<p>Yes, the formula you quote applies to some extend.</p>

<p>It can also be written in terms of Boltzmann's constant $k_\mathrm{B}$ as
$$v_\mathrm{rms} = \sqrt{\frac{3k_\mathrm{B}T}{m_\mathrm{m}}},$$
with $m_\mathrm{m}$ the mass of the molecule in question,
gives the mean ("root-mean-square") velocity of the gas molecules as a function of temperature $T$. Comparing this to the escape velocity
$$v_\mathrm{esc}=\sqrt{\frac{2GM_\mathrm{p}}{R_\mathrm{p}}},$$
where $M_\mathrm{p}$ and $R_\mathrm{p}$ are the mass and radius of the planet, respectively, gives an order-of-magnitude estimate of whether or not the planet will be able to maintain its atmosphere.</p>

<p>However, if the two velocities are equal, it would still mean that half of the atmosphere evaporated instantly, and that over time, a large fraction would disappear, although the time scale for this to happen may be very long. In order for the planet to really sustain its atmosphere, $v_\mathrm{esc}$ needs to be roughly 6 times larger than $v_\mathrm{rms}$.</p>

<p>Gravity is not the only factor determining the stability of the atmosphere, though. Radiation pressure from the star ("Solar wind") can easily rip a planet of its atmosphere. On the other hand, if the planet has an efficient magnetic field, this will shield it against the wind.</p>

<p><a href="http://en.wikipedia.org/wiki/Atmospheric_escape" rel="noreferrer">Other factors</a> are also involved. But you're right in that at the very least, the planet needs a certain gravitational field to keep its atmosphere.</p>

CC BY-SA 3.0
2015-03-25T08:51:23.787Z
Thanks, this is exactly what i was looking for. I'm looking to generate planet environments procedurally and for the sake of simplicity will omit the other factors for now.
Marcin
CC BY-SA 3.0
2015-03-26T00:01:17.337Z
Can you please elaborate on the source of the factor of 6?  Also, I assume the other equation for mean gas velocity can be used in place of the Boltzmann version?  I ran some numbers and it appears that a planet with properties of Mercury (assuming constant temp of 433.15K) would have almost enough gravity to trap water vapor (again ignoring the other factors).  Vesc = 4247.9996 m/s and Vrms = 732.1046 m/s, so about 5.8 factor.
Marcin
CC BY-SA 3.0
2015-03-26T09:43:18.043Z
@td-lambda: To be honest, I don't remember my source for this number, but googling something like [retain atmosphere times larger than the escape velocity](https://www.google.dk/#q=retain+atmosphere+times+larger+than+the+escape+velocity) leads me to values like [4-6](http://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/esc_vel_atmospheres.html), [6](https://www.geol.umd.edu/~jmerck/geol212/lectures/18.html), [6](http://csep10.phys.utk.edu/ojta/c1c/earthmoon/overview/atmosphere_tl.html), and [6](http://lifeng.lamost.org/courses/astrotoday/CHAISSON/AT308/HTML/MP0801.HTM)
pela
CC BY-SA 3.0
2015-03-26T09:55:29.027Z
One typically considers the rms velocity of gas particles, rather than the average (in principle, the average could be zero, for two particles moving in opposite directions). Since so many more or less unknown factors enter this problem, and since they are the same order of magnitude, it doesn't matter much, though. Wrt. Mercury, your calculation shows that it should be able to retain an atmosphere, and I suppose that the reason it doesn't is the fact that it's so close to the Sun (so that the Solar wind is powerful) while magnetic field is very low (~1% of the Earth's).
pela
CC BY-SA 3.0
2015-03-25T08:35:27.047Z
2015-03-25T08:35:27.047Z
pela
<p>How can you determine the gravitational force needed to keep a particular gas in the atmosphere of a planet (for example, carbon dioxide (CO<sub>2</sub>))?</p>

<p>I came across the following formula $\left(\frac{8RT}{πM}\right)^{0.5}$</p>

<p>here: <a href="http://www.tau.ac.il/~roichman/CVI/hw1/hw1.pdf" rel="noreferrer">http://www.tau.ac.il/~roichman/CVI/hw1/hw1.pdf</a></p>

<p>$M$ = Molecular mass </p>

<p>$R$ = Gas constant</p>

<p>$T$ = Temperature</p>

<p>Would this formula apply?</p>

<gravity><astrophysics><planetary-atmosphere>
Gravitational pull needed to keep a gas in atmosphere
Marcin
# Gravitational pull needed to keep a gas in atmosphere

(Asked by: username_0 on 2015-03-25)

How can you determine the gravitational force needed to keep a particular gas in the atmosphere of a planet (for example, carbon dioxide (CO2))?

I came across the following formula $\left(\frac{8RT}{πM}\right)^{0.5}$

here: <http://www.tau.ac.il/~roichman/CVI/hw1/hw1.pdf>

$M$ = Molecular mass

$R$ = Gas constant

$T$ = Temperature

Would this formula apply?

**Tags:** gravity, astrophysics, planetary-atmosphere


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## 1 answers

### Answer by username_1 on 2015-03-25. Score: 5 | Accepted answer

Yes, the formula you quote applies to some extend.

It can also be written in terms of Boltzmann's constant $k\_\mathrm{B}$ as
$$v\_\mathrm{rms} = \sqrt{\frac{3k\_\mathrm{B}T}{m\_\mathrm{m}}},$$
with $m\_\mathrm{m}$ the mass of the molecule in question,
gives the mean ("root-mean-square") velocity of the gas molecules as a function of temperature $T$. Comparing this to the escape velocity
$$v\_\mathrm{esc}=\sqrt{\frac{2GM\_\mathrm{p}}{R\_\mathrm{p}}},$$
where $M\_\mathrm{p}$ and $R\_\mathrm{p}$ are the mass and radius of the planet, respectively, gives an order-of-magnitude estimate of whether or not the planet will be able to maintain its atmosphere.

However, if the two velocities are equal, it would still mean that half of the atmosphere evaporated instantly, and that over time, a large fraction would disappear, although the time scale for this to happen may be very long. In order for the planet to really sustain its atmosphere, $v\_\mathrm{esc}$ needs to be roughly 6 times larger than $v\_\mathrm{rms}$.

Gravity is not the only factor determining the stability of the atmosphere, though. Radiation pressure from the star ("Solar wind") can easily rip a planet of its atmosphere. On the other hand, if the planet has an efficient magnetic field, this will shield it against the wind.

[Other factors](http://en.wikipedia.org/wiki/Atmospheric_escape) are also involved. But you're right in that at the very least, the planet needs a certain gravitational field to keep its atmosphere.

> **Comments:**
> **username_0**: Thanks, this is exactly what i was looking for. I'm looking to generate planet environments procedurally and for the sake of simplicity will omit the other factors for now.  
> **username_0**: Can you please elaborate on the source of the factor of 6?  Also, I assume the other equation for mean gas velocity can be used in place of the Boltzmann version?  I ran some numbers and it appears that a planet with properties of Mercury (assuming constant temp of 433.15K) would have almost enough gravity to trap water vapor (again ignoring the other factors).  Vesc = 4247.9996 m/s and Vrms = 732.1046 m/s, so about 5.8 factor.  
> **username_1**: @td-lambda: To be honest, I don't remember my source for this number, but googling something like [retain atmosphere times larger than the escape velocity](https://www.google.dk/#q=retain+atmosphere+times+larger+than+the+escape+velocity) leads me to values like [4-6](http://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/esc_vel_atmospheres.html), [6](https://www.geol.umd.edu/~jmerck/geol212/lectures/18.html), [6](http://csep10.phys.utk.edu/ojta/c1c/earthmoon/overview/atmosphere_tl.html), and [6](http://lifeng.lamost.org/courses/astrotoday/CHAISSON/AT308/HTML/MP0801.HTM)  
> **username_1**: One typically considers the rms velocity of gas particles, rather than the average (in principle, the average could be zero, for two particles moving in opposite directions). Since so many more or less unknown factors enter this problem, and since they are the same order of magnitude, it doesn't matter much, though. Wrt. Mercury, your calculation shows that it should be able to retain an atmosphere, and I suppose that the reason it doesn't is the fact that it's so close to the Sun (so that the Solar wind is powerful) while magnetic field is very low (~1% of the Earth's).  

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