# The Weight of Particles

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Hello everybody,

First off, hope you're all doing well. I was wondering what is the weight of all the particles in the Earth's atmosphere?

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I'm not sure you could actually know this beyond an estimation. I stand to be corrected of course.  How would you measure or calculate this? The particle densities and distributions would vary so greatly from location to location... all you could really do is estimate an average and calculate from there. Pollen counts fluctuate. Humidity varies.etc.. Also, why?

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I'm not sure you could actually know this beyond an estimation.

That's what I assumed as well.

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2 hours ago, Farid said:

Hello everybody,

First off, hope you're all doing well. I was wondering what is the weight of all the particles in the Earth's atmosphere?

What do you mean by “particles? Just dust? Dust and aerosols? All atoms and molecules?

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Atmosphere: ~5.5 quadrillion tons

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It is estimated that about 100,000 kg of dust falls to Earth from space every day.

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Atmosphere: ~5.5 quadrillion tons

That's almost as heavy as Earth itself.

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It is estimated that about 100,000 kg of dust falls to Earth from space every day.

Wow!

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Is it time to clarify weight vs mass?

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13 hours ago, swansont said:

Atmosphere: ~5.5 quadrillion tons

Quick calculation I get 5.79 quadrillion based on atmospheric pressure at sea level and the surface area of a sphere with a radius equal to the average radius of the Earth. I guess the difference is mostly due to the volume of the land mass above sea level?

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10 hours ago, Farid said:

That's almost as heavy as Earth itself.

No, and it shouldn’t be reasonable to expect that. The atmosphere is much smaller in extent and density.

A quadrillion is 10^15, so this is 5.5 x 10^18 kg, as opposed to 6 x 10^24 kg for the earth. So about a millionth, which they say in the article

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On 5/18/2020 at 12:38 PM, swansont said:

No, and it shouldn’t be reasonable to expect that. The atmosphere is much smaller in extent and density.

A quadrillion is 10^15, so this is 5.5 x 10^18 kg, as opposed to 6 x 10^24 kg for the earth. So about a millionth, which they say in the article

Quadrillion in some countries is 10^24. The same issue as with billion and milliard. The same word but different meaning depending where you are living.

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41 minutes ago, Sensei said:

Quadrillion in some countries is 10^24. The same issue as with billion and milliard. The same word but different meaning depending where you are living.

But, as I pointed out, that answer is unreasonable, which should tell you that you are using the antiquated/wrong value.

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On 5/17/2020 at 8:34 PM, Farid said:

That's almost as heavy as Earth itself.

First of all, weight is a force, measured in say Newtons.  I did the pressure * area thing and got 5.1x10^19 Newtons.

That's far heavier than Earth itself since Earth, unlike the atmosphere, is not resting on anything and is thus weightless.

This is my first test post and it didn't take the html <sup>19</sup>

Edited by Halc
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20 minutes ago, Halc said:

This is my first test post and it didn't take the html <sup>19</sup>

There is a superscript button in the toolbar at the top of the editor window

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31 minutes ago, Halc said:

First of all, weight is a force, measured in say Newtons.  I did the pressure * area thing and got 5.1x10^19 Newtons.

That's far heavier than Earth itself since Earth, unlike the atmosphere, is not resting on anything and is thus weightless.

How is that heavier than the earth? Can you explain your reasoning?

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3 hours ago, swansont said:

How is that heavier than the earth? Can you explain your reasoning?

Not saying Earth masses less than the atmosphere.

Earth is in freefall, hence has no weight. You can place Earth on top of me, in which case it would weigh about 900 Newtons, still considerably less than the atmosphere collective weight. A 20 digit number of Newtons is more weight than zero.

On the other hand, weight is a force, no? Force is a vector, and adding all the force applied by the atmosphere pretty much (not completely) cancels itself out, leaving many fewer digits of actual net force (weight?) on Earth.

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16 minutes ago, Halc said:

Earth is in freefall, hence has no weight.

So the comparison is moot. That’s why we were discussing mass.

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On 5/18/2020 at 7:38 AM, swansont said:

No, and it shouldn’t be reasonable to expect that. The atmosphere is much smaller in extent and density.

A quadrillion is 10^15, so this is 5.5 x 10^18 kg, as opposed to 6 x 10^24 kg for the earth. So about a millionth, which they say in the article

Was your original answer in metric tons? I assumed it was US tons.

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17 minutes ago, J.C.MacSwell said:

Was your original answer in metric tons? I assumed it was US tons.

They’re only different by ~10%

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19 minutes ago, swansont said:

They’re only different by ~10%

It wasn't clear from the link. I got 5.79 quadrillion US tons based on the Earth being Sea level...and figured the difference was the land volume above sea level. So it seemed about right being that much higher (.29 quadrillion US tons of atmosphere not existing in that volume)

But if it's metric or British tons that's about 10% more with no explanation for my being about that much under (that I can readily think of)

I know we are now talking mass and I was originally considering weight, but same line of thought.

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9 hours ago, J.C.MacSwell said:

It wasn't clear from the link. I got 5.79 quadrillion US tons based on the Earth being Sea level...and figured the difference was the land volume above sea level. So it seemed about right being that much higher (.29 quadrillion US tons of atmosphere not existing in that volume)

But if it's metric or British tons that's about 10% more with no explanation for my being about that much under (that I can readily think of)

I know we are now talking mass and I was originally considering weight, but same line of thought.

I had not thought the number given was exact, and they don't show a calculation, so it's hard to say where a discrepancy would come in.  I imagine the non-spherical nature of the earth (elevations, as you point out, and deformations) would play a role if one were to try to model this with sufficient precision.

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If you want a more accurate mass estimate, multiply the surface area by the pressure at an altitude of something like 400m, which is close to the average altitude where the atmosphere ends worldwide.

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1 hour ago, Halc said:

If you want a more accurate mass estimate, multiply the surface area by the pressure at an altitude of something like 400m, which is close to the average altitude where the atmosphere ends worldwide.

The atmosphere ends at 400m? I don't think so. I think they can still breathe in Mexico city, altitude 2240m (or the reason they can't is pollution, rather than lack of atmosphere)

"The mean height of land above sea level is about 797 m"
https://en.wikipedia.org/wiki/Earth

So you could take the pressure at 800m, weight it by 1/3 and use the pressure for that area, and use the sea level pressure for the other 2/3.

_____

Sea level pressure 101325 Pa (at 15ºC)

Pressure at 800 m 92100 Pa https://www.mide.com/air-pressure-at-altitude-calculator

Earth surface area 5.1 x 10^14 m^2 https://www.universetoday.com/25756/surface-area-of-the-earth/

Naive calculation (sea level P * area) is 5.2 x 10^19 N (divide by 9.8 to get 5.3 x 10^18 kg)

Weighted calc

1.566 x 10^19 N + 3.445 x 10^19 N = 5.0 x 10^19 N (1.1 x 10^19 lbs or 5.6 x 10^15 US tons)

Divide by 9.8 and you get 5.1 x 10^18 kg

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9 minutes ago, swansont said:

The atmosphere ends at 400m? I don't think so. I think they can still breathe in Mexico city, altitude 2240m (or the reason they can't is pollution, rather than lack of atmosphere)

"The mean height of land above sea level is about 797 m"
https://en.wikipedia.org/wiki/Earth

So you could take the pressure at 800m, weight it by 1/3 and use the pressure for that area, and use the sea level pressure for the other 2/3.

By 'ends' I mean the bottom.  If all the land and ice were spread evenly across the globe (like in waterworld), the bottom of the atmosphere would be something like 400m higher than current sea level, and pressure there would be no different than it is now.

You get a lower figure with your method, but I was taking into account the current volume of ice which displaces atmosphere in addition to what the land does.

I used 840 (reddit), more than 1/3 (more like 3/8), and used projected sea level rise from global warming charts to estimate ice volume.  400 is probably still a bit high.  360 maybe?

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41 minutes ago, Halc said:

By 'ends' I mean the bottom.  If all the land and ice were spread evenly across the globe (like in waterworld), the bottom of the atmosphere would be something like 400m higher than current sea level, and pressure there would be no different than it is now.

You get a lower figure with your method, but I was taking into account the current volume of ice which displaces atmosphere in addition to what the land does.

I used 840 (reddit), more than 1/3 (more like 3/8), and used projected sea level rise from global warming charts to estimate ice volume.  400 is probably still a bit high.  360 maybe?

You can see that the naive calculation and the approximation I used give answers differing by only about a part in 50. Further refinements should yield difference of that order, or smaller (unless there's something that has been missed entirely)

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