The <Jeans escape parameter> compares gravitational binding with thermal energy:
$$
\lambda(r)=\frac{GM_pm}{k_BTr}
=\frac{v_{\rm esc}^2}{2k_BT/m}.
$$
Hydrostatic thermal escape becomes strong when the Maxwellian tail is no longer exponentially small, roughly $\lambda\lesssim2$--$3$, and blow-off occurs for order-unity $\lambda$. Atomic hydrogen at an exobase radius $R_e$ therefore requires
$$
\boxed{T_e\gtrsim\frac{GM_pm_H}{k_BR_e}}
$$
for $\lambda\lesssim1$.

For a Neptune-mass planet this is about $3.4\times10^4\,\mathrm K$ at the optical radius $R_{\rm Nep}$. If the relevant base is the observed ultraviolet absorbing radius, approximately $10R_{\rm Nep}$, the corresponding value is about $3.4\times10^3\,\mathrm K$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 3 / b / ii / Solution

{c}
{title2=$\lambda$}
{wiki=Atmospheric_escape#Jeans_escape}

For particles of mass $m$ at radius $r$ and temperature $T$, the Jeans escape parameter is
$$
\lambda=\frac{GM_pm}{k_BTr}.
$$
Large $\lambda$ gives weak tail escape, while order-unity $\lambda$ indicates hydrodynamic blow-off.


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