---
title: "Calculate the escaping thermal tail"
description: "Derive the Jeans flux from a velocity-weighted Maxwellian, locate an illustrative exobase and distinguish particle fractions from surface fluxes."
canonical_url: "https://sandboxphysics.com/stories/atmospheric-escape/thermal-tails/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/atmospheric-escape/thermal-tails/"
last_modified: "2026-10-28"
---

[Planetary atmospheres from thermal tails to winds](https://sandboxphysics.com/stories/atmospheric-escape/)

AC13 · L2

# Calculate the escaping thermal tail

Sandbox Physics · 2026-10-28

The Jeans escape rate is the outward particle flux above escape energy, evaluated where further collisions have become rare. Its velocity weighting is the reason a simple count of fast particles gives the wrong answer.

## Before you begin

Read How a planet loses its atmosphere. You need spherical geometry and an integral as a sum over small velocity intervals; the angular and energy substitutions are shown below.

## Define the particle and the surface

Take a spherical surface of radius $r$ centered on a planet with gravitational parameter $GM$. Radius increases outward. A particle has mass $m$; the local number density $n$ is measured per cubic meter, and temperature $T$ is in kelvin. Introduce $c=\sqrt{kT/m}$, the one-component thermal speed scale. It is also the isothermal sound speed of this single-component ideal gas.

Assume that immediately below this surface collisions have maintained an isotropic Maxwellian with negligible bulk drift. Above it, approximate motion by collisionless central gravity. Ignore incoming particles from infinity. These assumptions describe an illustrative strong-binding exobase.

$$
f(\mathbf v)=\frac{n}{(2\pi c^2)^{3/2}}\exp\!\left[-\frac{v^2}{2c^2}\right],\qquad v_{\rm esc}=\sqrt{\frac{2GM}{r}}
$$

The velocity density integrates to n over all three-dimensional velocities. Escape requires total speed above the escape speed and a positive outward radial velocity.

Reading: Volkov et al. 2011, Thermally driven atmospheric escape, arXiv v1

## Weight each particle by how fast it crosses

In a short time $\Delta t$, particles with outward velocity $v_r$ cross the surface from a layer of thickness $v_r\Delta t$. Their contribution per area and time is therefore $v_r f\,d^3v$. Choose the polar angle from the outward radial direction. Then $v_r=v\cos\theta$ and the velocity-space volume element is $v^2\sin\theta\,dv\,d\theta\,d\phi$.

Integrate over the outward hemisphere: the azimuth contributes $2\pi$, and $\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=1/2$. The angular factor is therefore $\pi$. The remaining speed integral contains a third power, because the crossing speed contributes one extra factor of velocity.

$$
\Phi_{\rm J}=\frac{n\pi}{(2\pi c^2)^{3/2}}\int_{v_{\rm esc}}^{\infty}v^3e^{-v^2/(2c^2)}\,dv
$$

This is a number flux per area per time. The angular integral includes only outward trajectories.

$$
z=\frac{v^2}{2c^2},\quad v^3dv=2c^4z\,dz,\quad\lambda=\frac{GM}{rc^2}
$$

The lower integration limit becomes the binding parameter. An energy variable makes the exponential tail easy to integrate.

$$
\Phi_{\rm J}=\frac{nc}{\sqrt{2\pi}}\underbrace{\int_\lambda^\infty ze^{-z}dz}_{(1+\lambda)e^{-\lambda}},\qquad\dot M_{\rm J}=4\pi r^2m\Phi_{\rm J}
$$

Integration by parts gives the last factor. Multiplying by the sphere area and particle mass converts number flux to mass per second.

Reading: Volkov et al. 2011, Thermally driven atmospheric escape, arXiv v1

## Locate a surface where collisions run out

An exobase is a local collision criterion. Adopt an effective hard-sphere cross section $\sigma$ and mean free path $\ell=1/(\sqrt2 n\sigma)$. Compare it with a density scale height $H$, the distance over which density changes by a substantial fraction. Their ratio $\mathrm{Kn}=\ell/H$ is the local Knudsen number. This differs from a Knudsen number defined using a planet’s radius.

To locate an illustrative exobase, first use a static isothermal atmosphere below it. Balance the pressure gradient against gravity, substitute the ideal-gas pressure, then integrate between the reservoir radius and the candidate surface.

$$
\frac{d\ln n}{dr}=-\frac{GM}{c^2r^2},\quad n(r)=n_{\rm b}\exp\!\left[\lambda_{\rm b}\!\left(\frac{R_{\rm b}}r-1\right)\right],\quad H=\frac{c^2r^2}{GM}
$$

The logarithmic derivative is the relative density change per distance. Integrating its inverse-square dependence gives the density profile and local scale height.

Reading: Volkov et al. 2011, Thermally driven atmospheric escape, arXiv v1

## A worked strong-binding example

Use eight adopted Earth-mass units, a reservoir radius of two Earth radii, $T=1000\,\mathrm K$ and $n_{\rm b}=10^{15}\,\mathrm{m^{-3}}$. The fixed particle mass is $1.6735\times10^{-27}\,\mathrm{kg}$ and the effective cross section is $2\times10^{-19}\,\mathrm{m^2}$. These are declared synthetic inputs, not a retrieved atmosphere.

The base binding parameter is $30.3342$. Bracket the first root of the local condition $\mathrm{Kn}=1$ between the reservoir and the formal isothermal sonic radius. It lies at $r_{\rm ex}/R_{\rm b}=1.20447$, where $\lambda_{\rm ex}=25.1846$ and $n_{\rm ex}=5.80\times10^{12}\,\mathrm{m^{-3}}$. The flux-weighted escaping fraction is $3.02\times10^{-10}$. Substituting the local density, radius and thermal speed gives $\dot M_{\rm J}=9.96\times10^{-6}\,\mathrm{kg\,s^{-1}}$.

The display uses eight unbound paths so that you can see escape at all. Their total weight is $(1+\lambda_{\rm ex})e^{-\lambda_{\rm ex}}$; forty-eight bound paths carry the remaining weight. Counting the drawn amber dots would therefore grossly overestimate the real flux fraction. The path shapes follow central gravity, with accelerated display time.

## Know when to change the calculation

The hydrostatic profile tends to a nonzero density if extended to infinity. It is only used up to its first local exobase crossing, below the formal critical region. If no such crossing exists above a collisional base, the calculation does not manufacture an exobase.

A strong-binding threshold of six is used here to select a Jeans illustration. It is a teaching restriction on the approximation, not a universal boundary between escaping regimes. Kinetic calculations can include returning particles, non-Maxwellian populations and thermal conduction, and can differ from the simple Jeans value. A collisional bulk outflow instead needs continuity and momentum together; that is the next chapter.

Reading: Volkov et al. 2011, Thermally driven atmospheric escape, arXiv v1 · Cranmer 2004, New views of the solar wind with the Lambert W function, v1

## Exercises and explanations

1. Why is the escaping fraction of outward flux $(1+\lambda)e^{-\lambda}$ rather than the unweighted fraction of particles above escape speed?

   Read the explanation

   Fast outward particles sweep a thicker layer through the surface in the same time. Multiplying the velocity distribution by outward velocity changes the speed integral. The fraction refers to crossings, not to particles picked uniformly from a volume.

2. Double $n_{\rm b}$ at fixed gravity and temperature. Can you double the final Jeans mass loss without moving the exobase?

   Read the explanation

   Only if the evaluation surface is deliberately held fixed. With a physically located exobase, doubling density reduces the local Knudsen number at each radius, moving its crossing outward. Radius, binding and local density must all be recomputed.

3. Transfer the argument to a heavier molecule at the same temperature. Which two changes make escape harder?

   Read the explanation

   Its thermal speed falls as the inverse square root of mass, and its binding parameter increases in proportion to mass. The exponential tail factor is especially sensitive to the latter change. A realistic molecule also has a different collision cross section and possibly a different chemistry.

## Try the prediction

### [Locate the rare tail](https://sandboxphysics.com/labs/planetary-atmospheric-escape/)

1. Choose A rare escaping tail. Predict how increasing planet mass affects the exobase binding parameter.
2. Change only planet mass, export the calculation and inspect the exobase radius and flux weights.
3. Keep the temperature fixed and vary XUV irradiance. Explain why a prescribed Maxwellian temperature leaves this Jeans illustration unchanged.

[Previous: How a planet loses its atmosphere](https://sandboxphysics.com/stories/atmospheric-escape/leaving-a-planet/) [Next: Build a wind through its sonic point](https://sandboxphysics.com/stories/atmospheric-escape/critical-wind/)

## Sources and calculation scope

1. [Cranmer 2004, New views of the solar wind with the Lambert W function, v1](https://arxiv.org/pdf/astro-ph/0406176v1)

   7June2004; AJP 72,1397–1403. Full text, SectionsIII–IV, equations12–16: isothermal momentum equation, critical point and the two real Lambert-W branches. The solar ionized-hydrogen sound-speed convention is replaced by our explicitly fixed single-particle mass.

2. [Volkov et al. 2011, Thermally driven atmospheric escape, arXiv v1](https://arxiv.org/pdf/1009.5110v1)

   26September2010 preprint, published ApJL729,L24(2011). Full text, Section 2 and Figures1–5: hard-sphere mean free path, radius-based versus local scale-height Knudsen numbers, kinetic escape and conditional Jeans corrections. Their heating below the lower boundary differs from a globally maintained isothermal wind. The adopted Lab collision cross section is a teaching input, not their molecular-nitrogen value.