarf galaxy like Small Magellan Cloud around 200 thousand light years away and the nearest big spiral galaxy around 3 million light years away?

StupendousMan
2009-Jan-05, 01:39 PM
1.What would be the temperature of an object in interstellar space given that the nearest Sunlike star is about 5 light years away and normal intensity of cosmic rays and the CMB?


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.

Ken G
2009-Jan-05, 02:57 PM
Also, a lot of heating tends to come not from the closest star (if you are not in a solar system but instead at some random point in the galaxy), but rather from the nearest massive and hot star (which could be quite far away). Also, there are other heating mechanisms (you mentioned cosmic rays), so sometimes it is better to think of the temperature as being due to a balance between heating and cooling. In many situations, the cooling rate (say of a gas) depends on the density of that gas moreso than does its heating rate, so very low density gas has a tendency to inefficiently cool but efficiently heat, and in that case it can get very warm or downright hot. It depends a great deal on the local conditions, which is why the "interstellar medium" shows many different temperatures in different places.

Tucson_Tim
2009-Jan-05, 03:06 PM
Would the temperature of the object be about the same as the background microwave radiation from the big bang, 2.7K? Maybe slightly above?

m1omg
2009-Jan-05, 03:21 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.

Well, thanks but I am only 15 years old and bad at math.
I understand very many concepts, but please don't post unsolved complex formulas to me because I will not understand it.

Ken G
2009-Jan-05, 03:29 PM
Would the temperature of the object be about the same as the background microwave radiation from the big bang, 2.7K? Maybe slightly above?The CMB is kind of a minimum possible temperature, if there is no other heating. To see the importance of heating mechanisms other than the CMB, consider that most of the hydrogen in the universe is found in between galaxies, in a low density gas pervading clusters of galaxies, and this gas is extremely hot and highly ionized, essentially because its density is so low that it hardly cools at all. What heats it is still a matter of research!

m1omg
2009-Jan-05, 04:05 PM
The CMB is kind of a minimum possible temperature, if there is no other heating. To see the importance of heating mechanisms other than the CMB, consider that most of the hydrogen in the universe is found in between galaxies, in a low density gas pervading clusters of galaxies, and this gas is extremely hot and highly ionized, essentially because its density is so low that it hardly cools at all. What heats it is still a matter of research!

Well, Boomerang Nebula is colder, something can be colder than CMB in free space if it has some mechanism to cool, the Boomerang nebula expands so fast that it has lost almost all of its heat; http://en.wikipedia.org/wiki/Boomerang_nebula , it is only 1 K "hot", colder than CMB.

JohnD
2009-Jan-05, 04:53 PM
m1omg,
Maybe I should leave this to stupendous, but if you want to take an interest in astronomy you need to try a little harder at maths!

The equation he quoted is not very complex, though I'm unsure about the (1/4), which may be "raised to the power of a quarter" or ^0.25, a complex idea!

The Wiki (which I have criticised in the past, but anyway) has an article at:
http://en.wikipedia.org/wiki/Black_body that may interest you.
(A "black body" is an imaginary object that absorbs ALL radiation that falls on it. See article)
The article, by some fairly fierce algebra, derives a simple equation for the temperature of a planet, that will do as well for your object. See "The Result", about two thirds down the page.

For the Earth:
Tearth = Tsun SQR(SQR(1-alpha x Radius-sun) divided by 2 x distance sun-earth)

alpha is the albedo (how reflective the Earth is) and SQR stands for square root, of what's in the brackets.
The article gives you some values for Earth and Sun.
You could do so for your object, and get the answer with your pocket calculator.
Remember; in calculating, plus and minus must come before taking a square root, which must come before multiply or divide.
Write down each stage - don't try to store it in the calculator memory!
Clue: If your object is a black body then it's albedo is zero, and the (1 - alpha x Radius-sun) becomes "1". Then SQR(1)=1!

If that's tough for you, why not ask your maths teacher?
They won't do the sums for you, any more than I will, but you might get some satisfaction in acheivement by being led through the calculation!

Good luck!
John

grant hutchison
2009-Jan-05, 05:00 PM
... I'm unsure about the (1/4), which may be "raised to the power of a quarter" or ^0.25 ...That's right. Or you can think of it as the fourth root of the bracketed quantity. It's in there because of the fourth-power relationship between temperature and luminosity.

Grant Hutchison

Ken G
2009-Jan-06, 04:48 PM
Well, Boomerang Nebula is colder, something can be colder than CMB in free space if it has some mechanism to cool, the Boomerang nebula expands so fast that it has lost almost all of its heat; http://en.wikipedia.org/wiki/Boomerang_nebula , it is only 1 K "hot", colder than CMB.Good point, the CMB limit applies only to equilibrium objects, not ones that are expanding or being cooled via some other nonequilibrium means.