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Astronomy - Greenhouse Effect (integral troubles mainly)

  1. Nov 14, 2005 #1


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    Given equations:
    [tex]\frac{d T^4}{dr} = - \frac{3 \kappa \rho}{a c} F_{rad}[/tex]
    where [tex]a = 4 \sigma_{B} / c[/tex] with [tex]\sigma_{B}[/tex] being the Boltzmann constant.
    Also, define the optical depth [tex]\tau = \int \kappa \rho dr[/tex]
    Optical depth measured at ground level is [tex]\tau_{g}[/tex]
    Where [tex]\tau_{g} = \int^\infty_{ground} \kappa \rho dr[/tex] and the optical depth at the photosphere equals 2/3
    Also [tex]F_{rad} = \frac {L_{r}}{4 \pi r^2} = \frac {A \sigma_{B} T_{g}^{4}}{4 \pi r^2} = \sigma_{B} T_{g}^{4}[/tex]
    Now, I have to use the first equation given to find:
    [tex]T^4_{g} = T^4_{p} [1 + \frac{3}{4} (\tau_{g} - \frac {2}{3})][/tex]
    So what I did was take the integral of the first equation and try to work from there but I am having difficulty understanding what to do with the left hand side of the integral but here's what I have so far:
    1) [tex]\frac{d T^4}{dr} = - \frac{3 \kappa \rho}{a c} F_{rad}[/tex]
    2) [tex]\int^{photosphere}_{ground} d T^4 = - \frac{3}{a c} F_{rad} \int^{photosphere}_{ground} \kappa \rho dr[/tex] (because F_rad is constant)
    3) [tex](T^4_{g} - T^4_{p}) = - \frac{3}{a c} F_{rad} ( \int^\infty_{ground} \kappa \rho dr - \int^\infty_{photosphere} \kappa \rho dr)[/tex]
    4) [tex](T^4_{g} - T^4_{p}) = - \frac{3}{a c} F_{rad} (\tau_g - \frac{2}{3})[/tex]
    5) [tex](T^4_{g} - T^4_{p}) = - \frac{3}{4} T_{g}^{4} (\tau_g - \frac{2}{3})[/tex]
    I know I'm close and it's not complete and I'm stuck... Did I make a mistake anywhere or what? If anyone could help me out that'd be great.
    Last edited: Nov 14, 2005
  2. jcsd
  3. Nov 15, 2005 #2


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    Staff Emeritus
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    Could you be more explicit about what you're trying to show?
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