= Solution
The circular <Keplerian orbit> has speed $v_K=(GM/r)^{1/2}$. Equating it with the heated gas <sound speed> gives
$$
\boxed{r_g=\frac{GM}{c^2}.}
$$
This 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 $r\geq r_g$, this more than compensates the circular-orbit binding energy. Equivalently the hot hydrostatic scale height satisfies $H_{\mathrm{hot}}/r\sim c/v_K\gtrsim1$, 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 $\sqrt2v_K$. Detailed wind launching can change the numerical critical radius by factors of order unity. Using exactly the supplied numerical estimates, $c=10^6\,\mathrm{cm\,s^{-1}}$, and therefore
$$
r_g\simeq\frac{(7\times10^{-8})(2\times10^{33})}{10^{12}}\,\mathrm{cm}
=1.4\times10^{14}\,\mathrm{cm}=\boxed{14\,\mathrm{AU}}.
$$
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