Mass of the atmosphere

In uk.sci.astronomy message , Tue, 21 Nov
2006 23:16:59, Jonathan Silverlight
wrote:

Interesting! It doesn't explicitly say so, but I think that exercise
takes into account something I thought of today - air isn't an inactive
weight. Its pressure is due partly to the kinetic energy of the
molecules.
Thanks for the email. You should post it - other people might be
interested.

( Considering an isolated spherical Earth with an vertically-thin
atmosphere in equilibrium; corrections for the real world are small. )

The downward pressure of the air on the Earth (and on anything else) is
ENTIRELY due to the vertical momentum change of molecules hitting the
Earth and bouncing off[*]. For a given type of molecule, the kinetic
energy is proportional to the square of the total momentum (and that is
three times the square of the vertical component of momentum).

The pressure on unit area of the surface is exactly equal to, and
ultimately caused by, the gravitational force on the air above it. That
area of the surface is the ultimate support of that column of air.

The mass of the atmosphere in pounds is therefore given by the product
of the atmospheric pressure in PSI and the area of the Earth in square
inches.

Fancy calculations are not needed for getting a sufficiently accurate
result.
[*] Well, some don't bounce immediately; but, in equilibrium, those that
arrive and stick are balanced by those that unstick and leave.

--
(c) John Stockton, Surrey, UK. Turnpike v6.05 MIME.
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

In message id, Dr J R
Stockton writes
In uk.sci.astronomy message , Tue, 21
Nov 2006 23:16:59, Jonathan Silverlight
wrote:

Interesting! It doesn't explicitly say so, but I think that exercise
takes into account something I thought of today - air isn't an
inactive weight. Its pressure is due partly to the kinetic energy of
the molecules.
Thanks for the email. You should post it - other people might be
interested.

( Considering an isolated spherical Earth with an vertically-thin
atmosphere in equilibrium; corrections for the real world are small. )

The downward pressure of the air on the Earth (and on anything else) is
ENTIRELY due to the vertical momentum change of molecules hitting the
Earth and bouncing off[*].

The pressure on unit area of the surface is exactly equal to, and
ultimately caused by, the gravitational force on the air above it.
That area of the surface is the ultimate support of that column of air.

The mass of the atmosphere in pounds is therefore given by the product
of the atmospheric pressure in PSI and the area of the Earth in square
inches.

Now I'm even more confused :-) It can't be ENTIRELY due to momentum
change.
Suppose the atmosphere froze (another thought experiment, though "A Pail
of Air" is well worth reading :-)
The kinetic energy of the particles is much lower. What's the weight on
every square inch of the Earth, and hence the pressure? (Now it is the
same as the mass, AFAICS)
But I suspect this is what Patrick Tyson meant when he referred to a
fallacy.

In article ,
Jonathan Silverlight wrote:

Now I'm even more confused :-) It can't be ENTIRELY due to momentum
change.

But consider...

Suppose the atmosphere froze

Then the pressure would *not* be due to momentum change. The pressure
is due to momentum change when it's a gas, because the way you support
a gas is by changing the momentum of its molecules. That's not the
way you support a solid.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

In article , I wrote:

Then the pressure would *not* be due to momentum change. The pressure
is due to momentum change when it's a gas, because the way you support
a gas is by changing the momentum of its molecules. That's not the
way you support a solid.

I forgot to give an analogy. Consider a man holding some balls, and a
man juggling the same balls. They both (on average) weigh the same, and
are exerting the same (average) upward force on the balls. The juggler
is doing it by changing the momentum of the balls (i.e. giving impulses
to them).

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

In uk.sci.astronomy message , Wed, 22 Nov
2006 19:24:48, Jonathan Silverlight
wrote:
In message id, Dr J
R Stockton writes

The downward pressure of the air on the Earth (and on anything else)
is ENTIRELY due to the vertical momentum change of molecules hitting
the Earth and bouncing off[*].

Now I'm even more confused :-) It can't be ENTIRELY due to momentum
change.

It is. For a column of air of mass M, there is a gravitational force
Mg. The ONLY possible opposing forces are that of the bouncing together
of the molecules, which keeps the gas extended but not does not hold it
up /in toto/, and the bouncing on the ground, which stops the gas
sinking into the Earth.

Suppose the atmosphere froze (another thought experiment, though "A
Pail of Air" is well worth reading :-)
The kinetic energy of the particles is much lower. What's the weight on
every square inch of the Earth, and hence the pressure? (Now it is the
same as the mass, AFAICS)

There is another inter-particle force, which was negligible in the gas
state. The particles cannot interpenetrate, nor can they penetrate the
ground (which is why they bounced before); and they do not have the
energy needed to separate either. So the upper ones rest on the lower
ones, and the bottom ones rest on the ground.

The inter-particle repulsive force is now continuous; in the gas state
it was intermittent, effective only during bounces.

In the solid state, thermal vibration (generally) causes minor
expansion; and will I suspect make a SMALL contribution to the force
between the solid air and the ground.

____

I have had E-mail from Professor Pekka Kauppi of Helsinki - " In deed,
there is an error in this BBC text. ... Your numbers are correct! ...
So: Trees contain 250-300 Gt (gigatons), the atmosphere 800-830 Gt, and
the annual emissions of fossil carbon are about 7 Gt at present."

He observes also (my words) that changing DE-forestation of 0.5% p.a. to
Re-forestation of 0.5% p.a. would make a huge impact.

-

The main stores of Carbon are Air, Sea, Ground, and Biosphere. I wonder
how much Carbon is elsewhere? Food before/after use probably counts as
Biosphere; but ISTM that Fuels and Artefacts are not accommodated by the
four main categories.

--
(c) John Stockton, Surrey, UK. Turnpike v6.05 MIME.
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.

In message id, Dr J R
Stockton writes
In uk.sci.astronomy message , Wed, 22
Nov 2006 19:24:48, Jonathan Silverlight
wrote:
In message id, Dr J
R Stockton writes

The downward pressure of the air on the Earth (and on anything else)
is ENTIRELY due to the vertical momentum change of molecules hitting
Earth and bouncing off[*].

Now I'm even more confused :-) It can't be ENTIRELY due to momentum
change.

Snip

Thanks - very enlightening.


I have had E-mail from Professor Pekka Kauppi of Helsinki - " In deed,
there is an error in this BBC text. ... Your numbers are correct! ...
So: Trees contain 250-300 Gt (gigatons), the atmosphere 800-830 Gt, and
the annual emissions of fossil carbon are about 7 Gt at present."

He observes also (my words) that changing DE-forestation of 0.5% p.a. to
Re-forestation of 0.5% p.a. would make a huge impact.

-

The main stores of Carbon are Air, Sea, Ground, and Biosphere. I
wonder how much Carbon is elsewhere? Food before/after use probably
counts as Biosphere; but ISTM that Fuels and Artefacts are not
accommodated by the four main categories.

There's a table in "The New Solar System" chapter "Atmospheres of the
Terrestrial Planets" (1981 edition) that gives 5.6 x 10^16 moles (67 x
10^16 grams or 670 Gt, close enough to Prof. Kauppi's figure) 3.2 x
10^18 moles in the ocean, 5 x 10^21 moles in carbonate rocks, and 10^21
tons as organic carbon in sedimentary rocks.

In message , Jonathan Silverlight
writes

10^21 tons as organic carbon in sedimentary rocks.

That should be moles, of course! But it gives me the chance to note that
according to
http://fermat.nap.edu/openbook.php?record_id=1790&page=191 "the total
amount of carbon in the mantle is 4.2 x 10^23 mole, which is 42 times
the amount in crust, ocean, and atmosphere."

"Jonathan Silverlight" wrote
in message ...
That should be moles, of course! But it gives me the chance to note that
according to http://fermat.nap.edu/openbook.php?record_id=1790&page=191
"the total amount of carbon in the mantle is 4.2 x 10^23 mole, which is 42
times the amount in crust, ocean, and atmosphere."

Wow - just wait until we've extracted all that and converted it to CO2 -
we'll be just like Venus. Don't tell Brad for goodness sake!

In message , TeaTime
writes

"Jonathan Silverlight" wrote
in message ...
That should be moles, of course! But it gives me the chance to note that
according to http://fermat.nap.edu/openbook.php?record_id=1790&page=191
"the total amount of carbon in the mantle is 4.2 x 10^23 mole, which is 42
times the amount in crust, ocean, and atmosphere."

Wow - just wait until we've extracted all that and converted it to CO2 -
we'll be just like Venus. Don't tell Brad for goodness sake!


You don't need anything like that amount. The mass of Venus's atmosphere
is about 100 x that of ours, 4.8 x 10^23 grams or 10^22 moles of carbon
dioxide.