For $\lambda\gtrsim10$--$20$, escape is well described by the dilute high-velocity tail of a nearly hydrostatic exosphere. For $\lambda\lesssim2$--$3$, collisions couple the escaping gas into <hydrodynamic atmospheric escape>; the intermediate regime requires a kinetic or transonic calculation.

The hydrostatic-tail estimate follows by integrating a Maxwell distribution over outward velocities exceeding escape speed. The <Jeans escape flux> is
$$
\Phi_J=\frac{n_ev_{\rm th}}{2\sqrt\pi}(1+\lambda)e^{-\lambda},
\qquad
v_{\rm th}=\sqrt{\frac{2k_BT}{m}}.
$$
Therefore
$$
\boxed{\dot M_J
=4\pi R_e^2m\Phi_J
=4\pi R_e^2m\frac{n_ev_{\rm th}}{2\sqrt\pi}
(1+\lambda)e^{-\lambda}}.
$$
When $\lambda\leq1$, this is of order the free thermal supply $4\pi R_e^2\rho_ec_s$; quantitatively the flow is hydrodynamic and its conserved mass-loss rate is $\dot M=4\pi r^2\rho u$ through the transonic wind.

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 / iii / Solution

{c}

For exobase number density $n_e$ and $v_{\rm th}=\sqrt{2k_BT/m}$, the collisionless Jeans number flux is
$$
\Phi_J=\frac{n_ev_{\rm th}}{2\sqrt\pi}(1+\lambda)e^{-\lambda}.
$$


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