 Solution

Assume uniform reradiation over the planetary surface, unit thermal emissivity, and zero Bond albedo. The planet absorbs over its projected disc and emits over its full sphere. Radiative equilibrium and the Stefan–Boltzmann law giveUsing the liquid-water range $273.15\le T_p\le 373.15\mathrm{K}$ as the crude thermal criterion givesFor nonzero Bond albedo $A_B$, both radii acquire the factor $\sqrt{1-A_B}$. For example $A_B=0.3$ gives approximately $0.470$ to $0.877\mathrm{AU}$. An airless planet need not have uniform surface temperature; inefficient heat redistribution makes the dayside hotter and the nightside colder, so a single equilibrium temperature is an averaging approximation.

This simple habitable zone contains the Earth's orbit under the zero-albedo assumption, but also contains Venus's orbit, although its greenhouse effect produces an uninhabitably hot surface. Mars lies beyond the calculated outer boundary. These comparisons show the usefulness of the orbital scale and the limitations of a temperature-only criterion. The stipulated airless model also omits the atmospheric pressure needed for stable surface liquid water: this is an effective thermal window, not a sufficient habitability condition.