# Why does atmospheric pressure act on us?

Imagine that the air in the atmosphere was just somehow sitting there unpressurised. What would happen?

Well, Earth's gravity would be attracting all that air towards the centre. So the air would start to fall downwards.

The very bottom layer of air would be prevented from falling through the solid surface, as the air molecules rebound off the molecules of the surface. But the layer above that doesn't stop. So Earth's gravity forces the air in the lower part of the atmosphere to accumulate against the surface of the planet, becoming more and more dense.

As the air gets denser near the surface, it becomes more and more likely that air molecules collide. That's what air pressure is: the average force of all those air that would hit a surface you placed in the air. But the air pressure also acts on the air itself. So eventually the force of the air pressure at the bottom layer of air pushes up on the layer of air just above it enough to counteract the pull of Earth's gravity on that layer of air. And so you get another layer that is prevented from falling.

But the air above that is still being pulled down, and so more air is being squashed down into this second layer above the surface. This increases the force that the bottom layer needs to provide to the next layer; the air molecule collisions not only need to provide enough force to counteract the weight of the air immediately above it, but also to provide those molecules with enough momentum that when they in turn collide with the air in the third-bottom layer it can support the weight of that layer as well. So more air squeezes down to the surface until the pressure at the bottom layer is sufficient to support the weight of the 2 layers above that.

Obviously the atmosphere isn't actually split into discrete layers like this1, but hopefully it's a helpful way to think about it. You should be able to see how gravity squeezes the air down against the solid surface, until the pressure at the bottom is just enough to support the weight of all the air above it.

This is why air pressure drops off at higher altitude. As you go up, there is less air above squeezing down, so equilibrium with gravity is reached at a lower pressure.

So it's not literally that the air pressure you feel is the weight of the column of air above you. It's not that your head is somehow "holding up" a 100km column of air above it. But the air pressure of the air surrounding you must provide an equivalent force to the weight of all the air above it. If it did not then the weight of the air above would be partially unsupported, so gravity would squeeze it down further, increasing the pressure until it was equal to the weight of all of the air above.

This is also why the top of your head doesn't feel any difference in air pressure to the side of your body. Air pressure is the same in all directions, because the air molecules are really just zipping around in countless different directions, uncoordinated with each other. Those molecules colliding with things must supply enough average force in the upward direction to support the weight of the atmosphere, but when the pressure increases due to gravity it can't cause a coordinated force that is only upwards, so there is just as much force from air pressure on the side of your body as there is on your head.

1 And if you actually had the atmosphere of Earth spread out in a diffuse low pressure cloud and let it all fall under gravity the results would be much more exciting than I have described.

I think it is misleading to explain air pressure as the weight of the air column above our head acting per unit area.

Actually your head only feels those air molecules which are directly hitting the surface of your head. You don't feel any air molecules which are further away.

As explained in Wikipedia - Kinetic theory of gases - Pressure and kinetic energy:

... the pressure is equal to the force exerted by the atoms hitting and rebounding from a unit area of the gas container surface.

Using this approach the quoted article derives this formula for the gas pressure $$P$$: $$P = \frac{Nm\overline{v^2}}{3V},$$ where $$m$$ is the mass of a single gas molecule, $$\overline{v^2}$$ is the average square velocity of the molecules, and $$N$$ is the number of molecules per volume $$V$$.

From this formula you can see quantitatively, that gas pressure is caused by the speed of the hitting molecules, and it increases with the density of the gas ($$Nm/V$$).

Consider blocks of equal mass, stacked on each other.

The block at the top ‘feel’ the force of gravity and normal force from the block below. The block just below feels force of gravity (it’s weight) plus the force acting from the box above by Newton’s $$3^\mathrm{rd}$$ Law (if it ‘provides’ normal force on box A, box A acts on it with equal magnitude, but in opposite direction force). The result is the greater normal force.

If you continue with this stack and place yourself or any other ‘object’ below this stack, you will conclude that each block ‘contributes’ to the total force that is acting on you. Air molecules, although much smaller, must obey Newton’s Laws, too.

Your last assumption would be mostly correct if gravity field near air particles would be negligible.

Please note that atmospheric pressure varies due to various other factors, but this simple model should explain the principle.