Intensity measured from a point and surface intensity

[Chowie]

Homework Statement

An astronomer is trying to estimate the surface temperature of a star with a radius of 5.0*10^8m by modeling it as an ideal blackbody. The astronomer has measured the intensity of radiation due to the star at a distance of 2.5*10^13m and found it to be equal to 0.055W/m^2 . Given this information, what is the temperature of the surface of the star?
Express your answer in kelvins to two significant digits.

Homework Equations

I is proportional to Stefans constant * Temp to the power of 4

Surface intensity = Intensity measured from a distance * (Distance / Radius)^2

The Attempt at a Solution

With those two equations above the answer itself just falls out, however I only found the second equation while looking around for help on this question.

Can anyone explain to me in English why the second equation holds true? Is it already in its simplest form and I should just remember it?

Cheers

 

[eczeno]

i am not sure what distance/radius means, but i think i understand what you are talking about.

the intensity at a distance r from the star will be: I(r) = Is/r^2 , where Is is the surface intensity. this follows from the fact that the energy can be considered to be spread evenly over the surface of a sphere which is increasing in size. the surface area of the sphere is 4 pi r^2, so the area increases as r^2, so the total energy in any given area (since the total energy is constant) must fall as 1/r^2 as you move out.

hope this helps

 

[Chowie]

Distance/radius meant distance divided by radius, sorry for the ambiguity.

Your explanation although correct and very helpful would not produce the correct answer as it isn't accounting for the radius of the star. What I'm struggling with is the fact that when given this problem and not quite getting it I immediately googled the whole thing and was lead to this: http://answers.yahoo.com/question/index?qid=20081203004833AAHRsZD which is the correct answer but I wish he would have just explained his working.

This should be trivial stuff for someone at my level and it's causing a lot of anxiety, maybe is due to me trying to work backwards from the answer to find out how, apparently it's just simple geometry which is why it's so infuriating that I cannot comprehend it.

 

[cepheid]

Distance/radius meant distance divided by radius, sorry for the ambiguity.

Your explanation although correct and very helpful would not produce the correct answer as it isn't accounting for the radius of the star. What I'm struggling with is the fact that when given this problem and not quite getting it I immediately googled the whole thing and was lead to this: http://answers.yahoo.com/question/index?qid=20081203004833AAHRsZD which is the correct answer but I wish he would have just explained his working.

This should be trivial stuff for someone at my level and it's causing a lot of anxiety, maybe is due to me trying to work backwards from the answer to find out how, apparently it's just simple geometry which is why it's so infuriating that I cannot comprehend it.

Hello Chowie. We astronomers tend to use the word "flux" (F) to label the quantity that is measured in W/m², so this is just a head's up that that is the term I will be using.

For something that emits isotropically (the same in all directions) the flux from a source only depends on your radial distance r from that source. In other words, F is a function of r, F(r).

Now, the Stefan-Boltzmann law says that for an ideal blackbody of radius R, the surface flux (the rate at which energy is being emitted from a unit area on the surface e.g. from 1 square metre) is given by:

$$F(r = R) = \sigma T^4$$

Therefore, the luminosity (L) of the source (the total power that it emits in watts) is just this flux (power per unit area), multiplied by the total surface area. For a sphere this is just:

$$L = 4\pi R^2 F(R) = 4 \pi R^2 \sigma T^4$$

Now, imagine you are at a distance r > R from the source. Again, the light energy from the source is being emitted in all directions equally. Therefore, at this distance, this same amount of total energy is now spread out in space over a much larger sphere of radius r. Therefore, the power per unit area (flux) being received at a distance r is just the luminosity divided by this total area over which the light energy is now spread out.

$$F(r) = \frac{L}{4\pi r^2}$$

$$= \frac{4 \pi R^2 \sigma T^4}{4\pi r^2}$$

$$= \sigma T^4 \frac{R^2}{r^2} \propto \frac{1}{r^2}$$

What I have just discussed above is an explanation of why the inverse-square law for flux is true (for sources that emit isotropically)

 

[Chowie]

Thank you very much for this concise explanation, it has really furthered my understanding.