Question about atmospheric pressure

[brainfuel]

For a given mass, as radius decreases, would atmospheric pressure increase, since surface gravity would increase? I'm assuming that atmospheric composition doesn't change.

For a planet with radius = 1.1*Earth and mass = 1.3*Earth and atmospheric molar weight within 1% of Earth's, does a surface pressure of 108.75 kPa seem reasonable?

 

[IsometricPion]

If you assume the atmosphere is made of an ideal gas in isothermal equilibrium, and that it is not so thick that the gravitational acceleration changes significantly across it, the formula is: [tex] P(h)=Ce^{\frac{-\mu{}GM}{RTr^2}h} [/tex] where [tex] h [/tex] is the height, [tex] \mu [/tex] is the molar mass of the gass, [tex] G [/tex] is the universal gravitational constant, [tex] R [/tex] is the molar gas constant, [tex] T [/tex] is the temperature, [tex] r [/tex] is the radius of the planet, and [tex] C [/tex] is the pressure at the surface of the planet. Since one has to know the pressure at some height before one can determine this function, it cannot answer your question as to the appropriateness of the surface pressure you gave. However, other factors could be used to determine an appropriate surface pressure. Geochemically, the atmosphere has its current composition and volume due to the balance of the processes that add and remove gases from it (e.g. photosynthesis adds oxygen, oxidation (burning, rusting, etc.) removes it). Biologically, organisms (especially complex ones) usually only survive in a fairly narrow region of pressures and temperatures.

So, one must consider additional constraints in order to answer your questions. For example if one assumes that the total mass of the atmosphere is constant (with pressure as a function of height given by the equation above and gravity by the inverse square law), then reducing a planet's radius by a certain factor will increase the pressure at the surface by the (approximately) inverse square of that factor (e.g., [tex] r_{2}=\frac{r_{1}}{2} \Rightarrow C_{2}=4C_{1}[/tex]). So, for the surface pressure to be 108.75 kPa the mass of the atmosphere would have to be about the same as the Earth's (for 1.1 times the radius and 1.3 times the mass and the same molar weight it would have to be the same within 1%, though I have no reason to believe that having the same mass of atmosphere is a realistic constraint)

 

[Janus]

Another factor to consider is that escape velocity plays a role in how much atmosphere a planet will hold. In your example the escape velocity would be 1.09 times stronger than Earth's while the surface gravity would only increase by 1.036. Thus would tend to retain a thicker atmosphere.

 

[brainfuel]

By thicker do you mean "of greater height" (from troposphere to exosphere) or "more dense/viscous"? In either case, surface pressure would be greater, right?

 

[qraal]

By thicker do you mean "of greater height" (from troposphere to exosphere) or "more dense/viscous"? In either case, surface pressure would be greater, right?

Right. Atmospheric pressure is from the weight of all the air above the given area in question, but the density/viscosity is from the local properties (temp, pressure, composition) and can change with height - the exosphere is mostly H/He, for example. Increasing gravity will increase the weight, but also the density due to the pressure increase.

 

[Ken G]

Just to correct some of the above, if you assume the temperature doesn't change (so no increased greenhouse effect) and the total amount of air is the same (so no difference in the escape of gas), neither of which are likely to be true but you have to start somewhere, then the way you can figure out the surface pressure is to first find the effect on the height of the atmosphere. The "scale height" scales inversely to the surface gravity, so that's the inverse-square effect with radius that IsometricPion mentioned. However, that's not the answer for the pressure, that's just the height of the atmosphere. The pressure will depend on the density, and if you have to squeeze the same amount of air into a volume that scales like the height of the atmosphere times the surface area of the planet, the surface area gives you two more powers of radius. So if the height scales like r² and the surface area also scales like r², the volume filled by the air scales like r⁴, and the density scales like r^-4. So a 1% decrease in radius comes with a 4% increase in pressure, keeping the amount of air constant.

 

[IsometricPion]

It appears I did the wrong integral to obtain the inverse square proportionality I gave above. Having reworked the integration, I agree with Ken G. In addition, I would like to mention that the surface pressure is (to first order) proportional to the ratio of the new mass to the old, so for 1.1 times Earth's radius and 1.3 times Earth's mass and the same mass of atmosphere, the surface pressure is 1-4*(1.1-1)+(1.3-1) = 1-0.4+0.3 = 1-0.1 = 0.9 times that on Earth. The total mass of the atmosphere is proportional to the surface pressure. So, to have a surface pressure of 108.75 kPa its atmosphere would have to have 108.75/(0.9*101.3) or about 1.2 times the mass of Earth's (which is seems plausible).

 

[qraal]

Decreasing radius is one way to increase a planet's gravity. The idea is to collapse the core into some super-dense form. That's the handwavium part. Take Mars. Gravity is 0.378 Earth. To increase to Earth levels we want to drop the radius to (0.378)^0.5 of it's present size - from an average 3390 km to 2084 km. Means the surface area drops to 0.378 of its present value, so the surface pressure would rise by the reciprocal. Of course all the surface would be heavily volcanic from the gravitational violence committed, but that's what a radical remaking of the planet would require.