
View Full Version : How heavy the Earth’s atmosphere is?
The front page of our Forum again is fulfilled with crank.net museum’s exhibitors. So it is time to return to pleasant Physics…
There is my first question:
The probe grounded on Titan easily measured the weight of the entire Titan’s atmosphere. How it did that?
How heavy the atmosphere of Earth is?
wesmorris 020305, 12:12 AM Is its surface area known? If so, F = PA so the force of the entire atmosphere is mean atmospheric pressure at the surface times the surface area of Titan.
I guess that would only be an approximation though, as the pressure thing would be basically impossible to measure and well, surface area is difficult to calculate on an irregular surface. It would probably be a very useful approximation though if you were needing a good ball park of the weight of the atmosphere.
Dear wesmorris,
Good, very good. But what if there was a strong wind?
wesmorris 020305, 12:24 AM Dear wesmorris,
Good, very good. But what if there was a strong wind?
At that point, I edit my post to account for the fact that those questions came to mind after I'd already posted. :)
I'm still spoiled from all those "ideal" shapes etc. from college.
I'm still trying to think of another way to do it.
I'm still trying to think of another way to do it.
It will be wonderful if you will find another way, NASA will appreciate it very much... But you was on very good way already; all what you need now to think about why wind changes the pressure you need to calculate weight of atmosphere by your first manner?
P.S. And irregularities of surface do not spoil your approach: in your formula A is indeed "a regular surface area": it is a magic feature of pressure to squeeze planet with the same tension no meter what shape its surface is (recall Pascal’s law!)
The front page of our Forum again is fulfilled with crank.net museum’s exhibitors. So it is time to return to pleasant Physics…
There is my first question:
The probe grounded on Titan easily measured the weight of the entire Titan’s atmosphere. How it did that?
How heavy the atmosphere of Earth is?
The weight of the earth atmosphere is zero.
The weight of the earth atmosphere is zero.
OK, 1100F, Now suck it all up and squeeze it all into a bag and put it onto some scales. Now how much does it weigh? Has the atmosphere got a weight. I think it has.
OK, 1100F, Now suck it all up and squeeze it all into a bag and put it onto some scales. Now how much does it weigh? Has the atmosphere got a weight. I think it has.
Well, it would be more appropriate to can ask for the atmosphere mass instead of weight.
you can use for example the method proposed by wesmorris to find that it is about 5.1 10<sup>18</sup>kg
The weight of the earth atmosphere is zero. http://n93.cs.fiu.edu/measures/
The atmosphere weighs 15 pounds for every square inch. Converted to metric so I can think, that's a mass of 6,900 kilograms per square meter.
The surface area of the Earth ( http://www.vendian.org/envelope/dir1/earth_jupiter_sun.html ) is 510 trillion square meters. That makes the mass of the atmosphere 3.5 quintillion kilograms, and at 9.8 newtons per kilogram it weighs 35 quintillion (35 x 10^18) newtons.
That's a tad more than zero.
http://n93.cs.fiu.edu/measures/
The atmosphere weighs 15 pounds for every square inch. Converted to metric so I can think, that's a mass of 6,900 kilograms per square meter.
The surface area of the Earth ( http://www.vendian.org/envelope/dir1/earth_jupiter_sun.html ) is 510 trillion square meters. That makes the mass of the atmosphere 3.5 quintillion kilograms, and at 9.8 newtons per kilogram it weighs 35 quintillion (35 x 10^18) newtons.
That's a tad more than zero.
You are talking about the mass of the atmosphere.
The weight of the atmosphere is the gravitational force acted on the atmosphere by earth. The total force that is acted by gravitation on the atmosphere will still be zero
You are talking about the mass of the atmosphere.
The weight of the atmosphere is the gravitational force acted on the atmosphere by earth. The total force that is acted by gravitation on the atmosphere will still be zeroOnly if you're talking about the sum of the gravitational force vectors.
If you purely talking numbers and ignoring vectors then the atmosphere DOES have weight  otherwise there would be no pressure.
Weight = Mass x Acceleration (in this case gravity).
If there was no acceleration (i.e. no gravity) then you could put a 1,000,000kg block of concrete above my head and it would have no weight and I would feel no pressure from it.
However, on Earth there is gravity, and luckily I don't have a 1,000,000kg concrete block over my head.
It is the WEIGHT of the airmolecules that create the pressure  which is why pressure drops the higher you get into the atmosphere.
It is not the MASS of the airmolecules that does that.
Only if you're talking about the sum of the gravitational force vectors.
Weight is defined as the gravitational force vector.
If you purely talking numbers and ignoring vectors then the atmosphere DOES have weight  otherwise there would be no pressure.
Weight = Mass x Acceleration (in this case gravity).
If there was no acceleration (i.e. no gravity) then you could put a 1,000,000kg block of concrete above my head and it would have no weight and I would feel no pressure from it.
However, on Earth there is gravity, and luckily I don't have a 1,000,000kg concrete block over my head.
It is the WEIGHT of the airmolecules that create the pressure  which is why pressure drops the higher you get into the atmosphere.
It is not the MASS of the airmolecules that does that.
Well, what you show is that each part of the atmosphere has a weight that is not zero.
If you take a column of atmosphere over som surface, it will have a weight. But again, since the weight is defined as the gravitational force, the weight of the whole atmosphere will be zero.
wesmorris 020305, 08:46 AM Well, it would be more appropriate to can ask for the atmosphere mass instead of weight.
you can use for example the method proposed by wesmorris to find that it is about 5.1 10<sup>18</sup>kg
Well, given F=MA, weight's not a tough calculation given mass and our standard 9.8 m/s^2 convention for accelleration.
Oh I see your point.
Since the accelleration of air toward the surface is net zero... the atmosphere is weightless, but has a whole lotta mass.
Guys,
The discussion in this thread went in the wrong directions, leaving the really interesting issues aside…
1100f,
you are mistaken: weight is not a gravitational force! It is very important! The gravitational force exists and acts on body all the time (till it is into the gravitational field) but weight depends on motion of body and even can vanish at some type of motions (at free falling on center of gravity). Therefore, weight is not a gravitational force!
The direct and correct definition of weight in Physics is the following:
Weight of body is the force with which this body is pressing its support (bearing) or is pulling its suspension bracket/clip. The force of weight is applied not to the body, but to its support or suspension!
Therefore, weight always is caused by a combination of the gravitational force and all other forces acting on the body (usually – inertial ones, but it might be the Coulomb forces, EM forces and any other).
So, forget useless dispute based upon wrong definition of weight, especially in regard to a gas atmosphere, which in any gravitational field will remain to be … an atmosphere with a density distribution according to the Boltzmann’s formula.
So, the main question stays the same: What if Titan had a very windy weather?
P.S. wesmorris, why you change your way of thoughts so easy, instead of challenge the opponent's way first?
wesmorris 020305, 09:31 AM Because I was confused at the difference between weight and gravitational force. I was thinking of them as the same. I was thinking F=ma, so weight = MA. The term is very specific as you've pointed out, and highlights one of the reasons I'm no physicist. :) Hehe.
everneo 020305, 09:41 AM The force of weight is applied not to the body, but to its support or suspension!
Is it not the net force of weight of the atomosphere applied to the whole sphere (support) is 0, as 1100f said ?
But you were absolutely right in the case of our problem:
weight of any athmosphere indeed is equal to the product of its static preassure and area of support!
So, the problem now is the only one: how probe should measure the static pressure if there is a strong wind?
wesmorris 020305, 09:45 AM Is it not the net force of weight of the atomosphere applied to the whole sphere (support) is 0, as 1100f said ?
If that were true, atmospheric pressure at the surface would be zero.
Okay.. I'm going to think about static pressure with a wind.
The rotation of the body is known, so if you subtract the pressure added by the drag of the atmosphere on the body, you should be able to find static pressure. So you could get the mean wind, which could be looked at as the pressure vector tangential to the radius of the body. So you stick a fan up, see how fast it rotates, calculate pressure from that... substract it from the atmospheric pressure measurement and shazaam, static pressure.
everneo,
No, in case of pressure you can not add elementary forces acting on support as they are the free vectors: they are not, they are a coupled vectors, that are tied with their points of action. Isotropic pressure is squeezing body very much. Zero force would not effect on body at all. So, one can understand that doing like 1100f suggests we will come to simple ... nonsense.
But you were absolutely right in the case of our problem:
weight of any athmosphere indeed is equal to the product of its static preassure and area of support!
So, the problem now is the only one: how probe should measure the static pressure if there is a strong wind?
Maybe I'm being a bit simplistic, but wouldn't you just shield the measuring device from the wind  i.e. build it inside the probe, assuming the probe can equalise pressures (is open at some place to the environment).
Given that the weather will affect the readings of pressure (on Earth it can range from 970mb to 1040mb or more (probably far more) regardless of actual exposure to the physical effects of the weather, you would just need to read measurements over a long period of time.
Assuming the wind never dies down, though, I wouldn't know how you would be able to measure anything other than the localised atmospheric pressure.
sarkus,
you need to do the last , the final step, before you will advice to NASA your own good designs of measurer of the staticoressure: to search Internet and find out how the static preassure is meassured ... in science. When you find it out, come back and tell us what you have discovered. Then we will be able to conclude this thread with ... right solution. OK?
blobrana 020305, 02:25 PM Hum,
Just a small offtopic note to those that haven’t heard but it was recently found that the air is slightly heavier than we previously thought.
The composition of air is mostly nitrogen (around 72%) and oxygen (21%). The rest is a mixture of carbon dioxide, water vapour and argon. All straight forward and well understood, however, it is the amount of argon that is the critical factor here, because the more argon present, the denser the air.
The density of air under standard conditions is only 1.239 milligrams per cubic centimetres.
In 1969, it was calculated that the amount of argon in the air was around 0.917%. However, recently, a team from the Korea Research Institute of Standards and Science (KRISS) and the International Bureau of Weights and Measures (BIPM) in France, have determined that the actual amount of argon is 0.9332%, ± 0.0006.
Air density is increased.
So this in turn will affect buoyancy, or static pressure…
Anyway the presence of argon 40 was also discovered in the atmosphere of titan, though I haven’t seen the data; but it would be a similar process to work out the density of Titan’s air.
but i think they used the data from the decent time (allowing for the size of the parachutes) they could narrow down the column height of the air and work out its density.
You are talking about the mass of the atmosphere.I expressed the math in terms of how much the mass of the atmosphere weighs in newtons, which is a unit of weight. Read it again while you're thinking again.
The weight of the atmosphere is the gravitational force acted on the atmosphere by earth.Incorrect. Weight is the resistance to acceleration.
everneo 020405, 01:58 AM weight of any athmosphere indeed is equal to the product of its static preassure and area of support!
If the whole atmosphere gets heated uniformly, the isometric pressure on the surface too increases, that is weight of the atmosphere changes with temperature?
So, the problem now is the only one: how probe should measure the static pressure if there is a strong wind?
The main reason for wind is the difference in atmosphere temperature hence the difference in pressure.
If you take PA as the weight of the isometric atmosphere you have to bring in the temperature also into the picture.
everneo,
Wonderful!
If the whole atmosphere gets heated uniformly, the isometric pressure on the surface too increases, that is weight of the atmosphere changes with temperature?
Please, explain why you think so? (I guess, the answer will be interested to everybody here...)
everneo 020405, 04:56 AM Yuiry,
I was expecting an answer from you!
It was an ideal situation, ( but in reality we know that the atmosphere has different layers like troposhpere and lower & upper stratospheres with generally decreasing temperature from the earth surface to the upper spheres. Pressure reduces with decreasing temperature and increasing altitude. .. with all sort of thermodynamic, fluiddynamic properties ) where i assumed the tempeature is uniform throughout the atmosphere that has isometric pressure.
In that case, with uniform increase in temperature, the atmosphere tends to increase in thickness because of pressure, but it is bounded by the earth surface at lowest level and bounded by gravity at upper level. So, the uniform increase in temperature throughout the atmosphere increases the pressure at surface (lower boundary) significantly.
HallsofIvy 020405, 02:41 PM >“ Originally Posted by 1100f
>You are talking about the mass of the atmosphere. ”
>I expressed the math in terms of how much the mass of the atmosphere weighs in >newtons, which is a unit of weight. Read it again while you're thinking again.
Absolutely! "weight" is a force, not mass.
>“ Originally Posted by 1100f
>The weight of the atmosphere is the gravitational force acted on the atmosphere by >earth. ”
>Incorrect. Weight is the resistance to acceleration.
Now, THAT'S incorrect. MASS is the resistance to acceleration, which can take place even in deep space with no gravitational masses nearby. WEIGHT, is by definition, the gravitational force of the earth on a body (on the earth weight on another planet, moon, etc. would be that body's gravitational force).
Guys,
One more time:
weight is not a gravitational force! It is very important! The gravitational force exists and acts on body all the time (till it is into the gravitational field) but weight depends on motion of body and even can vanish at some type of motions (at free falling on center of gravity). Therefore, weight is not a gravitational force! In some conditions weight has the same value and direction as the gravitational force, in other conditions  it is not the same. The state of weightless is the best proof of that.
The direct and correct definition of weight in Physics is the following:
Weight of body is the force with which this body is pressing its support (bearing) or is pulling its suspension bracket/clip. The force of weight is applied not to the body, but to its support or suspension!Therefore, weight always is caused by a combination of the gravitational force and all other forces acting on the body (usually – inertial ones, but it might be the Coulomb forces, EM forces and any other).
So, forget useless dispute based upon wrong definition of weight, especially in regard to a gas atmosphere, which in any gravitational field will remain to be … an atmosphere with a density distribution according to the Boltzmann’s formula.
geistkiesel 020405, 09:09 PM Yuriy,
Wouldn't the mass of the heated atmosphere increase for the same reason the mass of an accelerated particle seems to increase when v > c?
Geistkiesel
Ask your friend MacM, he will answer you "scientifically"....
geistkiesel 020405, 11:39 PM If the whole atmosphere gets heated uniformly, the isometric pressure on the surface too increases, that is weight of the atmosphere changes with temperature?
The main reason for wind is the difference in atmosphere temperature hence the difference in pressure.
If you take PA as the weight of the isometric atmosphere you have to bring in the temperature also into the picture.
Everneo,
It is impossible to heat the atmosphere uniformly. Half the atmosphere is opposite to the sun, and is therefore cooln; further for the atmosphere exposed to the sun the heat absorption can never be uniform due, for one reason, to the tilt of the earth's axis.The total weight of the atmosphere must always vary, if indeed the temperature increases the weight.
Geistkiesel
geistkiesel 020405, 11:40 PM Ask your friend MacM, he will answer you "scientifically"....
It seems you are making this a personal mater.
Geistkiesel
everneo posted:
It was an ideal situation, ( but in reality we know that the atmosphere has different layers like troposhpere and lower & upper stratospheres with generally decreasing temperature from the earth surface to the upper spheres. Pressure reduces with decreasing temperature and increasing altitude. .. with all sort of thermodynamic, fluiddynamic properties ) where i assumed the tempeature is uniform throughout the atmosphere that has isometric pressure.
In that case, with uniform increase in temperature, the atmosphere tends to increase in thickness because of pressure, but it is bounded by the earth surface at lowest level and bounded by gravity at upper level. So, the uniform increase in temperature throughout the atmosphere increases the pressure at surface (lower boundary) significantly.
Actually in this post we have mentioned everything that is needed to be considered to come to the right conclusion in this thread. all what we need now  a little connection with Science...
1. The first of all we should recognize how the contemporary Physics describes the atmospheres of planets.
In the equilibrium state (the rested atmosphere, no winds, no gradients of temperature, etc) the distribution of gaseous atmosphere (in Physics term “atmosphere” is used in regard of many other situations: for instance, when vortex moves in liquid it caries on some amount of liquid, which is called as “the vortex’s atmosphere”; this atmosphere is streamlined by rest liquid) is described by famous Boltzmann’s formula:
D(h T) = D(0T)exp(mgh/kT)…………………(1)
D(xT) is the gas’ density on the level x (above ground x = 0), m is gas’ molecules’ mass, g =981 cm/s^2, k is the Boltzmann’s constant and T is absolute temperature.
So, the atmosphere has no “upper boundary”! But D essentially depends on T.
2. If we will change the temperature of atmosphere, T, the distribution of gases will change, but the pressure, Po, on the planet’s surface will remain the same – equal to the weight of atmosphere per unit of square of the planet’s surface. So, measuring this (the static or taken at total rest of atmosphere!) pressure on the ground of planet, we actually are measuring the weight of the planet’s atmosphere (it assumes that we know the area of the planet’s surface!). this pressure of a totally rested gas is called “the stagnation pressure” or “the static pressure”.
3. We can not apply the ideal gas laws of thermodynamics to the atmosphere as a whole – the atmosphere as a whole is not a closed thermodynamic system! But we can apply those laws to some parts of atmosphere if we are considering some fast enough processes with these parts (for instance, to kinematics of wind). Such processes usually are adiabatic thermodynamic processes and are very well ‘describable” by Poisson’s formulas and usual hydrodynamics of an ideal gas.
4. As we know, the major consequence of the stationary motion of an ideal gas is the Bernulli’s law:
5.
Po = P + D(0T)v^2 /2 ……………………..(2)
So, the measured pressure P will be differ (will be less) from the pressure Po on the value of so called “dynamical pressure” D(0T)v^2 /2. Of course, this pressure has nothing to do with weight of atmosphere!
Therefore, to measure Po, we should use such a tool that … measures namely pressure Po, not P. This tool was invented by Pitot and is called “the Pitot’s tube”.
Pitot’s tube measures the dynamical pressure in a gas’ flow. If we we can measure only P, not Po, we have to accompany it with measuring of temperature T and velocity of gas, v. Then knowing what gas we deal with, we will be able to find D(0T) and calculate Po.
superluminal 021705, 04:14 PM http://www.efunda.com/designstandards/sensors/pitot_tubes/pitot_tubes_theory.cfm
Thanks, superL, for good reference. I fixed some errors in my post after this material. Now it looks better.
superluminal 021705, 06:17 PM My pleasure.
