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Homology Help

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  1. Oct 27, 2014 #1
    I have a question in my astrophysics textbook that I need some help with.

    Given [tex]\frac{dT}{dr}\propto\frac{\kappa\rho{L}}{r^2T^3}[/tex] and [tex]\frac{dL}{dr}\propto{r^2}{\rho}\epsilon[/tex] show that [tex]L\propto {M^{5.4}}[/tex] and [tex]R\propto {M^{0.2}}[/tex] if [tex]\kappa\propto\rho{T^{-3.5}}[/tex] and [tex]\epsilon\propto\rho{T^{5}}[/tex].

    Using the equation [tex]\frac{dT}{dr}\propto\frac{\kappa\rho{L}}{r^2T^3}[/tex] and substituting the value for [tex]\kappa[/tex] and also [tex]\rho\propto{\frac{M}{R^3}}[/tex] I got the answer [tex]L\propto {M^{5.5}R^{-0.5}}[/tex]

    Then using [tex]\frac{dL}{dr}\propto{r^2}{\rho}\epsilon[/tex] and substituting [tex]\epsilon\propto\rho{T^{5}}[/tex] as well as [tex]\rho\propto{\frac{M}{R^3}}[/tex] and the result from above I got [tex]M^{2.5}\propto{R}[/tex]

    Obviously I have done something wrong. I'm a little slow in tex but if you would like my full working in order to help me just let me know.
     
  2. jcsd
  3. Oct 28, 2014 #2

    Ken G

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    Gold Member

    The problem you give is underconstrained. After substituting for epsilon and rho, your first equation gives L as a function of R, M, and T, yet you express it as purely a function of R and M. What happened to T? Also, you are told a second way to express L as a function of R, M, and T, so by equating the two, all you get is one equation involving R, M, and T, which cannot be solved for R as a function of M alone. You are missing a constraint on T. To find it, think physically about what you have already asserted about the star, and what is missing.

    The equations you give are the equation of radiative diffusion, and the equation of energy generation. What you are missing is the equation of hydrostatic equilibrium, which is dP/dr = -rho*g. That brings in the pressure P, which you have to get rid of using the ideal gas equation. You should see that T is proportional to M/R, and if you put that into what you have above, you should get the answer you want.
     
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