Some gases are lighter than others and rise. Why don't they continue going up, leave the atmosphere, and then enter outer space?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Related: physics.stackexchange.com/q/202138/17547 $\endgroup$– Lightness Races in OrbitCommented Aug 24, 2015 at 9:58
-
$\begingroup$ Related: earthscience.stackexchange.com/questions/736 $\endgroup$– caseyCommented Aug 25, 2015 at 19:02
-
$\begingroup$ They did. All the helium ad hydrogen gone from the atmosphere and keep escaping nowadays, too. $\endgroup$– GregCommented Aug 22, 2017 at 5:03
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
The atmosphere actually loses gases to outer space.
The average velocity $\bar v$ of gas molecules is determined by temperature $T$. However, not all the molecules travel with the same velocity. The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds $$\begin{align} f{\left(v\right)}&=4\pi\sqrt{{\left(\frac m{2\pi kT}\right)}^3}v^2\exp\left(-\frac{m{v^2}}{2kT}\right)\\[6pt] &=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right) \end{align}$$ where $m$ is the mass of the molecule, $k$ is the Boltzmann constant, $M$ is the molar mass of the gas, and $R$ is the molar gas constant.
Individual molecules may reach escape velocity $v_\mathrm e$ and thus be able to leave the atmosphere.
Escape velocity is the minimum velocity that is sufficient for an object to escape from the gravitational attraction of a massive body. For a planet, the escape velocity may be estimated by using the formula $${v_\mathrm e}=\sqrt{\frac{2Gm_\text{planet}}r}$$ where $G$ is the gravitational constant, $m_\text{planet}$ is the mass of the planet, and $r$ is the distance from the centre of mass of the planet.
Therefore, atmospheric escape depends on the mass of the planet, the temperature of the atmosphere, and the molar mass of the gas.
-
$\begingroup$ There is also the strength of both the solar wind and the magnetic field to take into account $\endgroup$ Commented Aug 24, 2015 at 2:33
-
$\begingroup$ By my calculation, the escape velocity of a particle about 6400 km from the center of the earth is around 11,000 m/s. That seems unreasonably fast for any particle to ever escape which I suppose is good proof of your answer, but does a particle ever get moving that fast while in the atmosphere? $\endgroup$– jheindelCommented Aug 27, 2015 at 18:44
-
7$\begingroup$ @jheindel Yes, escape velocity on Earth is 11.2 km/s (Mars: 5.0 km/s; Moon: 2.4 km/s). The average speed of a hydrogen atom (M = 1 g/mol) at T = 1000 K is about 5 km/s. (Note that the temperature in Earth’s exosphere can be well above 1000 K.) Particles from the high-speed tail of the Maxwell distribution may actually reach escape velocity. Including other processes, Earth loses hydrogen at a rate of 3 kg/s (helium: 50 g/s). See also D. C. Catling, K. J. Zahnle: The Planetary Air Leak, Scientific American, May 2009, 26 $\endgroup$– user7951Commented Aug 28, 2015 at 10:46
$\begingroup$
$\endgroup$
Actually some do. There is a problem of helium escaping the earth's atmosphere.
https://en.wikipedia.org/wiki/Atmospheric_escape
There are many other sources about this very problematic occurrence.
$\begingroup$
$\endgroup$
2
Lighter gases don't go up after they've mixed with the surrounding atmosphere. Buoyancy (like with a balloon filled with $\ce{He}$ rising or $\ce{CO_2}$ from a leak in a line dropping to the floor and filling the room) occurs only with bulk masses.
You need extreme g fields/gradients, like in a ultracentrifuge, to separate gases by weight.
There is a certain net flux of light compounds upwards, because e.g. $\ce{He}$ (from radioactive decay) comes into the atmosphere at the ground and escapes into space from the upper layers (see other answer).
-
2$\begingroup$ Above the turbopause (about 100km) in the heterosphere atmospheric gasses stratify according to molecular weight. $\endgroup$– caseyCommented Aug 23, 2015 at 17:34
-
$\begingroup$ Argument taken, but 100 km is already "space" ;-), with mean free path lenghts in the range of centimetres. A good amount of the hydrogen up there comes from photolysis of water, i think. $\endgroup$– KarlCommented Aug 23, 2015 at 18:54