What is planetary surface temperature given constant sub-surface temperature? If a planet of radius $R_1$ has a constant sub-surface temperature $T_0$ at $R_0<R_1$, what is the long-term equilibrium surface temperature $T_1$?  Say we assume constant thermal diffusivity of the planet material, surface emissivity $1.0$, no atmosphere, and no incoming radiation.  I figure the temperature profile is harmonic, i.e. $T=a+b/R$, for constants $a$ and $b$, and that we can use $(R_0,T_0)$ to eliminate one of these constants.  Is there enough information to obtain the other constant?
 A: You want to equate the conductive energy flux at the surface which is a constant times T1-T0 -given your geometry and conductivity you can determine the coupling factor, with the radiative heat flux sigma*T**4. You could also add in the CMB, which is simply the same sigma*T**4 using the CMB temperature. Then you simply have a nonlinear algebraic equation in T0 to solve. It should be solvable by iteration without much effort.
A: That is easy: an object at temperature T0 in a zero temperature environment (no incoming radiation) will radiate till its temperature is zero. So the long term equilibrium is zero.
The question as posed describes a hypothetical situation, there is no environment at zero temperature. The universe is filled with CMB radiation of finite temperature that drops in time due to the cosmic (accelerated) expansion. So the real situation is a wee bit more complicated...
A: Sorry for the confusion.  R0 is some given particular radius between 0 and R1, e.g. R0=5000, R1=6000 km.  R is the general radial variable, as in T=a+b/R.  T0 is the temperature at position R0, assumed constant throughout time.  
A: Assuming your planet is uniform and at equilibrium your planet will be losing heat due to radiation at a rate of
$$\dot Q=\sigma\,A\,T^4=4\,\sigma\,\pi\,R_1^2\,T_1^4$$
This $\dot Q$ must be coming from the interior to maintain equilibrium and pass through all of the crust. Thus
$$\frac{dT}{dR}=-\frac{\dot Q}{k\,A}=-\frac{\dot Q}{4 k\,\pi R^2}$$
$$\int1 \;dT=\int-\frac{\dot Q}{4 k\,\pi R^2}\;dR$$
$$T=\frac{\dot Q}{8 k\,\pi R}+C$$
Then you can plug in for $\dot Q$ to get:
$$T=\frac{\sigma\,R_1^2\,T_1^4}{2 k\,R}+C$$
Then you can use $(R_0,T_0)$ to find $C$
$$T=\frac{\sigma\,R_1^2\,T_1^4}{2 k}(\frac1{R}-\frac1{R_0})+T_0$$
Now you can find $T_1$
$$T_1=\frac{\sigma\,R_1^2\,T_1^4}{2 k}(\frac1{R_1}-\frac1{R_0})+T_0$$
$$\frac{\sigma\,R_1^2}{2 k}(\frac1{R_1}-\frac1{R_0})\,T_1^4 -T_1+T_0=0$$
While there are exact analytic solutions to this forth order polynomial they are longer than would be prudent to post here, and evaluating them would most likely result in more numerical inaccuracy than a simple numerical method on this equation.
A simple recursive method that should converge quickly would be:
$$T_1^\prime=\sqrt[4]{\frac{2k(T_1-T_0)}{\sigma\,R_1^2(\frac1{R_1}-\frac1{R_0})}} $$
For values of:
$$R_1=4000mi$$
$$R_0=3999mi$$
$$T_0=300K$$
$$\sigma=5.67\times10^{-8}\frac{W}{m^2\,K^4}$$
$$k=5\frac{W}{m\,K}$$
And an initial guess of $T_1=200K$ yeilds $T_1=70.8K$ after 3 iterations (70.79020354 after 10)
A: I wouldn't think the planet's radius would have any effect on its temperature, especially if the temperature, as you say, is fixed for given planet.
