We can solve this equation for $S_i$, which yields

$$S_i = S_{p,i} = \frac{1}{1 + \frac{1}{\sum_l \frac{A^{\gamma_p, p_{p,i}}}{aV_g} \sqrt{\frac{RT}{2\pi m_{p,i}}}}} \quad (A24)$$

The equilibrium vapor pressure cancels out in this equation, and the only element-specific parameters that remain are $x_{\gamma,i}$, $\gamma_{\gamma,i}$, and $m_{\gamma,i}$. We have little constraints on the evaporation coefficients at very high temperature, and we assume that all $\gamma_{\gamma,i}$ converge to the same value $\tilde{\gamma}$ (see main text for details). If we consider volatile elements of similar masses (e.g., for K and Rb we have $\sqrt{39} \approx 6$ and $\sqrt{85.5} \approx 9$, respectively) and we set $m_{\gamma,i}$ to a constant value $\tilde{m}$ in Equation (A24), all species of all elements will have approximately the same saturation (we use $\sum_l x_{\gamma,i} = 1$),

$$S \approx \frac{1}{1 + \frac{aV_g}{A\tilde{\gamma}} \sqrt{\frac{2\pi\tilde{m}}{RT}}} \quad (A25)$$

### A.3. Jeans escape from the lunar atmosphere

The mode of atmospheric escape of volatile species depends on the Jeans parameter,

$$\lambda_e = \frac{4G\rho\pi r^3 m}{3kT(r + Z_e)}, \quad (A26)$$

where $m$ is the mass of the vapor atom ($3.8 \times 10^{-26}$ kg for Na), $r$ and $\rho$ are the radius and mean density of the planetary body ($1737 \times 10^3$ m and $3350$ kg m$^{-3}$ for the Moon, respectively), $G$ is the gravitational constant ($6.67 \times 10^{-11}$ m$^3$ kg$^{-1}$ s$^{-2}$), $k$ is Boltzmann's constant ($1.38 \times 10^{-23}$ m$^2$ kg s$^{-2}$ K$^{-1}$), $T$ is the temperature (in K), and $Z_e$ is the altitude above ground of the exobase (in m). This parameter compares the gravitational energy to the thermal energy of atoms in the atmosphere. If $\lambda_e > \sim 6$, volatile loss proceeds through Jeans thermal escape where atoms/molecules in the high-energy tail of the Boltzmann distribution at the exobase can escape (Volkov et al. 2011). If $\lambda_e < \sim 2$, the mean thermal energy is comparable to the gravitational energy and volatile loss proceeds though hydrodynamic (blowoff) escape where atoms are lost in bulk to space (Volkov et al. 2011). For an Na atmosphere around the Moon at a temperature of 1500 K, we calculate $\lambda_e \sim 5$.

Let us consider a model atmosphere of uniform temperature (T in K) made of a single atomic vapor species (e.g., Na for the Moon) that follows the ideal gas law $P = NkT$, where $P$ is the pressure (in kg m s$^{-2}$), $k$ is Boltzmann's constant ($1.38 \times 10^{-23}$ m$^2$ kg s$^{-2}$ K$^{-1}$), and $N$ is the atom density (in atoms m$^{-3}$). The Jeans escape flux ($\phi_e$ in atoms m$^{-2}$ s$^{-1}$) at the exobase is given by

$$\phi_e = \frac{P_e}{\sqrt{2\pi mkT}} \left[ 1 + \frac{4m\pi r^3 \rho G}{3(r + Z_e)kT} \right] e^{-\frac{4m\pi r^3 \rho G}{3(r + Z_e)kT}}, \quad (A27)$$

where $P_e$ is the pressure at the exobase and $Z_e$ is the altitude above ground of the exobase (in m). We consider an atmosphere that is in hydrostatic equilibrium in a constant gravitational acceleration. The pressure is related to the altitude

$z$ (in m) through

$$P = P_0 e^{-\frac{4\pi r \rho G m z}{3kT}} \quad (A28)$$

The scale height ($H$ in m) of such an atmosphere is

$$H = \frac{3kT}{4\pi r \rho G m} \quad (A29)$$

The exobase is the location where the mean free path $l = kT/(\sqrt{2} sP)$ ($s = \pi d^2$ is the collision cross section for the atoms, in m$^2$, with $d$ the atom diameter; 380 pm for Na) is equal to the scale height of the atmosphere ($l = H$). We therefore have

$$P_e = \frac{4\pi r \rho G m}{3\sqrt{2} s} \quad (A30)$$

We can then estimate $Z_e$ using Equation (A28),

$$Z_e = H \ln \frac{P_0}{P_e} = \frac{3kT}{4\pi r \rho G m} \ln \frac{3\sqrt{2} s P_0}{4\pi r \rho G m} \quad (A31)$$

Introducing these expressions (Equation (A30) for $P_e$ and Equation (A31) for $Z_e$) into Equation (A27) yields

$$\phi_e = \frac{2r\rho G}{3s\sqrt{kT}} \left[ 1 + \frac{4m\pi r^3 \rho G}{3rkT + \frac{9k^2T^2}{4\pi r\rho Gm} \ln\left(\frac{3\sqrt{2} s P_0}{4\pi r\rho Gm}\right)} \right] \\ \times e^{-\frac{4m\pi r^3 \rho G}{3rkT + \frac{9k^2T^2}{4\pi r\rho Gm} \ln\left(\frac{3\sqrt{2} s P_0}{4\pi r\rho Gm}\right)}} \quad (A32)$$

At steady state (a valid assumption given that the hydrostatic timescale is $\sim 1$ hr compared to a 1000 yr lifetime for the free-radiating magma ocean), another equation relating $\phi_e$ to $P_0$ can be derived by equating the evaporative flux at the magma ocean/atmosphere boundary ($\phi_0$ in atoms m$^{-2}$ s$^{-1}$) with the Jeans escape flux,

$$\phi_0 r^2 = \phi_e (r + Z_e)^2. \quad (A33)$$

The evaporation flux from the top of the magma ocean into the atmosphere is governed by the Hertz-Knudsen equation (we assume that the atmosphere is at the magma ocean temperature),

$$\phi_0 = \gamma \frac{P_{\text{eq}} - P_0}{\sqrt{2\pi mkT}}, \quad (A34)$$

where $\gamma$ is the evaporation coefficient (0.26 for Na; Fedkin et al. 2006), $P_{\text{eq}}$ (in Pa or kg m$^{-1}$ s$^{-2}$) is the equilibrium vapor pressure of the element in the atmosphere, and $P_0$ is the actual vapor pressure at the interface between the magma ocean and atmosphere. This equation formalizes the fact that the vapor must be undersaturated for a net evaporative flux to be present. The equilibrium vapor pressure of an element like Na depends on temperature following the formula given in the main text $P_{\text{Na,eq}}(T) = \Gamma_{\text{NaO}_0.5} X_{\text{NaO}_0.5} e^{-\Delta G_{\text{Na}}/RT} / f_{O_2}^{1/4}$, where $\Gamma_{\text{NaO}_0.5}$ is the activity coefficient of NaO$_{0.5}$ in the melt, $X_{\text{NaO}_0.5}$ is its mole fraction in the melt, $\Delta G_{\text{Na}}$ is the standard Gibbs free energy of the vaporization reaction NaO$_{0.5,l} \leftrightarrow Na_v + 0.25O_{2,v}$, and $f_{O_2}^{1/4}$ is the oxygen fugacity (activity coefficient, standard Gibbs free energy, and oxygen fugacity would all have some temperature
dependence). Injecting the expression for $Z_e$ given above (Equation (A31)), we have

$$ \phi_e = \gamma \frac{P_{eq}(T) - P_0}{\sqrt{2\pi mkT}} \left( r + \frac{3kT}{4\pi r \rho Gm} \ln \frac{3\sqrt{2}s P_0}{4\pi r \rho Gm} \right)^2. \quad (A35) $$

We have two Equations (A32) and (A35) in two unknowns ($\phi_e$ and $P_0$). This system cannot be solved analytically, but it can be solved numerically. The result of the numerical solution shows that for larger bodies the surface pressure is always close to the equilibrium vapor pressure, so Equation (A32) yields the approximate expression

$$ \phi_e \approx \frac{2r\rho G}{3s} \sqrt{\frac{m\pi}{kT}} \left[ 1 + \frac{4m\pi r^3 \rho G}{3rkT + \frac{9k^2 T^2}{4\pi r \rho Gm} \ln\left(\frac{3\sqrt{2}s P_{eq}}{4\pi r \rho Gm}\right)} \right] \times e^{-\frac{4m\pi r^3 \rho G}{3rkT + \frac{9k^2 T^2}{4\pi r \rho Gm} \ln\left(\frac{3\sqrt{2}s P_{eq}}{4\pi r \rho Gm}\right)}}. \quad (A36) $$

The total escape flux per unit time integrated over the surface area of the exobase is

$$ \begin{aligned} \phi_{e,tot} &= \phi_e 4\pi(r + Z_e)^2 \\ &= \phi_e 4\pi \left( r + \frac{3kT}{4\pi r \rho Gm} \ln \frac{3\sqrt{2}s P_0}{4\pi r \rho Gm} \right)^2, \end{aligned} \quad (A37) $$

which can be approximated as

$$ \phi_{e,tot} \approx \phi_e 4\pi \left( r + \frac{3kT}{4\pi r \rho Gm} \ln \frac{3\sqrt{2}s P_{eq}}{4\pi r \rho Gm} \right)^2. \quad (A38) $$

At 1700 K, the calculated $\phi_{e,tot}$ value is $3.0 \times 10^{29}$ atoms s$^{-1}$, translating into $5.0 \times 10^5$ moles s$^{-1}$. We used this formula to calculate the amount of Na lost from a BSE-like proto-Moon for a duration of ~1000 yr at 1700 K and obtained a total Na loss of $1.57 \times 10^{16}$ moles. Compared with initial inventory of Na in the Moon ($8.5 \times 10^{21}$ moles), we find that the fraction of Na lost from the Moon by Jeans escape would be entirely negligible, which is consistent with the conclusion of Saxena et al. (2017).

## ORCID iDs

Nicolas Dauphas https://orcid.org/0000-0002-1330-2038

Nicole X. Nie https://orcid.org/0000-0003-1828-5377

Marc Blanchard https://orcid.org/0000-0003-2897-9575

Zhe J. Zhang https://orcid.org/0000-0001-8076-5094

Justin Y. Hu https://orcid.org/0000-0002-3705-8148

Merlin Meheut https://orcid.org/0000-0003-2037-2365

Channon Visscher https://orcid.org/0000-0001-6627-6067

Robin Canup https://orcid.org/0000-0002-2342-3458

Timo Hopp https://orcid.org/0000-0002-4056-5431

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