<p>Even in the exosphere, at 1800 deg centigrade, the only components of the Earth's atmosphere that can currently attain escape velocity are hydrogen (3 kg per sec), and and helium (next to nothing). The notion that the Earth's atmosphere would be stripped away by solar wind but for the protective magnetic shield, is a myth. So there is no 'characteristic time of loss of the Earth's atmosphere' in the sense of slow decay. There <em>is</em> certainty that, as the sun approaches the red giant stage, several billion years from now, there will be catastrophic heating of the atmosphere such that oxygen and nitrogen will be stripped away, leaving Earth to overheat in runaway carbon-dioxide. Unless, that is, if gravitational mayhem, caused by galactic collision with Andromeda, flings Earth substantially further from the sun. </p>

<p>Oxygen, nitrogen and inert gasses <em>can</em> escape from bodies the size of the moon, Venus and Mars because the thermal process, 'Jeans Escape', in less massive bodies incurs substantially lower escape velocities. </p>

CC BY-SA 3.0
2016-10-08T07:56:27.723Z
Exosphere? 1800 C? Huh?
Gimelist
CC BY-SA 3.0
2016-10-08T10:18:54.607Z
@Michael It is very rare gas, it doesn't mean that the satellites would evaporate :-) Protons coming from the Sun can be considered as 1million K hidrogen gas. The physics of the *rare* gases is very different. They are rarer as the best vacuum what can be produced on Earth.
peterh
CC BY-SA 3.0
2016-10-08T11:47:20.823Z
Thank you very much the answer! I suspect, the *"there is no characteristic time of loss of the Earth's atmosphere"*, means here essentially, that it would be *much* too long for a meaningful calculation (compared to the other significant processes, for example the Sun's depletion of Hydrogen)?
peterh
CC BY-SA 3.0
2016-10-09T12:27:42.800Z
Just to add to the explanation of the ultra-hot exosphere, the gas at this altitude (which extends beyond the International Space Station) is so thin it is almost not there. Individual atoms or molecules can be in the order of a metre apart, and very seldom physically interact. So although the temperature is so high, the amount of heat is not far from zero. Think of spark from an angle grinder on iron - then think of of sparks millions of times smaller. They would land on your skin and you wouldn't even notice them.
Gordon Stanger
CC BY-SA 3.0
2016-10-08T06:49:57.120Z
2016-10-08T06:49:57.120Z
Gordon Stanger
<p>Somewhere I've read it is in the order of some tens of billions of years, but unfortunately I can't find any reference with it in Google.</p>

<p>I suspect, the losses are coming mainly from analyzing the statistical distribution of the upper atmosphere, extending with the interaction of the Solar Wind. The result is probably a finite element model and not an analytical solution. Am I right? What is the result?</p>

<atmosphere><atmosphere-modelling><moon><upper-atmosphere>
What is the characteristic time of the loss of the Earth atmosphere, how can it be calculated?
peterh
# What is the characteristic time of the loss of the Earth atmosphere, how can it be calculated?

(Asked by: username_0 on 2016-10-07)

Somewhere I've read it is in the order of some tens of billions of years, but unfortunately I can't find any reference with it in Google.

I suspect, the losses are coming mainly from analyzing the statistical distribution of the upper atmosphere, extending with the interaction of the Solar Wind. The result is probably a finite element model and not an analytical solution. Am I right? What is the result?

**Tags:** atmosphere, atmosphere-modelling, moon, upper-atmosphere


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## 1 answers

### Answer by username_1 on 2016-10-08. Score: 4 | Accepted answer

Even in the exosphere, at 1800 deg centigrade, the only components of the Earth's atmosphere that can currently attain escape velocity are hydrogen (3 kg per sec), and and helium (next to nothing). The notion that the Earth's atmosphere would be stripped away by solar wind but for the protective magnetic shield, is a myth. So there is no 'characteristic time of loss of the Earth's atmosphere' in the sense of slow decay. There *is* certainty that, as the sun approaches the red giant stage, several billion years from now, there will be catastrophic heating of the atmosphere such that oxygen and nitrogen will be stripped away, leaving Earth to overheat in runaway carbon-dioxide. Unless, that is, if gravitational mayhem, caused by galactic collision with Andromeda, flings Earth substantially further from the sun.

Oxygen, nitrogen and inert gasses *can* escape from bodies the size of the moon, Venus and Mars because the thermal process, 'Jeans Escape', in less massive bodies incurs substantially lower escape velocities.

> **Comments:**
> **username_2**: Exosphere? 1800 C? Huh?  
> **username_0**: @Michael It is very rare gas, it doesn't mean that the satellites would evaporate :-) Protons coming from the Sun can be considered as 1million K hidrogen gas. The physics of the *rare* gases is very different. They are rarer as the best vacuum what can be produced on Earth.  
> **username_0**: Thank you very much the answer! I suspect, the *"there is no characteristic time of loss of the Earth's atmosphere"*, means here essentially, that it would be *much* too long for a meaningful calculation (compared to the other significant processes, for example the Sun's depletion of Hydrogen)?  
> **username_1**: Just to add to the explanation of the ultra-hot exosphere, the gas at this altitude (which extends beyond the International Space Station) is so thin it is almost not there. Individual atoms or molecules can be in the order of a metre apart, and very seldom physically interact. So although the temperature is so high, the amount of heat is not far from zero. Think of spark from an angle grinder on iron - then think of of sparks millions of times smaller. They would land on your skin and you wouldn't even notice them.  

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