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Calculating Stellar radius from bolometric flux

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lquinnl
#1
Aug6-12, 06:51 AM
P: 26
1. The problem statement, all variables and given/known data

The monochromatic flux emitted from unit surface area of a black body is given by

[tex] F_{\lambda}(T) = \frac{2h\pi c^{2}}{\lambda^{5}} \frac{1}{exp(hc/\lambda KT)-1} Wm^{-2}m^{-1} [/tex]

If the distance to star X is 620 parsecs, calculate:

(a) the radius of star X, in Solar radii; [3]

(b) the bolometric luminosity of star X, in units of Solar luminosity. [1]

2. Relevant equations

OK so I know if we integrate the given equation between 0 and infintiy we get the bolometric flux for F(T). When we perform this integration we end up with the stefan boltzmann equation:

[tex]F(T) = \sigma T^{4} Wm^{-2}[/tex]

this enables me to calulate the bolometric surface flux, great.

Now i want the radius of star X. I am aware that

[tex]F^{bol} = (\frac{R}{d})^{2}\sigma T^{4}[/tex]

where R is the stellar radius and d the distance to the star.

My problem comes here. I can't calculate the flux from the stefan boltzmann equation and use it in this last equation because I will always have the distance to the star equal to the radius, which cannot be correct.


Any help would be greatly appreciated
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phyzguy
#2
Aug6-12, 07:00 AM
P: 2,179
Everything you've said is correct except the final conclusion. The flux received here on Earth is hugely less than the flux emitted at the surface of the star, because the factor R/d is much much less than 1. Why do you conclude that it is equal to 1? Do you know how to calculate the bolometric flux, given a brightness measure, such as magnitude?
lquinnl
#3
Aug6-12, 07:26 AM
P: 26
I know the apparent magnitude is related to flux by the relationship

[tex]m=-2.5log_{10}(F)+ constant[/tex]

am i to somehow use this to calculate the bolometric flux at the given distance then use this flux on the left hand side of

[tex]F^{bol} = (\frac{R}{d})^{2}\sigma T^{4}[/tex]

Although i do not know the magnitude of the star, apparent or absolute.

phyzguy
#4
Aug6-12, 07:44 AM
P: 2,179
Calculating Stellar radius from bolometric flux

You must be given some additional information about star X, no?
lquinnl
#5
Aug6-12, 07:50 AM
P: 26
A star X is observed in the V-band filter (central wavelength 550 nm, bandwidth
88 nm) using a telescope with a diameter of 40cm. The telescope and camera detect
50% of the incident photons and, during a 10 sec exposure, 5500 photons are detected.

I have calculated the monochromatic flux of the star.

So from here should i calculate the magnitude?
phyzguy
#6
Aug6-12, 08:00 AM
P: 2,179
No need to calculate the magnitude. You should be able to calculate the emitted flux from the equation you wrote in the first post. Then you should be able to calculate the flux here on Earth from the information you are given. What is their ratio?

Unless I am mistaken or you have been given additional information, you will need to assume some temperature for the star, since you are only given the flux at a single wavelength.
lquinnl
#7
Aug6-12, 08:29 AM
P: 26
OK so I have:

Monochromatic flux detected from Earth:

[tex]F_{\lambda} = \frac{\Delta E}{\Delta A \Delta \lambda \Delta t}[/tex]

The measured spectrum of the star peaks at a wavelength of 650 nm. Assuming the
star radiates as a black body,

[tex]T = \frac{2.898 \times 10^{-3} mK}{6.5\times 10^{-9}m}[/tex]

Bolometric surface flux:

[tex]F^{bol} (T)= \sigma T^{4}[/tex]


But surely I cannot compare Bolometric surface flux to Monochromatic flux detected on Earth because they are over different wavelength ranges.
voko
#8
Aug6-12, 08:30 AM
Thanks
P: 5,678
I would guess the wavelength given is at the peak of the energy spectrum.

edit: I see it is not.
phyzguy
#9
Aug6-12, 08:43 AM
P: 2,179
Quote Quote by lquinnl View Post
OK so I have:

Monochromatic flux detected from Earth:

[tex]F_{\lambda} = \frac{\Delta E}{\Delta A \Delta \lambda \Delta t}[/tex]
OK, looks good.

Quote Quote by lquinnl View Post
The measured spectrum of the star peaks at a wavelength of 650 nm. Assuming the
star radiates as a black body,

[tex]T = \frac{2.898 \times 10^{-3} mK}{6.5\times 10^{-9}m}[/tex]
right idea, but is 650nm = 6.5*10^-9 ???

Quote Quote by lquinnl View Post
Bolometric surface flux:

[tex]F^{bol} (T)= \sigma T^{4}[/tex]


But surely I cannot compare Bolometric surface flux to Monochromatic flux detected on Earth because they are over different wavelength ranges.
Why would you want to? You have calculated Flambda here on Earth. Your first post gave the formula for Flambda at the star's surface. Compare these two.
lquinnl
#10
Aug6-12, 09:25 AM
P: 26
Ooops so 650nm should be 6.5 x10^-7 m.

So compare

[tex]F_{\lambda}(T) = \frac{2h\pi c^{2}}{\lambda^{5}} \frac{1}{exp(hc/\lambda KT)-1} Wm^{2}m^{-1}[/tex]

with

[tex]F_{\lambda}(T) = \frac{\Delta E}{\Delta A \Delta \lambda \Delta t}[/tex]

then use this ratio, call it Z ,to calculate the Fbolometric at the distance of 620 parsecs.

we know that F bol at star surface is:
[tex]F^{bol}(at surface) = \sigma T^{4}[/tex]


and because

[tex]F^{bol}(atdistance) = (\frac{R}{d})^{2} \sigma T^{4}[/tex]

we have:
[tex]F^{bol}(at surface) \times Z = (\frac{R}{d})^{2} \sigma T^{4}[/tex]

then rearrange for R

Sound good?
phyzguy
#11
Aug6-12, 09:29 AM
P: 2,179
No! Forget the total, integrated flux, what you're calling Fbol, because you don't know this value at the Earth. Just compare the flux in the measured wavelength band to the calculated flux in that same band from your formula for Flambda. This ratio will be (R/d)^2
lquinnl
#12
Aug6-12, 09:58 AM
P: 26
Quote Quote by phyzguy View Post
No! Forget the total, integrated flux, what you're calling Fbol, because you don't know this value at the Earth. Just compare the flux in the measured wavelength band to the calculated flux in that same band from your formula for Flambda. This ratio will be (R/d)^2
So the ratio of this

[tex] F_{\lambda}(T) = \frac{2h\pi c^{2}}{\lambda^{5}} \frac{1}{exp(hc/\lambda KT)-1} Wm^{-2}m^{-1} [/tex]

to this:


[tex]F_\lambda = \frac{\Delta E}{\Delta A \Delta t \Delta \lambda}[/tex]

is

[tex] (\frac{R}{d})^{2}[/tex]
phyzguy
#13
Aug6-12, 10:24 AM
P: 2,179
Right. Do you see why? The first is the flux at the star's surface. The second is the measured flux at the Earth. Since the flux drops of as (1/r)^2, the ratio is equal to the ratio of the radii squared.
lquinnl
#14
Aug6-12, 11:16 AM
P: 26
Yeah that makes sense now!

Thank you very much for sticking with me and helping out with this problem. I'm trying to revise for an exam and I was hittting a brick wall with that!

Thanks again


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