I was wondering if a planet that has a higher escape velocity but lower surface gravity lose its atmosphere more faster/slower than a planet that has a higher surface gravity but lower escape velocity? Let's assume both planets have magnetic fields.

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    $\begingroup$ This question may be more suited for Physics SE or Space SE. Did you check if it wasn't already asked there? $\endgroup$ – Babika Babaka Aug 1 '16 at 12:34
  • $\begingroup$ Would not the two variables you mention be related by the mass of the planet and the radial distance from the center of the planet? i.e. $v_e = sqrt(2gr)$ $\endgroup$ – Serban Tanasa Apr 28 '17 at 14:13

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.

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    $\begingroup$ Since you are summarizing non-thermal 'escape' mechanisms, there is also a non-zero contribution from chemical reactions with the lithosphere. For example, most of the carbon dioxide in the young Earth's atmosphere dissolved in water and ended up as carbonate minerals. Had the Earth been too light to keep its nitrogen, but large enough to keep its carbon dioxide, the presence of water might have removed what existed of the carbon dioxide atmosphere. $\endgroup$ – kingledion Apr 27 '17 at 19:20
  • $\begingroup$ @kingledion That's a good point; I had been focusing on space-based escape. I seem to recall having a pdf on lithospheric reactions. I'll see if I can dig it up. $\endgroup$ – HDE 226868 Apr 27 '17 at 19:30
  • $\begingroup$ @kingledion I made some edits. $\endgroup$ – HDE 226868 Apr 28 '17 at 13:35

It's impossible to say based on only those parameters.

First of all, Escape Velocity is not a constant value a planet just has, it depends on how far from the planet's center of mass the initial point is. Usually, the initial point is assumed to be at the planet's surface. With that in mind, your described situation doesn't make sense. If the gravity at the surface is lower on planet A, the escape velocity from that surface will also be lower. This can be seen if you compare the formulas to calculate the two values:

Escape Velocity: $ v_{e}={\sqrt {\frac {2GM}{r}}} $

Surface gravity: $ g(r)=-{\frac {GM}{r^{2}}} $

Where G is the universal gravitational constant, M the mass of the body to be escaped, and r the distance from the center of mass of the mass M to the object. for surface gravity, all of the mass of the planet is assumed to be inside radius $ r $.

To have a lower escape velocity from the surface, planet A would have to have lower $M / r$ than planet B, by some combination of smaller $M$ and/or bigger $r$. If you apply those to the formula for gravity, it will always give you a smaller value as well.

If you look instead at the escape velocity at a certain distance $x$ from the planet's center of mass that is the same for both planets and higher than radius $r$ of the bigger planet, you could have a higher escape velocity for planet A despite it having a lower surface gravity if the planet had more mass in total, but was much less dense. Unfortunately, that doesn't say anything about the atmosphere, since $x$ might be well outside the upper atmosphere for the smaller planet, making that value all but irrelevant.

It gets even more complicated because the amount of gas in the atmosphere and its density also plays a big role. Venus for example has slightly lower surface gravity (8.87 vs 9.81 $m/s^{2}$), but the pressure is 90 times as high as Earth's at the surface and the atmosphere extends 220-350 km above the surface as opposed to 100 km for Earth. This probably means Venus loses atmosphere faster than Earth, since the escape velocity at 300 km above the surface of Venus is lower than that 100 km above Earth's surface. However, that's not going to be a relief for any unlucky observer on Venus for a very long time...

When you start comparing wildly different planets with different compositions and atmospheres, you need to plug all of those into the calculations to say which one might be losing atmosphere faster.

  • $\begingroup$ Not only the radius, also the mass is a variable. The escape velocity depends on $M/r$, the surface gravity on $M/r^2$. Therefore it is possible to make one larger and the other smaller. For example, if you make the mass 4 times as large, and the radius 2 times as large, the surface gravity will not change, but the escape velocity will grow by about 40%. If you make the mass only 3 times as large, you'll see a 22% increase of escape velocity, and a 25% decrease in surface gravity. $\endgroup$ – celtschk Aug 1 '16 at 14:45
  • $\begingroup$ Duh, I will fix my answer when I get home. I did a quick check with some different numbers and skipped a more thorough check. Guess it's not saving me any time after all. Maybe I should delete all that and just go with "Atmosphere composition counts for a lot too." $\endgroup$ – Cyrus Aug 1 '16 at 15:01
  • $\begingroup$ @celtschk You didn't use term, but this shows how density plays a role. That will account for the different masses. The two planets in the OP's question would have to have different densities. Atmospheric composition also counts for a lot. So does temperature. But the question focused on escape velocity and surface gravity. Good answer for the question. $\endgroup$ – a4android Aug 2 '16 at 4:13
  • $\begingroup$ @a4android: Indeed, if you rewrite the formulas to densities, you find $v_e\propto r\sqrt{\rho}$ and $g\propto r\rho$. So at constant density, $r$ changes both values in the same way, but at constant radius, a different density affects the two values differently. BTW, another nice formula you can derive from those formulas is $v_e=\sqrt{2|g|r}$. That's exactly the velocity that you would need in a homogeneous gravitational field of strength $|g|$ to reach a height of $r$. $\endgroup$ – celtschk Aug 2 '16 at 6:54
  • $\begingroup$ Does this mean a rocket at a surface gravity |g| it needs velocity ve to reach radius r? I want to confirm this because it is relevant to something I am tinkering with elsewhere. I like the way this shows how density effects the two values. Thanks for the additional information. $\endgroup$ – a4android Aug 3 '16 at 5:39

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