<p>All planets lose a small amount of atmospheric gases due to some atoms/molecules (even neutral ones) having an energy high enough to escape the gravitational field of the planet. This tends to affect lighter elements more as they have higher velocities at the same temperature.
This kind of thermal 'outgassing' is in principle the same as that producing the 'solar wind'.</p>
<p>The ejection of ions in the polar regions mentioned in this link are due to a slightly different mechanism. They are caused by electric fields (plasma polarization fields) in the ionosphere. But these in turn are due to the electrons not being gravitationally bound and thus being able to escape from the earth, hence creating a net electric field. They are only able to escape in the region of open magnetic field lines though i.e. near the poles. Again, also this is lastly a result of some particles being able to escape the gravitational field.</p>
<p>It is not too difficult to calculate the amount of gas that can escape the gravity field of the planet. Let's consider atomic hydrogen (mass m=1.6*10<sup>-24</sup>g} in case of the earth. The density at the height of about 400:km (above which the atmosphere can be considered collisionless) is about <span class="math-container">$n=10^5/cm^3$</span><br />
<a href="https://i.sstatic.net/f68wl.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/f68wl.gif" alt="upper atmosphere density profile" /></a></p>
<p>The temperature can be quite variable between about <span class="math-container">$600K$</span> and <span class="math-container">$2000K$</span> depending on daytime and solar activity. Let's take a medium value of <span class="math-container">$T=1200K$</span> here. The height of 400 km would then make the radius at that level <span class="math-container">$R=6778 km$</span> and thus the total area of the sphere <span class="math-container">$A=4\pi R^2$</span>. The escape flux from one square centimeter of that sphere is (assuming a Maxwell velocity distribution and considering that only half of the atoms (going upwards) can escape)</p>
<p><span class="math-container">$$F_{esc} = \frac {n\sqrt{\pi}}{4 v_0}  \int_{v_{esc}}^{\infty} dv * v*exp[-(\frac {v}{v_0})^2] =$$</span>
<span class="math-container">$$= \frac{\sqrt{\pi} n v_0}{8} *exp[-(\frac{v_{esc}}{v_0})^2]$$</span></p>
<p>where <span class="math-container">$v_0$</span> is the thermal speed</p>
<p><span class="math-container">$$v_0=\sqrt{\frac{2kT}{m}}$$</span></p>
<p>with <span class="math-container">$k$</span> the Boltmann constant, and</p>
<p><span class="math-container">$$v_{esc}=\sqrt{\frac{2GM}{R}}$$</span></p>
<p>the escape speed from the planet with mass <span class="math-container">$M$</span> (=6*10<sup>27</sup>g for earth) with radius R and G the gravitational constant.</p>
<p>With the values as given this results in an escape flux of H atoms of <span class="math-container">$F_{esc}=4*10^7/cm^2/sec$</span> and thus a total hydrogen mass loss of <span class="math-container">$F_{esc}*A*m=0.4 kg/sec$</span>. If you look at the Wikipedia page <a href="https://en.wikipedia.org/wiki/Atmospheric_escape#Earth" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Atmospheric_escape#Earth</a> , they give 3kg/sec for the earth's hydrogen loss, which is a bit more, but then this figure could be based on different physical parameters, especially as this depends very strongly on the temperature. With T=2000 K, the result would be already 4.4 kg/sec according to the above calculation. For other elements than hydrogen, the loss rate is much lower though due to smaller thermal velocities. Even at 2000 K , Helium for instance is only lost at a rate of 4*10<sup>-3</sup> kg/sec and atomic oxygen even at about 10<sup>-18</sup> kg/sec. This is all assuming a Maxwellian velocity distribution in the first place, which may or may not be an accurate approximation in the high velocity tail of the distribution.</p>
<p>In case of the solar plasma, there is the complication that the electrons are gravitationally  practically unbound because of their small mass. This means they will escape freely until a state of equilibrium is reached where the sun becomes positively charged by such an amount that their escape rate is equal to that of the (much heavier) ions (i.e. both the electrons and the ions must have the same net potential energy). So ignoring the gravitational potential energy of the electrons, in equilibrium the equation must hold (<span class="math-container">$U_G$</span>=gravitational potential energy, <span class="math-container">$U_E$</span>=electric potential energy)</p>
<p><span class="math-container">$$U_E= U_G-U_E$$</span></p>
<p>where the left hand side is the net potential energy for an electron and the right hand side for the ion (<span class="math-container">$U_E$</span> is negative on the right hand side because the positively charged sun will repel rather than attract the ions).</p>
<p>This equation obviously gives us</p>
<p><span class="math-container">$$U_E=\frac {U_G}{2}$$</span></p>
<p>that is the electric field induced by the electron escape (plasma polarization field) effectively reduces gravity by 1/2 for the ions. In the equations for the calculation of the escape flux above, we have therefore to effectively reduce the solar mass M by a factor 1/2  (which reduces the escape velocity by a factor <span class="math-container">$1/\sqrt{2}$</span>). Doing this and assuming an ion density of <span class="math-container">$n=10^8/cm^3$</span> at <span class="math-container">$R=740000 km$</span>, one can match the observed flux/density of the solar wind at the earth with a temperature of <span class="math-container">$T=2.5*10^6 K$</span>, which again seems a reasonable value. The total mass loss in this case follows from the above as <span class="math-container">$8*10^8 kg/sec$</span> which is also consistent with the value one can find quoted elsewhere.</p>

CC BY-SA 4.0
2021-02-19T08:47:21.560Z
Thanks for your words. Do you think it is possible to quantify the "outgassing" of a planet, or at least do some back-of-the-envelope approximation on the order of magnitude?
B--rian
CC BY-SA 4.0
2021-02-20T19:27:58.540Z
@B--rian  See my edited answer
Thomas
CC BY-SA 4.0
2021-02-18T21:30:30.117Z
2021-02-23T08:22:08.413Z
2021-02-23T08:22:08.413Z
Thomas
<p>From going through the literature the paper by Wang et al. 2021 is citing, I am nearly certain that the term &quot;Earth wind&quot; must be a recent invention, perhaps by those authors themselves.</p>
<p>It is however correct to call the solar wind a 'wind'. This is because a wind is a pressure-driven bulk motion of a collectively coupled gas. The solar wind, at its base is driven by the enormous pressure gradients via the 1 million Kelvin hot solar corona.<br />
While Earth's primary, hydrogen-rich atmosphere might have experienced such a phenomenon, it is incorrect in naming the current atmospheric escape as 'wind'.</p>
<p>But now vocabulary aside, current atmospheric escape rates at the terrestrial planets are governed by various ionic escape processes, the most important of them being polar cusp escape. So the data you are actually looking for is data on polar cusp escape from various orbiters. Fortunately, <a href="https://ui.adsabs.harvard.edu/abs/2018A%26A...614L...3G/abstract" rel="nofollow noreferrer">Gunell et al., (2018)</a> have given a recent compendium on this (see their table A.1), and attempted a simple modelling of the data. There is no data available on polar escape for Mars and Venus, as this concept doesn't apply to those planets. The other escape processes are on average smaller for hydrogen, but larger for oxygen ions.</p>
<p>Similarly, I think there is no data available on the escape of gases from the gas giants. The magnetospheres of Jupiter and Saturn have been studied, and their particle populations characterized, but I think it is unclear how much of the magnetispheric content gets ultimately lost. There is no reliable data available on polar escape for Uranus and Neptune, as those planets were only ever visited by one singular flyby (that of Voyager 2).</p>

CC BY-SA 4.0
2021-02-15T22:22:21.530Z
Thanks for the definition of wind and putting it into context. I wish I could upvote twice.
B--rian
CC BY-SA 4.0
2021-02-18T21:22:37.883Z
It is misleading to say the solar wind is pressure driven. Already the lower solar corona is practically a collisionless plasma. The plasma density there is about 10^5/cm^3 and the collision cross section at these temperatures is about 10^-19 cm^2 , so the mean free path is 10^11 cm which is about the size of the sun. In the upper corona this is already 1000 times as large i.e. of solar system dimensions. The 'solar wind' just reflects those particles of the velocity distribution function that have a speed high enough to escape the gravitational field of the sun.
Thomas
CC BY-SA 4.0
2021-02-18T21:34:37.140Z
Sorry, made a typo above. The density in the lower solar corona is about 10^8/cm^3. 10^5/cm^3 would be the upper corona. The mean free path I mentioned is for the former value not the latter.
Thomas
CC BY-SA 4.0
2021-02-19T12:09:19.033Z
@Thomas: Surely any exosphere has $\lambda_{mfp} \approx R_{object}$. Surely you are also aware that EM-fields can extend the hydro-approximation to signficiantly lower densities than allowed in an unmagnetized system. How else would there be such a thing as a collisionless MHD shock at the terrestrial bow-shock? Furthermore, Parker's original formulation is a pressure-driven one, that's where the notion originates, and one that works well for mass-loss rates of stars of all masses. Purely kinetic escape rates underestimate the measured mass loss rates.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-02-20T19:46:29.230Z
@AtmosphericPrisonEscape In his original theory (see http://adsabs.harvard.edu/full/1965SSRv....4..666P ) Parker makes explicitly two incorrect assumptions 1) that all particles are gravitationally bound, and 2) that the corona is dominated by particle collisions i.e the principles of thermodynamics can be applied when in fact the physical conditions could not be further away from this. His approach is thus flawed from the outset. It doesn't make any sense trying to describe a non-LTE situation by principles of LTE. See also the edit in my own answer.
Thomas
CC BY-SA 4.0
2021-02-20T20:13:49.993Z
@Thomas: Non-LTE for line populations and the validity of the hydrodynamic equations are two different things that you are confusing here. MHD models work due to the 100 times higher collisional cross section of ions vs. neutral particles, which translates into a validity of the MHD approach up to ~100 $R_{\odot}$. The value you quoted in your answer is 3 orders of magnitude too low. Winds are furthermore driven way below the exosphere, deep in the collisional regime at the $\tau=1$ surface. None of this changes that it's wrong to call a bunch of kinetic particles a 'wind'.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-02-20T22:14:03.393Z
@AtmosphericPrisonEscape The pressure in Parker's theory is a purely kinematical pressure. He sets p=nkT. There is no electric or magnetic field going into the pressure. And his equation is assumed to hold in the whole corona and beyond, although kinematic collisions actually don't happen anywhere above the base of the corona. What we are seeing is in fact just a free expansion of gravitationally unbound particles. Whether one should or could call this a 'wind' is just a matter of semantics.
Thomas
CC BY-SA 4.0
2021-02-15T15:29:20.943Z
2021-02-15T15:29:20.943Z
AtmosphericPrisonEscape
<p>In February 2021, a new form of &quot;electrified gas breeze&quot; was announced, see blog by Tony Phillips entitled <a href="https://spaceweatherarchive.com/2021/02/12/a-new-form-of-space-weather-earth-wind/" rel="nofollow noreferrer">A New Form of Space Weather: Earth Wind</a> which might a possible explanation for traces of water on the moon:</p>
<blockquote>
<p>“Earth wind” comes from the axes of our planet. Every day, 24/7, fountains of gas shoot into space from the poles. The leakage is tiny compared to Earth’s total atmosphere, but it is enough to fill the magnetosphere with a riot of rapidly blowing charged particles. Ingredients include ionized hydrogen, helium, oxygen and nitrogen.</p>
<p>Once a month, the Moon gets hit by a blast of Earth wind. It happens around the time of the full Moon when Earth’s magnetic tail points like a shotgun toward the lunar disk.</p>
</blockquote>
<p>I suspect, that all planets with atmospheres should display this kind of planetary winds (even without having a strong magnetic field), but I am sure it was never termed <em>Venus wind</em> or <em>Jupiter wind</em> since that naturally refers to processes <em>within</em> the atmosphere of the repsective planet. Does anybody know (publications) on the topic of gas jets errupting from solar planets with atmosphere? What order of magnitude would such a particle flow have for Venus or Jupiter?</p>
<h3>References</h3>
<ul>
<li>H. Z. Wang <em>et al.</em>: <a href="https://iopscience.iop.org/article/10.3847/2041-8213/abd559" rel="nofollow noreferrer">Earth Wind as a Possible Exogenous Source of Lunar Surface Hydration</a> 2021 ApJL 907 L32</li>
<li><a href="https://astronomy.stackexchange.com/questions/10189/why-did-venus-not-lose-its-atmosphere-without-a-magnetic-field">Why did Venus not lose its atmosphere without a magnetic field?</a></li>
<li><a href="https://astronomy.stackexchange.com/questions/639/why-has-venuss-atmosphere-not-been-stripped-away-by-solar-wind">Why has Venus&#39;s atmosphere not been stripped away by solar wind?</a></li>
</ul>

CC BY-SA 4.0
2021-02-15T14:27:17.553Z
I have yet to read the paper, but it doesn't sound to me like a 'wind'. A wind is a hydrodynamic phenomenon, driven by pressure gradients. This sounds more like something that is known as polar magnetospheric escape, which consists of ballistic ions following open field lines into space.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-02-15T14:39:33.093Z
It is indeed about particles being expelled from the atmosphere (by magnetic fields). I am wondering about similar effects on Jupiter, but also trying to find an approximate for the mass loss rate in Venus.
B--rian
CC BY-SA 4.0
2021-02-15T16:44:11.447Z
sounds more like PR than something really new. Pickup-ions and atmospheric escape are something long known in the tail of magnetospheres
planetmaker
<the-moon><jupiter><venus><solar-wind>
Solar wind, Earth wind and planetary winds?
B--rian
# Solar wind, Earth wind and planetary winds?

(Asked by: username_0 on 2021-02-15)

In February 2021, a new form of "electrified gas breeze" was announced, see blog by Tony Phillips entitled [A New Form of Space Weather: Earth Wind](https://spaceweatherarchive.com/2021/02/12/a-new-form-of-space-weather-earth-wind/) which might a possible explanation for traces of water on the moon:

> “Earth wind” comes from the axes of our planet. Every day, 24/7, fountains of gas shoot into space from the poles. The leakage is tiny compared to Earth’s total atmosphere, but it is enough to fill the magnetosphere with a riot of rapidly blowing charged particles. Ingredients include ionized hydrogen, helium, oxygen and nitrogen.
>
> Once a month, the Moon gets hit by a blast of Earth wind. It happens around the time of the full Moon when Earth’s magnetic tail points like a shotgun toward the lunar disk.

I suspect, that all planets with atmospheres should display this kind of planetary winds (even without having a strong magnetic field), but I am sure it was never termed *Venus wind* or *Jupiter wind* since that naturally refers to processes *within* the atmosphere of the repsective planet. Does anybody know (publications) on the topic of gas jets errupting from solar planets with atmosphere? What order of magnitude would such a particle flow have for Venus or Jupiter?

### References

* H. Z. Wang *et al.*: [Earth Wind as a Possible Exogenous Source of Lunar Surface Hydration](https://iopscience.iop.org/article/10.3847/2041-8213/abd559) 2021 ApJL 907 L32
* [Why did Venus not lose its atmosphere without a magnetic field?](https://astronomy.stackexchange.com/questions/10189/why-did-venus-not-lose-its-atmosphere-without-a-magnetic-field)
* [Why has Venus's atmosphere not been stripped away by solar wind?](https://astronomy.stackexchange.com/questions/639/why-has-venuss-atmosphere-not-been-stripped-away-by-solar-wind)

**Tags:** the-moon, jupiter, venus, solar-wind
> **Comments:**
> **username_1**: I have yet to read the paper, but it doesn't sound to me like a 'wind'. A wind is a hydrodynamic phenomenon, driven by pressure gradients. This sounds more like something that is known as polar magnetospheric escape, which consists of ballistic ions following open field lines into space.  
> **username_0**: It is indeed about particles being expelled from the atmosphere (by magnetic fields). I am wondering about similar effects on Jupiter, but also trying to find an approximate for the mass loss rate in Venus.  
> **username_2**: sounds more like PR than something really new. Pickup-ions and atmospheric escape are something long known in the tail of magnetospheres  

---

## 2 answers

### Answer by username_1 on 2021-02-15. Score: 4 | Accepted answer

From going through the literature the paper by Wang et al. 2021 is citing, I am nearly certain that the term "Earth wind" must be a recent invention, perhaps by those authors themselves.

It is however correct to call the solar wind a 'wind'. This is because a wind is a pressure-driven bulk motion of a collectively coupled gas. The solar wind, at its base is driven by the enormous pressure gradients via the 1 million Kelvin hot solar corona.  
While Earth's primary, hydrogen-rich atmosphere might have experienced such a phenomenon, it is incorrect in naming the current atmospheric escape as 'wind'.

But now vocabulary aside, current atmospheric escape rates at the terrestrial planets are governed by various ionic escape processes, the most important of them being polar cusp escape. So the data you are actually looking for is data on polar cusp escape from various orbiters. Fortunately, [Gunell et al., (2018)](https://ui.adsabs.harvard.edu/abs/2018A%26A...614L...3G/abstract) have given a recent compendium on this (see their table A.1), and attempted a simple modelling of the data. There is no data available on polar escape for Mars and Venus, as this concept doesn't apply to those planets. The other escape processes are on average smaller for hydrogen, but larger for oxygen ions.

Similarly, I think there is no data available on the escape of gases from the gas giants. The magnetospheres of Jupiter and Saturn have been studied, and their particle populations characterized, but I think it is unclear how much of the magnetispheric content gets ultimately lost. There is no reliable data available on polar escape for Uranus and Neptune, as those planets were only ever visited by one singular flyby (that of Voyager 2).

> **Comments:**
> **username_0**: Thanks for the definition of wind and putting it into context. I wish I could upvote twice.  
> **username_3**: It is misleading to say the solar wind is pressure driven. Already the lower solar corona is practically a collisionless plasma. The plasma density there is about 10^5/cm^3 and the collision cross section at these temperatures is about 10^-19 cm^2 , so the mean free path is 10^11 cm which is about the size of the sun. In the upper corona this is already 1000 times as large i.e. of solar system dimensions. The 'solar wind' just reflects those particles of the velocity distribution function that have a speed high enough to escape the gravitational field of the sun.  
> **username_3**: Sorry, made a typo above. The density in the lower solar corona is about 10^8/cm^3. 10^5/cm^3 would be the upper corona. The mean free path I mentioned is for the former value not the latter.  
> **username_1**: @Thomas: Surely any exosphere has $\lambda_{mfp} \approx R_{object}$. Surely you are also aware that EM-fields can extend the hydro-approximation to signficiantly lower densities than allowed in an unmagnetized system. How else would there be such a thing as a collisionless MHD shock at the terrestrial bow-shock? Furthermore, Parker's original formulation is a pressure-driven one, that's where the notion originates, and one that works well for mass-loss rates of stars of all masses. Purely kinetic escape rates underestimate the measured mass loss rates.  
> **username_3**: @AtmosphericPrisonEscape In his original theory (see http://adsabs.harvard.edu/full/1965SSRv....4..666P ) Parker makes explicitly two incorrect assumptions 1) that all particles are gravitationally bound, and 2) that the corona is dominated by particle collisions i.e the principles of thermodynamics can be applied when in fact the physical conditions could not be further away from this. His approach is thus flawed from the outset. It doesn't make any sense trying to describe a non-LTE situation by principles of LTE. See also the edit in my own answer.  
> **username_1**: @Thomas: Non-LTE for line populations and the validity of the hydrodynamic equations are two different things that you are confusing here. MHD models work due to the 100 times higher collisional cross section of ions vs. neutral particles, which translates into a validity of the MHD approach up to ~100 $R_{\odot}$. The value you quoted in your answer is 3 orders of magnitude too low. Winds are furthermore driven way below the exosphere, deep in the collisional regime at the $\tau=1$ surface. None of this changes that it's wrong to call a bunch of kinetic particles a 'wind'.  
> **username_3**: @AtmosphericPrisonEscape The pressure in Parker's theory is a purely kinematical pressure. He sets p=nkT. There is no electric or magnetic field going into the pressure. And his equation is assumed to hold in the whole corona and beyond, although kinematic collisions actually don't happen anywhere above the base of the corona. What we are seeing is in fact just a free expansion of gravitationally unbound particles. Whether one should or could call this a 'wind' is just a matter of semantics.  

---

### Answer by username_3 on 2021-02-18. Score: 1

All planets lose a small amount of atmospheric gases due to some atoms/molecules (even neutral ones) having an energy high enough to escape the gravitational field of the planet. This tends to affect lighter elements more as they have higher velocities at the same temperature.
This kind of thermal 'outgassing' is in principle the same as that producing the 'solar wind'.

The ejection of ions in the polar regions mentioned in this link are due to a slightly different mechanism. They are caused by electric fields (plasma polarization fields) in the ionosphere. But these in turn are due to the electrons not being gravitationally bound and thus being able to escape from the earth, hence creating a net electric field. They are only able to escape in the region of open magnetic field lines though i.e. near the poles. Again, also this is lastly a result of some particles being able to escape the gravitational field.

It is not too difficult to calculate the amount of gas that can escape the gravity field of the planet. Let's consider atomic hydrogen (mass m=1.6\*10-24g} in case of the earth. The density at the height of about 400:km (above which the atmosphere can be considered collisionless) is about $n=10^5/cm^3$  
[![upper atmosphere density profile](https://i.sstatic.net/f68wl.gif)](https://i.sstatic.net/f68wl.gif)

The temperature can be quite variable between about $600K$ and $2000K$ depending on daytime and solar activity. Let's take a medium value of $T=1200K$ here. The height of 400 km would then make the radius at that level $R=6778 km$ and thus the total area of the sphere $A=4\pi R^2$. The escape flux from one square centimeter of that sphere is (assuming a Maxwell velocity distribution and considering that only half of the atoms (going upwards) can escape)

$$F\_{esc} = \frac {n\sqrt{\pi}}{4 v\_0} \int\_{v\_{esc}}^{\infty} dv \* v\*exp[-(\frac {v}{v\_0})^2] =$$
$$= \frac{\sqrt{\pi} n v\_0}{8} \*exp[-(\frac{v\_{esc}}{v\_0})^2]$$

where $v\_0$ is the thermal speed

$$v\_0=\sqrt{\frac{2kT}{m}}$$

with $k$ the Boltmann constant, and

$$v\_{esc}=\sqrt{\frac{2GM}{R}}$$

the escape speed from the planet with mass $M$ (=6\*1027g for earth) with radius R and G the gravitational constant.

With the values as given this results in an escape flux of H atoms of $F\_{esc}=4\*10^7/cm^2/sec$ and thus a total hydrogen mass loss of $F\_{esc}\*A\*m=0.4 kg/sec$. If you look at the Wikipedia page <https://en.wikipedia.org/wiki/Atmospheric_escape#Earth> , they give 3kg/sec for the earth's hydrogen loss, which is a bit more, but then this figure could be based on different physical parameters, especially as this depends very strongly on the temperature. With T=2000 K, the result would be already 4.4 kg/sec according to the above calculation. For other elements than hydrogen, the loss rate is much lower though due to smaller thermal velocities. Even at 2000 K , Helium for instance is only lost at a rate of 4\*10-3 kg/sec and atomic oxygen even at about 10-18 kg/sec. This is all assuming a Maxwellian velocity distribution in the first place, which may or may not be an accurate approximation in the high velocity tail of the distribution.

In case of the solar plasma, there is the complication that the electrons are gravitationally practically unbound because of their small mass. This means they will escape freely until a state of equilibrium is reached where the sun becomes positively charged by such an amount that their escape rate is equal to that of the (much heavier) ions (i.e. both the electrons and the ions must have the same net potential energy). So ignoring the gravitational potential energy of the electrons, in equilibrium the equation must hold ($U\_G$=gravitational potential energy, $U\_E$=electric potential energy)

$$U\_E= U\_G-U\_E$$

where the left hand side is the net potential energy for an electron and the right hand side for the ion ($U\_E$ is negative on the right hand side because the positively charged sun will repel rather than attract the ions).

This equation obviously gives us

$$U\_E=\frac {U\_G}{2}$$

that is the electric field induced by the electron escape (plasma polarization field) effectively reduces gravity by 1/2 for the ions. In the equations for the calculation of the escape flux above, we have therefore to effectively reduce the solar mass M by a factor 1/2 (which reduces the escape velocity by a factor $1/\sqrt{2}$). Doing this and assuming an ion density of $n=10^8/cm^3$ at $R=740000 km$, one can match the observed flux/density of the solar wind at the earth with a temperature of $T=2.5\*10^6 K$, which again seems a reasonable value. The total mass loss in this case follows from the above as $8\*10^8 kg/sec$ which is also consistent with the value one can find quoted elsewhere.

> **Comments:**
> **username_0**: Thanks for your words. Do you think it is possible to quantify the "outgassing" of a planet, or at least do some back-of-the-envelope approximation on the order of magnitude?  
> **username_3**: @B--rian  See my edited answer  

---

