Aliens around Tau Ceti

Imagine replacing the planet by a huge thin sphere with the star at its centre. The sphere has the same radius as the planet's orbital radius $R_{p}$ and we imagine that the star and the sphere are in thermal equilibrium. We will calculate the radii of the sphere when it has a temperature of 273 (freezing point of water) and 373 K (boiling point of water). Since the supplied data leads us to hope that the size of the planet does not affect the answer, it is not too much of a stretch to believe that the difference between these radii is the answer we are looking for.

There are three key ideas. First of all the power radiated by a black body is proportional to the fourth power of its absolute temperature and directly proportional to its area and so to the square of its radius. Secondly we assume that the star and the sphere are in thermal equilibrium. This means that sphere must emit as much power as it receives (other wise it would heat up or cool down and we would not be in equilibrium!). Lastly note that the sphere will emit half its energy back towards the star, and the other half will be radiated away from the outside of the sphere into the rest of the universe. In other words both its inside and outside surface must be included in the calculation - this is the source of the factor of 2.

Putting all this together using s and p as suffices for 'star' and 'planet' we have

$T_{s}^4R_{s}^2=2 T_{p}^4 R_{p}^2\\\implies R_{p}=\frac{R_{s}T_{s}^2}{2T_{p}^2}$

and from this we get the thickness of the habitable shell

$\frac{T_{s}^2 R_{s}}{2}\left(\frac{1}{T_{freezing}^2}-\frac{1}{T_{boiling}^2}\right)\\=0.5 \times 5344^2 \times 0.79 \times 695500 \times \left(\frac{1}{273.15^2}-\frac{1}{373.15^2} \right)=\boxed{48808109\text{ km}}$

According to the formula for black body radiation, and if we ignore albedo and emissivity, the temperature of a planet, Tp, depends only on the surface temperature of the sun, Ts, the radius of the sun, R, and the distance between the centers of the sun and the planet, D.

One further constraint, that of liquid water, implies that the surface temperature of the planet must be between 0 and 100 C, or 273 and 373 K. Then, Tp = Ts (R/2D)^0.5. Solving for D, D = 0.5 R (Ts/Tp)^2. R = 695500 * 0.79 = 549445. At the high end, Tp = 373, and the quantity in parentheses squared is 205.265. At the low end, this quantity is 383.184.

To find the thickness of the shell, we multiply the difference between these two numbers by 0.5 R, getting 48878410. This is closer than .01% of the listed correct answer, which means it agrees to 4 significant digits.