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I wonder if the escape velocity of gas molecules is the same as the one of rockets, which is 7.8km/s. And do oxygen escape at all or not? Are they too heavy?

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  • $\begingroup$ 7.8 km/s is for orbit, and escape velocity is sqrt(2) times larger. If it's moving at orbital speed, it could remain suspended up there if neutral, if we ignore the Newton's cannonball issue (which is possible since it can start out in the exosphere to begin with). However, it's unlikely to stay neutral for long, and it will be acted upon by the magnetic field. $\endgroup$ – Alan Rominger Jul 26 '14 at 17:21
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Yes, the general process of losing gas molecules is possible. It's called atmospheric escape.

The escape speed mechanism you mentioned is just one way in which it happens. Since lighter molecules move faster, they tend to be nearer the escape velocity. The Earth has lost (and continues to lose) hydrogen over the course of its history due to this. Oxygen is heavier and isn't as susceptible to this process, though I won't claim it doesn't happen to some degree.

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The speed of a molecule is given by: $$V_{rms}=\sqrt{\frac{3k_BT}{m}}{\tag1}$$

with v in m/s, T in kelvins, and m is the molecule mass (kg). The most probable speed is 81.6% of the rms speed, and the mean speeds 92.1% (isotropic distribution of speeds).

The mass of an oxygen molecule is $5.313\times 10^{-26}$ kg. This gives $v_{rms}$ a value of about 460 m/s at 0°C.

There will not be a lot of molecules able to reach the speed necessary to escape the gravity of Earth.

Added:
The temperature at 100 km above the ground is about 200 K, about the temperature of dry ice. The $v_{rms}$ of an oxygen molecule at this temperature would be about 390 m/s.

The escape velocity remains the same, independent from where you start.

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  • $\begingroup$ Could you also add info about the temp in the upper atmosphere as well as escape speed? $\endgroup$ – BMS Jul 26 '14 at 16:41
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Escape velocity on the surface of earth is 11.2 km/s which is 11200 m/s. RMS velocity is given by

      Vrms = √(3RT/M) 

where M = molecular mass of oxygen = 32 g/mol = 0.032 kg/mol (converting to SI unit). If oxygen must escape earth then Vrms = Vesc That gives T = 160000 Kelvin temperature.

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    $\begingroup$ Please do elaborate $\endgroup$ – QuIcKmAtHs Feb 15 '18 at 7:48
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Yes, the escape speed will be the same as for a rocket launched from the same altitude, which, depending on how high up you want to consider an escape from, may or may not be approximately as much as on the surface, which is actually 11.2 km/s, not 7.8 which is orbital speed.

The reason for this is escape speed is independent of mass, due to the fact that gravity's action upon an object can be conceived of in purely kinematic terms - that is, in terms of the geometric parameters of the object's motion - position, velocity, acceleration alone, with no reference to the dynamical parameter, mass, which describes the responsivity, actually the resistivity, of a physical object to an applied force (responsivity would be $\frac{1}{m}$, so massless objects are infinitely responsive to applied forces, while heavy ones show little response.). In particular, that gravity accelerates all objects the same way, independent of their mass, so that mass need not show up in considering the kinematic evolution of an object in a gravitational field (but does show up as a parameter in the description of the establishment of that field by an object). This is a deep fact about gravity which serves as the basis of Einstein's general theory of relativity and permits it to be described in purely geometric terms as due to a modification of the geometry of spacetime.

As a result, a molecule, a rocket, Elon Musk's car, and the manhole blasted from the Nevada Test Site in 1957 (which, sadly, almost surely vaporized due to ram pressure against the atmosphere, i.e. it burned up on exit as opposed to the more familiar, but symmetrical, burn up on entry due to similarly high velocities) as a result of an underground nuclear test - all would have the exact same escape speed.

More specifically, the escape speed of any object, no matter the mass, from a spherically-symmetric gravitator, which the Earth approximates, of mass $M$ is

$$v_\mathrm{esc} = \sqrt{\frac{2GM}{r}}$$

where $r$ is the radial distance from the center (Technically this would not be 100% correct for within atmosphere because it treats the whole mass including atmosphere as below the escaping object, but that atmosphere mass is negligible compared to the Earth's mass.). For Earth, $r = 6371\ \mathrm{km}$ and $M = 5972\ \mathrm{Yg}$ (that is, $5.972 \times 10^{24}\ \mathrm{kg}$), and with those parameters you get a $v_\mathrm{esc}$ of 11.2 km/s. At the conventionally-stated top of the atmosphere, usually taken as 100 km, so effectively $r = 6471\ \mathrm{km}$ the $v_\mathrm{esc}$ is about 11.1 km/s. At 408 km, about the altitude of the ISS, $v_\mathrm{esc}$ is 10.8 km/s. Thus gas molecules at each location will need to have at least this much speed to escape. In practice, they may need more because of collision with other molecules can dissipate their energy. The relevant parameter is the mean free path of the molecules which is the average distance it can travel before colliding with another molecule, and the number of collisions on the way out along with the masses of those molecules (assuming elastic collisions, which isn't quite right since molecules can break or get their electrons bumped into higher orbitals, processes which absorb energy making it inelastic). So it's a rather complicated calculation to get the actual speed required for a truly successful molecular escape with high probability (a molecule could "get lucky" with a large free corridor in front of it but that's not as likely) and also the needed starting altitude. But the minimum speed (i.e. assuming a perfect free corridor all the way out) is the usual escape speed from the molecule's starting height.

Now when a molecule has actually escaped, which is how it seems your original question is worded, that is, it is very far from the Earth, its speed will have slowed by whatever amount is needed to supply the remaining energy required to fight the last of the gravity well to exit (and anything lost due to collisions). If the molecule had additional energy above and beyond this, which is almost certain, it will have residual speed. And this residual speed could be rather variable as to its exact quantity.

As for who can escape, that depends on the mass. At any given temperature, and so a given average kinetic energy, things with more mass will move more slowly, so oxygen will have a harder time escaping than nitrogen, which in turn will have a MUCH harder time escaping than hydrogen - which is why there is very little free hydrogen in Earth's atmosphere, but giant planets like Jupiter and Saturn are practically made of the stuff, due to having huge gravity wells (which are also supplied by the hydrogen itself). This is also important in considering possible conditions on extrasolar planets. In particular, it appears from all the data we've gathered so far with missions like Kepler to be a fairly robust conclusion that if a planet has about more than 5 or 6 Earth masses or so, its observed density drops dramatically, meaning that it has accrued what is almost surely a large hydrogen envelope (though proving that definitively is hard without spectroscopic measurements). There are exceptions, of course, but it's a sigmoid-shaped (stretched out S-shaped) graph with the center of the "knee" of the sigmoid (the S) is at this mass level. This is because there is finally enough gravity to achieve the necessary hold at that point, more or less (in particular the escape speed is $\sqrt{5}$ times more, not counting change in the radius, and thus the kinetic energy per unit mass to escape is 5 times more, so a molecule 5 times less massive than, say, nitrogen, will be just as hard to escape as nitrogen is, and that's about 6 amu (or 6 "Daltons", Da) for a diatomic molecule, $\mathrm{H}_2$ is 2 amu, so we're rather close.).

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