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I realize that many factors affect this answer so assume the following:

  1. Star type: G2 star (like our Sun)
  2. Composition: O2, N2, CO2, H2O atmosphere
  3. Magnetic field: Two different magnetic field strengths (none or minimal & Earth strength)
  4. Distance: two different mean orbital radius (Earth distance & Mars distance & if you feel ambitious do Venus too :) )
  5. Duration: I'm looking for geologic times scales $\to$ 1-2 billion years or more)

I believe that planetary mass is a better parameter to use than surface gravity.

Earth's mass with a magnetic field is sufficient at our global temperature. Mars' mass (about 10% of Earth's) without a magnetic field is insufficient at its colder temperature.

Would the Earth without a magnetic field work? Would Mars with a magnetic field work?

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    $\begingroup$ Mars and Mercury have approximately the same surface gravity, but surface gravity doesn't affect the escape velocity of a planet. It is the escape velocity (combined with other factors) that dictates which gases a planet can hold onto. Escape velocity is a function of planetary mass only, surface gravity isn't in the equation. $\endgroup$ – Jim2B Apr 2 '15 at 16:37
  • $\begingroup$ The current belief is that Mars lost it's atmosphere as it lost it's magnetic field $\endgroup$ – bowlturner Apr 2 '15 at 17:24
  • $\begingroup$ @bowlturner there is a relatively simple relationship between surface temperature, escape velocity, and thermodynamic momentum distributions from which this can be calculated but other effects (like the solar wind splitting water into oxygen & hydrogen and then stripping the hydrogen) make this a difficult thing to figure. Magnetic fields protect against the solar wind. $\endgroup$ – Jim2B Apr 2 '15 at 17:53
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    $\begingroup$ Jim2B is taking World Building to the literal extreme. Probably has an iron-proto core in his garage. $\endgroup$ – corsiKa Apr 2 '15 at 17:56
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    $\begingroup$ @corsiKa you are correct, for decades I've been collecting and analyzing all sorts of information related to solar system and planetary formation (many megabytes of collected information). I've got a set of heuristics to put them together into solar systems. My spreadsheet includes the temperature / escape velocity equations to see what sort of gases it can hold onto but I have nothing to account for the magnetic fields. $\endgroup$ – Jim2B Apr 2 '15 at 17:59
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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.

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  • $\begingroup$ Does stack exchange syntax include the ability to build tables? $\endgroup$ – Jim2B Apr 3 '15 at 1:14
  • $\begingroup$ @Jim2B Sort of, but it's tough. I'm not too good at it. $\endgroup$ – HDE 226868 Apr 3 '15 at 1:32
  • $\begingroup$ I was digging through an old spreadsheet and found some doodles on this. I found the minimum molar masses each planet could retain for 1 billion years or more. I guess I'll try to write an answer from this. Since I didn't reference my work, I can't be certain that some of the steps aren't more than an empirical fit. $\endgroup$ – Jim2B Apr 3 '15 at 1:49
  • $\begingroup$ My doodles followed the same process you did, then I made a ratio of Vesc / Vrms. I then insert this ratio into another formula that somehow calculates the duration of the gas in the atmosphere - this is the one I'm not certain is a "real" formula. $\endgroup$ – Jim2B Apr 3 '15 at 2:04
  • $\begingroup$ How the heck do you get the fancy formula editing into your posts? I see nothing in the edit window to let me do that. $\endgroup$ – Jim2B Apr 3 '15 at 4:02
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For people who are real gluttons for punishment, here is all you would care to know about Atmospheric Escape Mechanics.

The article discusses many different mechanisms of gas losses. The easiest to model is the Thermal Escape mechanism (described in detail below). However, two other processes may contribute substantially to atmosphere loss in the absence of a magnetic field; Pickup and electric field acceleration.

I've found no mathematical treatment of these last two but will describe them.

Pick-up

Pick-up is the process whereby hydrogen ions from the solar wind directly impinge upon the gas molecules of the atmosphere of planets with no or weak magnetic fields. When the ions impact molecules or atoms in the atmosphere, they impart momentum allowing them to escape. This is the dominant non-thermal loss mechanism for Mars' atmosphere and bodies with thin atmospheres.

Electric Field Acceleration

The dominant loss process for Venus' atmosphere is through electric force field acceleration. As electrons are less massive than other particles, they are more likely to escape from the top of Venus's ionosphere.[3] As a result, a minor net positive charge develops. That net positive charge, in turn, creates an electric field that can accelerate other positive charges out of the atmosphere. As a result, H+ ions are accelerated beyond escape velocity.

from the wikipedia article on Atmospheric Loss Mechanisms

This loss mechanism is the dominant non-thermal loss mechanism on bodies with thick atmospheres.

Thermal escape

In most cases, thermal escape is the dominant atmospheric loss mechanism.

As @HDE226868 posted, calculate surface temperature of the planet using solar luminosity, albedo, and distance from the sun. Then calculate the Vrms of the gases. Then compare to the body's escape velocity.

         | Meas   Calc
----------------------
Tmercury | 700 K  438 K  
Tvenus   | 735 K  185 K  
Tearth   | 313 K  254 K  
Tmoon    | 390 K  268 K  
Tmars    | 293 K  210 K  
Tio      | 130 K   95 K  
Teuropa  | 125 K   92 K  
Tcallisto| 165 K  115 K  
Ttitan   |  93 K   85 K  
Ttriton  |  38 K   35 K  
Tpluto   |  55 K   36 K  

NOTE1: the problem with this is it calculates the average surface temperature, whereas thermal escape mechanics rely most heavily on the highest / day side temperature. The shown measured temps are the "high temperatures".

NOTE2: I've got the albedo of these bodies but they tend to make the temperatures diverge further from the measured values than using HDE's approximation of 0.3 for an average albedo.

I use a ratio of 300 as the cut-off for gases. Values above this number indicate the planet can retain the gas for billions of years.

Next calculate the escape velocity of the body. These are the values that I get:

             Escape
Body       Velocity
Mercury  =  4,250 m/s  
Venus    = 10,361 m/s  
Earth    = 11,178 m/s  
Moon     =  2,375 m/s  
Mars     =  5,021 m/s  
Io       =  2,560 m/s
Europa   =  2,035 m/s
Callisto =  2,444 m/s
Titan    =  2,641 m/s 
Triton   =  1,456 m/s 
Pluto    =  1,246 m/s 

The formula to find lightest weight gas the planet can hold onto is as follows:

Vesc must be larger than some calculated multiple of Vrms (I based the form of this calculated multiple off the half-life formula) $$ ln\left (1 \times 10^{9} \div 9 \right )^2 = \frac{Vesc}{Vrms} $$

$$ ln\left (1 \times 10^{9} \div 9 \right ) = \left ( \frac{2GM \times m}{r \times 3RT} \right ) $$

$$ \large m = \frac{ln\left (1 \times 10^{9} \div 9 \right ) 3RTr}{2GM} $$

The 1e9 value is the number of years you want the gas to stick around, this represents 1 billion years. I believe the natural log portion of the equation to be my own empirical fit to the problem.

$ \large m $ - Molar mass of compound

$ R $ - Universal gas constant $ 8.3144621 \frac{J}{mol K} $

$ T $ - Temperature (K)

$ r $ - Planet's radius in meters (my statements above are wrong, it does play a factor)

$ G $ - Gravitational constant $ (6.67 \times 10^{-11} {N} \left (\frac{m}{kg} \right ) ^2 ) $

$ M $ - Planet's mass in kg

The gases each body can retain over geologic periods are:

         Molar
Body      Mass   Gases
Mercury  = 114 ~ Br2 + I2 only; all other gases escape; No ices  
Venus    =  20 ~ N2 and heavier  
Earth    =   9 ~ CH4 and heavier  
Moon     = 203 ~ I2 only; all other gases escape; No ices
Mars     =  36 ~ F2 and heavier  
Io       =  58 ~ Kr + Cl2 only; all other gases escape; Ices of NH3, H2O, CO2, Br2, etc.
Europa   =  89 ~ Kr + Cl2 only; all other gases escape; Ices of NH3, H2O, CO2, Br2, etc.
Callisto =  81 ~ Kr + Cl2 only; all other gases escape; Ices of NH3, H2O, CO2, Br2, etc.
Titan    =  39 ~ N2 and heavier; Ices of CH4, NH3, H2O, CO2, O2, etc.
Triton   =  53 ~ None; all gases escape; Ices of NH3, H2O, CO2, N2, O2, etc.
Pluto    = 104 ~ None; all gases escape; Ices of NH3, H2O, CO2, N2, O2, etc.

Another twist to this is the fact that various molecules achieve much longer longevity when it is colder than their "snow line". Our solar system's snow line for water (the point at which it remains solid and doesn't evaporate/sublimate) occurs at the distance of our asteroid belt. Beyond this distance, solar system bodies can retain their ices.

Hypothetical Planets

It's been a long road but I think I finally have my answer. A simple swap between $ \large m $ and $ M $ generates the equation which determines what mass is required to retain a given gas if you use the assumptions below. First the equation:

$$ M = \frac{ln\left (1 \times 10^{9} \div 9 \right ) 3RTr}{2G \large m} $$

Now the assumptions 1. Replacement planets use the same density as Earth.
2. Replacement planets use the same albedo as Earth.
3. Planets need a Vesc/Vrms ratio of 400 to hold onto a gas for 4.5 billion years.
4. Planets need to retain gaseous water to maintain human habitability.
5. The daytime "hot" temp is 1.15x the temperature average.
6. Planets have a strong magnetic field so only thermal loss is important.

Which simplifies the equation to:

$$ M = \frac{1,200 \times 8.3144621 \frac{J}{mol K}Tr}{36 \times (6.67 \times 10^{-11} {N} \left (\frac{m}{kg} \right ) ^2)} $$

Then what I find is:

Orbit of  Min Mass  Vesc  Surface G
Venus     0.55e     9,159   0.82
Earth     0.43e     8,437   0.75
Mars      0.32e     7,647   0.68

Remember the mass is the important thing, so the planet could possess a much lower density and, therefore, a much lower surface gravity should the world builder so desire.

Interestingly, if Mars were just about 3x its current mass, it might have held onto a substantial atmosphere and been a pleasant place to live.

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  • $\begingroup$ @ArtOfCode thanks for the edits. Is there some sort of tutorial about the advanced syntax of SE? $\endgroup$ – Jim2B Apr 3 '15 at 14:27
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    $\begingroup$ The edit I've put in here is MathJax. We have our own FAQ guide to using the SE subset of it, or you can look at Mathematics.SE's more comprehensive guide. There's also the official docs. $\endgroup$ – ArtOfCode Apr 3 '15 at 14:35
  • $\begingroup$ From the research I've done it seems that we should use the exosphere temperature, not the surface temperature... but I don't see any information on how to calculate it. Any ideas? $\endgroup$ – 2012rcampion Apr 3 '15 at 23:10
  • $\begingroup$ Yeah, that's a problem too. I don't think there's a simple formula to calculate that. But I'd be happy to see what anyone else posts. Also, since I've got my spreadsheet already set up, I plan to add another section to this answer describing the what planetary masses we need in the orbits of Venus, Earth, & Mars in order to retain an atmosphere. But I'm tired and probably heading to bed early tonight. $\endgroup$ – Jim2B Apr 4 '15 at 0:12
  • $\begingroup$ I think the formula to calculate m is displayed wrong. I'll have to get a pencil and paper and work out the formula derivation to ensure it is correct. $\endgroup$ – Jim2B Apr 4 '15 at 3:14
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Around 0.5 Earth masses, or 40 Lunar masses.

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    $\begingroup$ Welcome to Worldbuilding, user52119! If you have a moment please take the tour and visit the help center to learn more about the site. You may also find Worldbuilding Meta and The Sandbox (both of which require 5 rep to post on) useful. Have fun! $\endgroup$ – FoxElemental Jun 17 '18 at 23:32
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    $\begingroup$ This may be true (at least under certain assumptions), but it would be stronger with some kind of citation or explanation. Also, any assumptions should be included. For example, is this assuming a surface temperature the same as Earth's? $\endgroup$ – Brythan Jun 17 '18 at 23:56
  • $\begingroup$ @user52119 Your non-answer response has been flagged as low quality. Stack Exchange answers should not be in the form of quick phrases or short sentences. You are required to offer more explanation and rationale. Also, since you plop down numbers, you should cite your sources so later viewers can check your veracity. $\endgroup$ – elemtilas Jun 18 '18 at 2:10

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