# Astronomy - Greenhouse Effect (integral troubles mainly)

1. Nov 14, 2005

### NIQ

Given equations:
$$\frac{d T^4}{dr} = - \frac{3 \kappa \rho}{a c} F_{rad}$$
where $$a = 4 \sigma_{B} / c$$ with $$\sigma_{B}$$ being the Boltzmann constant.
Also, define the optical depth $$\tau = \int \kappa \rho dr$$
Optical depth measured at ground level is $$\tau_{g}$$
Where $$\tau_{g} = \int^\infty_{ground} \kappa \rho dr$$ and the optical depth at the photosphere equals 2/3
Also $$F_{rad} = \frac {L_{r}}{4 \pi r^2} = \frac {A \sigma_{B} T_{g}^{4}}{4 \pi r^2} = \sigma_{B} T_{g}^{4}$$
Now, I have to use the first equation given to find:
$$T^4_{g} = T^4_{p} [1 + \frac{3}{4} (\tau_{g} - \frac {2}{3})]$$
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) $$\frac{d T^4}{dr} = - \frac{3 \kappa \rho}{a c} F_{rad}$$
2) $$\int^{photosphere}_{ground} d T^4 = - \frac{3}{a c} F_{rad} \int^{photosphere}_{ground} \kappa \rho dr$$ (because F_rad is constant)
3) $$(T^4_{g} - T^4_{p}) = - \frac{3}{a c} F_{rad} ( \int^\infty_{ground} \kappa \rho dr - \int^\infty_{photosphere} \kappa \rho dr)$$
4) $$(T^4_{g} - T^4_{p}) = - \frac{3}{a c} F_{rad} (\tau_g - \frac{2}{3})$$
5) $$(T^4_{g} - T^4_{p}) = - \frac{3}{4} T_{g}^{4} (\tau_g - \frac{2}{3})$$
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.
Thanks!

Last edited: Nov 14, 2005
2. Nov 15, 2005

### SpaceTiger

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