So, I know that atmospheric pressure is a result of both the atmospheric mass and the force of gravity acting on it. If you were to decrease a planet's gravity, but keep the atmospheric mass the same, obviously the lower gravity would result in decreased atmospheric pressure for the same mass of air. But would the air molecules also all spread out more, decreasing the air density as well? I'm wondering if breathing on a lower gravity planet would feel like breathing at a higher altitude on earth, even if the atmospheric mass was the same.
3 Answers
In an isothermal atmosphere, the density is given approximately by $$\rho = \rho_0\exp(-\mu gh/k_BT)\ ,$$ where $\mu$ is the mean mass of a particle, $h$ is the height above ground level and $g$ is the surface gravity.
The total mass per unit area, $\sigma$, is given by the integral of this expression from $h=0$ to $h=\infty$. $$\sigma = \frac{\rho_0 k_B T}{\mu g} $$
If the temperature and total atmospheric mass is fixed and the radius of the planet is also fixed, then $\sigma$ is constant and the density at ground level is therefore proportional to the surface gravity. As the mass of the planet decreases the atmospheric density will therefore decrease with the mass.
Yes the air density would decrease. The density is proportinal to $ e^{-g*h/T} $ where g is the local force of gravity h is the height and T is the tenperature. Lower gravity gives lower pressure and density. More can be found here
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$\begingroup$ I wanted to add that you can think of this intuitevly as the molecules being squeezed together less hard $\endgroup$ Commented Jul 29, 2022 at 1:21
In addition to other answers, since the question already states that pressure is going to decrease, it should be interesting to note that if temperature doesn't change, volume is inversely proportional to pressure (that is Boyle's law) and therefore density is proportional to pressure.
Then the answer is yes, breathing in a lower gravity planet with the same atmosphere would feel like breathing at higher altitude on Earth.