What is the minimum planetary mass to hold an atmosphere over geologic time scales?

First, kudos to you for realizing that planetary mass is not the only thing influencing how a planet (or even if!) a planet holds on to its atmosphere. Distance is also an important factor. Thanks for not putting it too close to the central star. I know that this is a terrestrial planet, so it wouldn't be a hot Jupiter, but conditions there would be just as brutal. In fact, we can calculate just how brutal they would be by calculating the planetary equilibrium temperature: $$T = \left(\frac{L_{\odot}(1-a)}{16 \sigma \pi D^2} \right)^{\frac{1}{4}}$$ We can approximation that $L_{\odot} \approx L_{\text{Sun}}=3.846 \times 10^{26}$. As another approximation, $a=0.3$. We also know that $\sigma=5.670 \times 10^{-8}$. Plugging this all in, $$T = \left(\frac{3.846 \times 10^{26}(1-0.3)}{16 \times 5.670 \times 10^{-8}\pi D^2} \right)^{\frac{1}{4}}=9.85856 \times 10^7 \times D^{-\frac{1}{2}}$$ At $D_V$, $D_E$, and $D_M$, this comes out to $$T_V=299.986 \text{ K}$$ $$T_E=254.547 \text{ K}$$ $$T_M=207.515 \text{ K}$$ As far as approximations go, those are pretty similar to what we see, give or take a few dozen Kelvin (with the exception of Venus, which got screwed over by greenhouse gases). Mars' approximation is actually accurate to within a few Kelvin. Earth is the only one which is off, and that's only by about 30 Kelvin. That's pretty good. Tempted though I am to add in a fudge factor, I grudgingly admit that the model works for Mars, and there are a whole bunch of things on Earth (cough cough water, land and humans) which influence its results.

Using kinetic energy, we can relate the root mean square speed of a particle to its temperature via $$v=\sqrt{\frac{3kT}{m}}$$ At each of the radii, we have a different relation: $$v_V=1.11 \times 10^{-10}\left(\frac{m}{\text{kg}}\right)^{-1/2}\text{ m/s}$$ $$v_E=1.03 \times 10^{-10}\left(\frac{m}{\text{kg}}\right)^{-1/2}\text{ m/s}$$ $$v_M=9.27 \times 10^{-11}\left(\frac{m}{\text{kg}}\right)^{-1/2}\text{ m/s}$$ If the root mean square speed is greater than escape velocity, then some of the atmosphere will escape. $$v_{\text{escape}}=\sqrt{\frac{2GM}{r}}$$ I ran the numbers for each planet and gas. I assumed that $m_{\text{O}_2}=5.3\times10^{-26}\text{ kg}$, $m_{\text{N}_2}=4.7\times10^{-26}\text{ kg}$, $m_{\text{CO}_2}=7.3\times10^{-26}\text{ kg}$, and $m_{\text{H}_2\text{O}}=3\times10^{-26}\text{ kg}$. I then have a grid of values for the minimum mass, $M_{\text{min}}$, found by setting $v_{\text{escape}}$ equal to each of the speeds calculated above. The results are given relative to Earth masses (in $10^{-3}M_{\oplus}$): $$ \begin{array}{|c|c|c|c|}\hline \text{} & \text{Venus} & \text{Earth} & \text{Mars}\\\hline \text{O}_2 & 1.86 & 1.60 & 1.30\\\hline \text{N}_2 & 2.10 & 1.80 & 1.46\\\hline \text{CO}_2 & 1.35 & 1.16 & 0.941\\\hline \text{H}_2\text{O} & 3.28 & 2.83 & 2.29\\\hline \end{array} $$ As you can see, these are all a few orders of magnitude below the mass of Earth, and are roughly the mass of the Moon - maybe less by a factor of several. As an order-of-magnitude estimate, this makes sense, given that the Moon only has an extremely tenuous atmosphere.

posted over 9 years ago

HDE 226868‭

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