It is my understanding that, in its gaseous state, oxygen molecules move fast enough to achieve escape velocity. On Earth, we see this more clearly with helium. Regardless of what's happening on Earth, my question is about Mars.

Given the mass/gravity of Mars, would it be able to hold free oxygen (O2) in a usable amount for a meaningful amount of time?

For purposes of this question, a usable amount would be Earth-like levels. A meaningful amount of time would be "worth the expense of getting it there for that length of time."

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    $\begingroup$ Re It is my understanding that, in its gaseous state, oxygen molecules move fast enough to achieve escape velocity. Your understanding is incorrect. This question is based on a false premise. $\endgroup$ – David Hammen Mar 19 at 0:09
  • $\begingroup$ @DavidHammen Well, my understanding may not be correct. Not a first for me. That is also why I ask questions. $\endgroup$ – Leezard Mar 19 at 0:19

Quick answer: yes.

The question is tricky because there are many (pun not intended) moving parts. Planets can lose atmosphere through several processes. The classic one you mention, Jeans escape, depends on gas molecules gaining escape velocity. This depends on the temperature of the exosphere which depends on a lot of things: it is not entirely trivial to estimate. There is also hydrodynamic escape when there is enough heating (mostly by UV) so that light gases begin to flow upward (possibly dragging other ones with them), non-thermal losses (charge escape and polar wind), impact loss (splashes when big impactors hit), and sputtering when there is no magnetic field (charged particles in the solar wind carry away atoms).

In the case of current Mars measurements suggest that the oxygen atom loss is around (talk 1): 1 to $4\cdot 10^{25}$ per second (Phobos-2 Taus and Aspera), $<10^{25}$ sputtering contribution, 2 to $6\cdot 10^{25}$ (deduced From Ionosphere), 0.3 to $4 \cdot 10^{26}$ (direct photochemical O escape), $0.3\dot 10^{25}$ (global MHD model), (talk 2): $4\cdot 10^{25}$ (with high variability). So a rough median guess is some $10^{25}$ atoms per second. That is about a quarter kg per second.

Now, were we to add a lot of mass to the atmosphere there would be proportionally more mass loss. As a crude estimate, suppose the surface pressure increases 166 times to one earth atmosphere. That would require an equal increase in mass (ignoring any nonlinearities and that an Earth atmosphere has a different molecular weight). So that would presumably increase the mass loss by 166 times - almost 42 kg per second. Which is not much at all. By this kind of super-crude calculation everything looks fine. Doing it slightly more carefully produces the same answer for at least Jeans escape, although it shows that especially the inevitable temperature changes will boost things a lot. The formation of a hot exosphere is also likely to make the loss much higher. But on a human timescale the atmosphere will remain.

A rough rule from Martyn Fogg's excellent terraforming book is that if the thermal velocity is $v_t=\sqrt{3kT/\mu}$ and the escape velocity is $v_e=\sqrt{2GM/R}$, then the exosphere loses $1/e$ of the gas over a time of days if $v_e/v_t=3$, decades when $v_e/v_t=4$, tens of millennia when $v_e/v_t=5$ and billions of years for $v_e/v_t=6$. For $\mu=5.31\cdot 10^{-26}$ kg, $T=1000$ K, we get $v_t=883.1916$ m/s and $v_e=5022.1$ m/s. $v_e/v_t=5.6863$ in this example, so we should expect a terraformed atmosphere to remain over millions of years at least.


Ignoring escape, by whatever means, let's look at how much mass is needed to at least temporarily provide an atmosphere with a partial oxygen pressure of 0.21 atmospheres (the partial pressure of oxygen at sea level on Earth) at Mars surface. This is a simple calculation:

$$m_{\text{oxygen}} = \frac{A_\text{Mars}\,p_\text{oxygen}}{g_\text{Mars}}$$


  • $A_\text{Mars}$ is Mars's surface area,
  • $p_\text{oxygen}$ is the desired partial pressure of oxygen,
  • $g_\text{Mars}$ is gravitational acceleration at the surface of Mars, and
  • $m_\text{oxygen}$ is the quantity of oxygen needed to accomplish this.

The result is 830 exagrams ($8.3\times10^{18}$ grams) of oxygen. Another way to look at it: It's two trillion times the mass of the International Space Station. Yet another way to look it: It's the mass of oxygen in 1.9 million comets made of pure ice, each with a diameter of 1 km.

Now how are you reasonably going to accomplish even a thousandth of that?

  • $\begingroup$ I assume the reasonable (and quite physically allowed) approach is to use self-replicating robots/solar sails to move much of the Kuiper belt object to Mars. That said, the question seems to be more about whether it would stay for long. $\endgroup$ – Anders Sandberg Mar 19 at 16:21

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