Worked exercise — Hydrogen escape from the ionosphere

Problem

In the ionosphere at $T \approx 1000$ K, the fastest molecules can escape Earth’s gravity. The escape velocity is $v_f = 11{,}2$ km/s. Estimate the fraction of H$_2$ molecules with $v > v_f$, and compare with nitrogen.

Solution

The tail of the Maxwell-Boltzmann distribution above a speed $v_f$ scales as $e^{-m v_f^2/(2 k_B T)}$. It is convenient to compute the exponent: this is what tells us whether the fraction is extremely small, or merely very small. For H$_2$ (molecular mass $m \approx 2\cdot 1{,}66\cdot 10^{-27}$ kg $= 3{,}32\cdot 10^{-27}$ kg): $\frac{m v_f^2}{2 k_B T}= \frac{3{,}32\cdot 10^{-27}\cdot (11{,}2\cdot 10^3)^2}{2\cdot 1{,}38\cdot 10^{-23}\cdot 1000}\approx 15{,}1$ So a fraction $\sim e^{-15{,}1} \approx 3\cdot 10^{-7}$ of the molecules exceeds the escape velocity: small, but not zero. Over billions of years, the exponential tails empty the atmosphere of its hydrogen: this is why today there is almost no free hydrogen in the air, even though it is the most abundant element in the universe. For N$_2$ the molecular mass is $14$ times larger: the exponent becomes $\approx 211$, and the fraction $e^{-211}$ is enormously small, indistinguishable from zero in any physical sense (not enough nitrogen molecules have ever escaped in total, in the whole history of the universe). The atmosphere firmly retains nitrogen.

The moral is twofold. First: small exponential numbers, integrated over astronomical timescales, can change the composition of an atmosphere. Second: two gases at the same temperature have different distributions for purely kinetic reasons, and this decides the fate of a planet. $\ev{\text{H}_2:\ \text{exponent}\approx 15{,}1\Rightarrow \text{fraction}\sim e^{-15{,}1}\approx 3\cdot 10^{-7};\quad \text{N}_2:\ \approx 211\Rightarrow e^{-211}\approx 0}$

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Topics: Kinetic theory of gases Concepts: Maxwell-Boltzmann distribution Skills: Fermi estimates · Limiting-case and speculative-physics analysis