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I'm asking for clarification here. If Earth had the same atmospheric mass per square unit of ground but the Earth had suddenly gained mass so it had twice the gravity at the surface, would the Earth now have twice the atmospheric pressure just because of the doubling of gravity? I know pressure is defined as force per area, but I'm not sure if air pressure works the same way. When I look up gravity and air pressure on the Internet it just has information on air pressure with height, but I'm not looking for that.

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  • $\begingroup$ In your scenario, would Earth's mass change without its size changing (i.e. change of its density)? If the size changes, and you don't change the total mass of the atmosphere, there will of course be less atmospheric mass for the same area. $\endgroup$
    – jcaron
    Dec 10, 2020 at 14:05

7 Answers 7

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Atmospheric air pressure occurs because the atmosphere is made of stuff that has mass, and therefore has weight due to gravity. The pressure of an atmosphere on a horizontal surface represents the total mass of a column of atmosphere from that surface all the way up into space. Of course atmospheric pressure works in all directions; considering only the vertical force allows us to understand how that pressure arises. As we travel up that column, pressure decreases because there is less mass above to weigh down on the column below.

As we travel up the column, gravity decreases (inversely proportional to square of distance from Earth's center) but we can ignore this because the atmospheric "column" isn't a uniform section - it is a tapered section (converges to an imaginary point at the Earth's center) which exactly cancels out the decrease due to gravity because the cross-section of our atmospheric slice will increase in area as the square of distance from the Earth's center.

If you double the surface gravity, all other things being equal, you will double the weight of that same mass of air, so you will double the pressure at the surface. Doubling the pressure will double the density - the atmosphere will "crowd" closer to the surface and the pressure vs altitude profile will look somewhat different.

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    $\begingroup$ True, the gravitational acceleration doesn't drop by much over 100 km. But actually we can ignore that small change. As Rob Jeffries noted in this comment on a recent Astronomy answer, "the increase in area compensates for the decrease in gravity", since gravity is proportional to $1/r^2$ and the area of a layer of atmosphere is proportional to $r^2$. $\endgroup$
    – PM 2Ring
    Dec 8, 2020 at 23:42
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    $\begingroup$ @PM2Ring That is a cute, but useful, observation. I'll chalk it all up to the divergence of a radial field being $\nabla \cdot (g(r)\hat r) = \frac 1 {r^2}\frac d {dr} (r^2g(r))=0$ The $r^2$ in the derivative forces the $1/r^2$ field, while the $1/r^2$ out front corrects for area of the (atmo)sphere. $\endgroup$
    – JEB
    Dec 9, 2020 at 0:05
  • $\begingroup$ Huh nice never knew it balances out like that $\endgroup$ Dec 10, 2020 at 14:32
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Air pressure has everything to do with height because it is caused by the force exerted from the volume of air above weighing down.

That said, if Earth had twice the gravitational field strength at the surface, by the inverse square law this would mean that the Earth doubled in mass. This would double the gravitational force on all objects if they remain at the same distance r from the Earth's centre as before:

$$F_{g,earth} = G \frac{M_{earth}m}{r^2}, \space \therefore F_{g,double} = 2F_{g,earth} = G \frac{2M_{earth}m}{r^2} $$

Therefore the force exerted by layers of air above would double. $P \propto F$, therefore atmospheric pressure $P_{atm}$ too would double.

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The phenomenon that determines atmospheric pressure is very similar to Pascal's Law for incompressible liquids. But air isn't an incompressible liquid: as a gas its density is quite dependent on pressure and temperature, for instance.

A doubling of $g$ would nonetheless double also the atmospheric pressure.

I know pressure is defined as force per area, but I'm not sure if air pressure works the same way.

The definition of pressure isn't dependent on the type of fluid.

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If you double the gravity the air pressure will be doubled.

$F = mg$

we can replace $m$, mass, with density and volume, so,

$F = \rho V g $

Now let's write the volume as area time height.

$F = \rho h A g$

Now we can calculate the pressure

$P = \frac{F}{A} = \rho h g$

You can see that pressure is directly proportional to $g$. Note that $h$ is height of the air above your head.

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  • $\begingroup$ +1 Instantaneously, this is true. However, due to the increased $g$, after some time the density of the air would increase while the height of the air column would decrease. $\endgroup$ Dec 9, 2020 at 18:09
  • $\begingroup$ For the atmosphere the formula $p=\rho gh $does not work. The density decreases with height so the relationship is not linear but exponential. And this is an approximation too. $\endgroup$
    – nasu
    Dec 9, 2020 at 23:52
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The atmospheric pressure is directly related to gravity. Double the gravity and you should have double the pressure, at least on average.

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    $\begingroup$ This is only approximately true when the atmosphere is very shallow compared to the radius and gravitational acceleration is nearly equal across the entire atmosphere. On Titan, for example, gravity decreases by about half across an atmosphere that's ~1000 km deep on a body only 2500 km in radius. A higher gravity would compress the atmosphere deeper into the gravity well and cause a greater than linear increase in pressure. $\endgroup$ Dec 9, 2020 at 3:06
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    $\begingroup$ That’s why I said “at least on average” Under the ocean about every 33 feet down is an atmosphere. $\endgroup$ Dec 9, 2020 at 4:08
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The force per unit area at the surface is the weight of the column of air above the unit area. If you double gravity, the weight of the column doubles.

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Without gravity the atmospheric pressure would be zero. That is, there would be no atmosphere.

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