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Jeans escape is collisionless atmospheric escape by particles in the high-speed tail of a thermal distribution at the exobase. The Jeans escape parameter $\lambda =GMm/(k_{B}T{r}_e)$ controls the exponential suppression, and the Jeans escape flux is proportional to $(1+\lambda )e^{-\lambda }$. There is no exact positive-temperature threshold below which the tail vanishes. For $\lambda$ of order a few or smaller, a purely hydrostatic tail model can fail and hydrodynamic atmospheric escape becomes important.

Three physically distinct mechanisms are as follows.

Thermal-tail escape, a bulk hydrodynamic wind, and nonthermal particle energization are three distinct channels; a detected tail need not distinguish them alone. Jeans escape and hydrodynamic atmospheric escape are both thermal mechanisms in the broad sense, but have different distribution functions and dynamical assumptions.

In the collisionless exosphere, an atom escapes if its outward trajectory has positive total mechanical energy. Neglect tides and stellar forces and use Newtonian gravity at exobase radius $r_e$:With thermal speed $v_{\mathrm{th}}=\sqrt{2k_B{T}_e/m}$, the Jeans escape parameter isA thermal distribution always has an escaping tail; Jeans escape is exponentially suppressed for ${\lambda }_e\gg 1$, with Jeans escape flux proportional to $(1+{\lambda }_e)e^{-{\lambda }_e}$. Efficient escape requires ${\lambda }_e$ of order a few or smaller, with an order-unity energetic estimate $k_B{T}_e\sim G{M}_{p}m/r_e$.

Assume a Neptune-like mass and radius, $r_e\simeq R_p$, and atomic hydrogen. Using the supplied rounded constants givesHenceUsing the mean kinetic energy $3k_{B}T/2$ instead gives an order-unity coefficient and $T\simeq 2.3\times 10^4\mathrm{K}$. An expanded exobase has weaker binding and lowers the estimate by $R_p/r_e$. A comet-like tail can also be shaped by radiation pressure and stellar-wind interactions; it does not by itself measure $T_e$ or prove that a hydrostatic Jeans model is valid. At ${\lambda }_e\sim 1$, $H/r_e\sim 1$ and hydrostatic equilibrium fails as a global description: substantial mass loss must usually be treated as hydrodynamic atmospheric escape.

The exobase is defined by mean free path $\ell \sim H$, not by a universal pressure. For a neutral hydrostatic gas with collision cross-section ${\sigma }_c$,For example, explicitly assuming ${\sigma }_c\sim 10^{-19}\text{–}10^{-20}{\mathrm{m}}^2$ gives $P_e\sim 2\times 10^{-7}\text{–}2\times 10^{-6}\mathrm{Pa}$, or $2\times 10^{-12}\text{–}2\times 10^{-11}\mathrm{bar}$. These are representative extremely dilute neutral-exobase pressures, with orders of magnitude varying with composition, cross-sections and expansion. The supplied constants contain no collision information, so they cannot uniquely determine an exobase pressure; ionization or a non-hydrostatic density profile changes this estimate.

Comparable hydrostatic thermal escape requires comparable Jeans escape parameter $\lambda =G{M}_{p}m/(k_{B}T{R}_{\mathrm{exo}})$. For the same escaping species and $R_{\mathrm{exo}}\simeq R_p$,Thus Jupiter at 5 au needs an exobase temperature at least about thirty times Earth's at the same irradiation to have comparable Jeans escape flux.

For the inner planet, the usable EUV power is $\eta \pi R_{\mathrm{exo}}^2{F}_{\mathrm{EUV}}$. If the binding energy per unit escaping mass is $G{M}_p/R_p$, energy-limited atmospheric escape givesThe time to lose a fraction $x$ of the planetary mass is thereforeThis neglects Roche-lobe reduction, radiative cooling, changes in radius and flux, and the planet's orbital evolution. If the gas is lifted only from $R_{\mathrm{exo}}$, replace $R_p$ in the denominator by $R_{\mathrm{exo}}$.

The Jeans escape parameter compares gravitational binding with thermal energy:Hydrostatic thermal escape becomes strong when the Maxwellian tail is no longer exponentially small, roughly $\lambda \lesssim 2$--$3$, and blow-off occurs for order-unity $\lambda$. Atomic hydrogen at an exobase radius $R_e$ therefore requiresfor $\lambda \lesssim 1$.

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