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 $m_{{\text{O}}_2}=5.3\times {10}^{-26}\text{ kg}$, $m_{{\text{N}}_2}=4.7\times {10}^{-26}\text{ kg}$, $m_{{\text{CO}}_2}=7.3\times {10}^{-26}\text{ kg}$, and $m_{{\text{H}}_2\text{O}}=3\times {10}^{-26}\text{ kg}$. 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 ${10}^{-3}{M}_{\oplus }$): $$\begin{array}{|cccc|}\hline & \text{Venus}& \text{Earth}& \text{Mars}\\ {\text{O}}_2& 1.86& 1.60& 1.30\\ {\text{N}}_2& 2.10& 1.80& 1.46\\ {\text{CO}}_2& 1.35& 1.16& 0.941\\ {\text{H}}_2\text{O}& 3.28& 2.83& 2.29\\ \hline\end{array}$$ 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 almost 10 years ago

HDE 226868‭

601 reputation 61 463 80 49