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The circular Keplerian orbit has speed $v_K=(GM/r{)}^{1/2}$. Equating it with the heated gas sound speed givesThis is the photoevaporative gravitational radius, where thermal and orbital binding energies have the same order of magnitude. A circular orbit has specific mechanical energy $-GM/(2r)$. Heating adds thermal energy and, in a fluid outflow, available specific enthalpy of order $c^2$. For example, if $c$ is the adiabatic sound speed, an ordinary ideal gas has enthalpy $c^2/(\gamma -1)$; for $\gamma =5/3$ this is $3c^2/2$. At $r\ge r_g$, this more than compensates the circular-orbit binding energy. Equivalently the hot hydrostatic scale height satisfies $H_{\mathrm{hot}}/r\sim c/v_K\gtrsim 1$, so a thin bound surface layer cannot be maintained. With continued irradiation, the gas can expand into a thermal wind: photoevaporation removes disk material.

The condition is a thermal binding scale, rather than an assertion that the sound speed equals the ballistic escape speed, which is $\sqrt{2}{v}_K$. Detailed wind launching can change the numerical critical radius by factors of order unity. Using exactly the supplied numerical estimates, $c=10^6\mathrm{cm}{\mathrm{s}}^{-1}$, and therefore

The wind removes $2\pi rW(r)dr$ of mass per unit time from an annulus; $W$ is already the surface-density loss term in the supplied mass conservation equation, so there is no additional two-face factor. Integrating the photoevaporation profile givesThe convergence at infinity is important: the loss is concentrated near the photoevaporative gravitational radius. Using $r_g=1.4\times 10^{14}\mathrm{cm}$ yields ${\dot{M}}_w\simeq 2.46\times 10^{17}\mathrm{g}{\mathrm{s}}^{-1}$, or about $3.9\times 10^{-9}{M}_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$. The initial disk mass is $2\times 10^{30}\mathrm{g}$, so the wind-only depletion time isThis estimate treats the heated area and wind normalization as fixed and neglects additional removal through stellar accretion. Once the disk shrinks, its wind rate and geometry need not remain constant.

When a protoplanetary disk's inward viscous supply approaches its wind-loss rate, photoevaporation can prevent replenishment of the inner disk. That region drains on a viscous timescale, leaving a gap or hole. A negative density in a formal steady state calculation signals the breakdown of that approximation, not a physical continuation. An exposed outer edge can subsequently disperse rapidly.

The thermal binding scale defined by equating heated gas sound speed to circular Keplerian orbit speed. It is $r_g=GM/c_s^2$. The circular-orbit binding energy is $GM/(2r)$ per unit mass, while the gas specific enthalpy is of order $c_s^2$. A heated disk surface at radii comparable to or larger than $r_g$ can therefore drive photoevaporation. The actual launching radius can differ by a factor of order unity; $c_s$ is not the ballistic escape speed.