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Why is the pressure on the ground equal to the weight of the atmospheric column on it? On the table top, the coins are stacked together, and the pressure on the table top is equal to the weight of the coin column, because the coins are in contact with each other. The atmospheric molecules are not in contact with each other. There is a large space between them. How can the pressure on the ground equal the weight of the atmospheric column? Is there any experimental proof?

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  • $\begingroup$ Thought experiment / useful homework problem: Suppose the pressure at the base of a column of air (or your favorite other fluid) were different from the weight of that fluid. (Of course you have to do pressure times area to compare to the weight force.) Construct a free-body diagram and predict the motion of the system. $\endgroup$ – rob Dec 14 '19 at 6:36
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    $\begingroup$ Does this answer your question? Why does atmospheric pressure act on us? $\endgroup$ – Sandejo Dec 14 '19 at 6:53
  • $\begingroup$ Why not? The pressure produced by a column of fluid is dependent on the amount of fluid that is above the point where the pressure is measured, even if that fluid is a gas. $\endgroup$ – David White Dec 14 '19 at 7:27
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The atmospheric molecules are in contact with each other. They bounce off one another constantly, and at high speed. In this way, each molecule communicates to those below it the pressure generated by the weight of those above it.

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  • $\begingroup$ How can the impact force equal the molecular weight? $\endgroup$ – enbin zheng Dec 14 '19 at 12:26
  • $\begingroup$ How would be a horizontal collision different than a vertical one? $\endgroup$ – Wolphram jonny Dec 14 '19 at 16:50
  • $\begingroup$ @Wolphramjonny The colliding vertical wall does not contain weight, does it? $\endgroup$ – enbin zheng Dec 14 '19 at 17:32
  • $\begingroup$ @enbinzheng it is irrelevant if it does, so you can assume either $\endgroup$ – Wolphram jonny Dec 14 '19 at 17:44
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So, what has happened is that you have over simplified your explanation for the stack of coins, and now you're having trouble simplifying the atmosphere in the same way. The actual explanation is the interplay of two principles so similar that they are regularly confused by undergraduates.

The first question is, why is any given coin not accelerating upward? The answer is given by Newton's first law, it is in a state of force balance, where the up-force is exactly the same as the down-force. The second question is, what does Newton's third law now state? It states that the up-force on any one coin must be compensated by an equal down-force on the coin underneath it.

That direct physical contact is totally irrelevant can be seen with magnets. If you have a magnet shaped like a washer with its north-south axis along the axis of rotational symmetry, then many of them together can be stuck together into a cylinder with a hole cut out of the middle. But if you stick a dowel rod through the middle and orient two magnets opposite, they will repel. If the dowel rod sticks vertically out of a table, then one of the magnets will float in mid-air, and it will not touch the lower magnet. But Newton's third law still requires the lower magnet to feel the same magnetic force that the upper magnet does, just opposite: and Newton's first law still requires that magnetic force to oppose the force of gravity for the floating magnet, otherwise it would be accelerating.

Air molecules in the atmosphere are constantly bouncing every single way, so force balance is not a perfect description. However, even though it doesn't hold for individual air molecules, it holds on average for a large volume containing many of them: in this case what we are looking at is not an unchanging individual momentum but a total momentum now also able to follow by exchanging particles with other nearby volumes. Still the total momentum in the volume must stay constant, and then conservation of total momentum requires the lower volume that pushes up to balance the upper volume.

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  • $\begingroup$ Maybe the air molecules don't collide, and there's a distant force between them, similar to the repulsion between magnets. Air molecules support each other through such forces, so they transfer the weight of air molecules to the ground. $\endgroup$ – enbin zheng Dec 14 '19 at 14:52
  • $\begingroup$ It doesn't matter. If there is no net acceleration of the air downwards then there is constant momentum and thus an effective force balance. $\endgroup$ – CR Drost Dec 15 '19 at 1:21
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On the table top, the coins are stacked together, and the pressure on the table top is equal to the weight of the coin column, because the coins are in contact with each other. The atmospheric molecules are not in contact with each other.

Even the coins are not in a total contact with each other. If you can magnify the space between each two successive coins, you can see that too many electron shells of one coin tend to repel the electron shells of the other keeping a very small void between the last atomic shells of each coin. The difference is just, in atmospheric atoms/molecules, this void is of greater magnitude.

The relevant forces (gravitational weights) are conveyed through these intense electric fields produced between the coins or atmosphere atoms regardless of whether they are in complete contact. Remember that if you bring two $S$ poles of two magnets close to each other, your hands feel force, though there is still a gap between the two magnets and they are not in contact yet.

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  • $\begingroup$ The space between air molecules is too large. How can we transfer weight? After all, coins are solid. $\endgroup$ – enbin zheng Dec 14 '19 at 12:41
  • $\begingroup$ Maybe molecules don't transfer weight by hitting each other. There are also forces between molecules at great distances. $\endgroup$ – enbin zheng Dec 14 '19 at 12:44
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Pressure is proportional to both the number of collisions per unit time and the speed of the collisions. Molecules at the top have more potential energy than those at the bottom. This creates a redistribution of either velocities or density, because as molecules move and collide the ones at the top will gain speed and transfer more momentum to the ones on the bottom. For an ideal gas $P=\rho kT/M$ (with M the molecular mass), so a change in pressure will necessarily be associated with a change in density, temperature or both.

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  • $\begingroup$ Is the collision of the bottom molecule exactly equal to the total weight of the top molecule? $\endgroup$ – enbin zheng Dec 14 '19 at 22:39
  • $\begingroup$ I am not sure what do you mean by that, but the force between the molecules is not equal to their weight $\endgroup$ – Wolphram jonny Dec 15 '19 at 0:15
  • $\begingroup$ but do not mix the two descriptions. One is macroscopic and talks about weight. It is the result of the microscopic description which is not directly apparent about weight. $\endgroup$ – Wolphram jonny Dec 15 '19 at 0:18

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