Jeans escape

Builds on Maxwell–Boltzmann distribution Escape velocity

Definition

Consider a planet of mass $M$ with an atmosphere whose exobase, the level above which the gas is collisionless, lies at radius $r_c$ . Let the gas there consist of molecules of mass $m$ with number density $n_c$ and a Maxwell–Boltzmann distribution of velocities at temperature $T$ . Let $G$ be the gravitational constant and $k_B$ the Boltzmann constant. Jeans escape is the loss of the molecules at the exobase that move upward faster than the escape velocity $v_{\text{esc}} = \sqrt{2GM/r_c}$ . Let $v_0 = \sqrt{2 k_B T / m}$ be the most probable speed and $\lambda = v_{\text{esc}}^2 / v_0^2 = G M m / (k_B T r_c)$ the escape parameter. The rate of escape per unit area is the Jeans flux $\Phi_J = n_c \, \frac{v_0}{2\sqrt{\pi}} \, (1 + \lambda) \, e^{-\lambda}.$

Remarks

Check: let $f$ be the Maxwell–Boltzmann density of velocities $v$ with total $n_c$ . Then $\Phi_J$ is the integral of $v_z f(v)$ over all velocities with $v_z > 0$ and $\abs{v} > v_{\text{esc}}$ . In spherical velocity coordinates this integral evaluates to the formula above. Example: $\lambda$ is proportional to $m$ , so a molecule four times heavier has four times the escape parameter. For $\lambda$ of order $10$ to $20$ , the factor $e^{-\lambda}$ then makes its flux many orders of magnitude smaller. Non-example: in the hydrodynamic outflow of a heated atmosphere, $\lambda$ is small and the gas streams out as a fluid; this is not Jeans escape. Neither are escape processes driven by ions or by photochemistry.

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Tag	`1KDVH03Q3G`
URI	`m2://1KDVH03Q3G`
Kind	definition
Writing	W0 stub
MSC	85A20, 82C40
Defines	Jeans escape
Uses	Maxwell–Boltzmann distribution, Escape velocity
Changed	3 November 2026

Jeans escape

Definition

Remarks

Reading path

Further reading

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