I wrote an answer on Physics Stack Exchange that discussed this a bit. I'll present a shorter and more focused version here.

There are two main processes by which a planet can lose atmosphere: Jeans escape (for lighter particles) and dissociation/non-thermal escape (for heavier particles).

Jeans escape

The speeds of particles in an atmosphere are not uniform, but follow a statistical Maxwell-Boltzmann distribution. This means that some particles will have speeds greater than the escape velocity of the body at a given point, and these particles may escape the atmosphere. This motion is due to temperature, and so it is called thermal escape, or Jeans escape. The Jeans flux of escaping particles of mass $m$ at a distance $r$ from the center of the planet is $$\phi_J(m)\propto n_c\sqrt{\frac{2kT}{m}}\left(1+\frac{GMm}{kTr}\right)\exp\left(-\frac{GMm}{kTr}\right)\tag{1}$$ where $T$ is temperature, $M$ is the mass of the planet, and $k$ and $G$ are constants. Now, escape velocity at a distance $r$ is given by $$v_e=v_e(r)=\sqrt{\frac{2GM}{r}}\implies \frac{GM}{r}=\frac{v_e^2}{2}$$ We can therefore make some substitutions into $(1)$ and get $$\phi_J(m)\propto n_c\sqrt{\frac{2kT}{m}}\left(1+\frac{v_e^2m}{2kT}\right)\exp\left(-\frac{v_e^2m}{2kT}\right)\tag{2}$$ Let's say that the planet is Earth-like and the particles are molecular hydrogen (and thus likely to escape). We then have $$T\simeq288\text{ K},\quad m=3.3\times10^{-27}\text{ kg}\implies \frac{m}{2kT}\simeq4.15\times10^{-7}\text{s}^{-2}\text{ m}^{-2}$$ Therefore, $$\phi_J(m)\propto 1552n_c\left(1+4.15\times10^{-7}v_e^2\right)\exp\left(-4.15\times10^{-7}v_e^2\right)$$ Graphing this shows that planets with higher escape velocities have lower particle fluxes, and thus should retain more of their atmospheres. Conceptually, this should make sense. If two sets of particles have the same speed distribution, more will have speeds above the lower escape velocity than above the higher escape velocity.

Surface gravity on the whole does not matter because there are other variable parameters in play, as Cyrus wrote. Escape velocity at a certain radius (not necessarily at the surface) is the quantity you want to look at here.

Non-thermal escape

A planet can lose heavier atoms and molecules (such as $\text{O}_2$ and $\text{O}^+$) through processes involving the solar wind and the magnetosphere, which dissociate (i.e. break up) heavier molecules into ions and lighter molecules. This is one way Earth loses $\text{O}^+$ ions, mostly in the polar regions. The same can happen on other planets, including Mars and Venus. Obviously, mass and escape velocity matter here, but properties of the magnetic field (if the planet has one) may be much more important. We can assume that otherwise similar two planets with different escape velocities/surface gravities would lose gas roughly equally in this way.

Lithospheric chemical reactions

kingledion mention interactions between the lithosphere and the atmosphere, so I figured I might as well add something in about that. My reference is Kasting et al. (1993). The carbonate-silicate cycle removes carbon dioxide from the atmosphere via weathering: $$\text{CaSiO}_3+2\text{CO}_2+\text{H}_2\text{O}\to\text{Ca}^{++}+2\text{HCO}_3^-+\text{SiO}_2\tag{3}$$ This is an abiotic process, i.e. occurring without life. If there is life, organisms may take some of the byproducts and make shells out of calcium carbonate, making the entire cycle $$\text{CaSiO}_3+\text{CO}_2\to\text{CaCO}_3+\text{SiO}_2$$ As was the case with non-thermal atmospheric escape, none of this depends on planetary surface gravity or escape velocity - at first glance. However, the total amount of certain gases does depend on escape velocity, because more massive planets will keep more of these gases, so there is an impact of the mass.

posted over 7 years ago

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