Relationship of temperature and radius.

AI Thread Summary
The discussion centers on the relationship between temperature and radius in stars, specifically regarding a main sequence star transitioning to a supergiant. Initially, the star has a temperature of 25,000 K and a radius 10 times that of the sun, dropping to 5,000 K as a supergiant. Participants explore using luminosity calculations and the Morgan-Keenan spectral classification to determine the star's radius in solar radii, with one participant arriving at an incorrect answer. They discuss the relationship between temperature and radius, concluding that if total energy output remains constant, luminosities must also be equivalent in both phases. This leads to the understanding that the luminosity remains the same despite changes in temperature and radius.
darkar
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Here's the question,

Initially, a main sequence star has a surface temperature of 25000 K and a radius 10 times that of the sun. Its temperature drop to 5000 K as it becomes a supergiant. What is its radius in term of solar radii?
a) 25
b) 50
c) 100
d) 250
e) 500

Well, I tried to do it using luminosity , L=4πR²σT⁴. Since the luminosity is not given, i use the Morgan-Keenan spectral classification to get the luminosity of the star at each phase, that's is at temperature of 25000 K has luminosity of 20,000 and for 5000 K has luminosity of 0.4. I got my final answer as 1.18 solar radius, which mean my answer is wrong. How should i approach the problem? and is that the spectral classification is only for main star sequence? and my friend use the relationship of temperature^4 is inversely proportional to 1/(radius)^2 and he got the answer for 250. BUt doesn't that mean that the luminosity is the same, is this true?

Thanks!
 
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If the total energy output is considered the same, then one can establish an equivalence, much like your friend did.

r12*T14 = r22*T24

or r1*T12 = r2*T22

or r2 = r1*(T1/T2)2

So what does this say about the luminosities?
 
I see... so the luminosities in that two phases are the same.

Thanks!
 
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