# Higher Gravity vs Higher Air Density

Planets that are larger and have higher gravity tend to have thicker atmospheres, since it is easier to hold onto gases.

Higher air pressure leads to a higher air density, and therefore, a greater the air resistance.

So, on a planet twice the mass of Earth with a proportionally thickened atmosphere, would a leaf fall faster or slower?

• @EliasRowanAlbatross Gravity is the primary factor that affects fall rates.... Nov 11 '19 at 0:41
• @EliasRowanAlbatross If you knew you had no idea, why did you guess and get it wrong? Nov 11 '19 at 3:05
• To clarify, are you asking about 1) a rocky planet; 2) twice the mass of Earth; 3) with the same density as Earth; 3) with the same atmospheric composition as Earth; 4) with the same surface temperature as Earth; 5) with the mass of the air column increased proportionally to surface gravity? (i.e., total mass of the atmosphere increased proportional to surface gravity * surface area) Nov 11 '19 at 9:33
• I was working on a universe simulator for some time and procedurally-generating atmospheric pressure is a rabbit hole I pulled my hair out for two weeks trying to understand. Basically, you'll be stumped if you don't realise that different atmospheres can have different volumes by cosmic chance (gases in the planetary nebula etc), so when you start applying thermodynamics and pressure calculations, the volume term will keep popping-out. In the end, I randomized the atmospheric volume and composition, then worked out the pressure from there using ideal gas equations and gravity relations. Nov 11 '19 at 11:45
• Bouyant force has barely anything to do with a leaf falling. What's the connection between the second and third paragraph of this question?
– JiK
Nov 11 '19 at 12:44

1. Twice the mass of Earth; $$M = 2 M_e$$
2. The same bulk density as Earth; $$\rho = \rho_e$$
3. The same atmospheric composition as Earth;
4. The same surface temperature as Earth; $$T = T_e$$
5. An air column proportional to surface gravity; i.e. the total mass of air above any square meter of surface on the planet is greater than on Earth by the same proportion as surface gravity; $$\sigma_{air} = \sigma_e \frac{g}{g_e}$$. With the previous assumptions, this ends up meaning that the total mass of the atmosphere is double that of Earth.

The mass of a sphere is its volume times its density: $$M = V*\rho$$ The volume is given by: $$V = \frac{4}{3} \pi R^3$$ By setting the mass equal to double the mass of Earth, we can find the radius: $$\frac{4}{3} \pi R^3\rho_e = 2 \frac{4}{3} \pi R_e^3\rho_e$$ $$R^3 = 2 R_e^3$$ $$R = \sqrt{2} R_e$$

# Surface gravity

Surface gravity is calculated from the formula: $$g = G \frac{M}{R^2}$$ Substituting from above, $$g = G \frac{2 M_e}{(\sqrt{2} R_e)^2}$$ $$= \sqrt{2} G \frac{M_e}{R_e^2}$$ $$= \sqrt{2} g_e$$

# Air pressure and density

We decided that the mass of the air column was proportional to surface gravity:

$$\sigma_{air} = \sigma_e \frac{g}{g_e} = \sqrt{2} \sigma_e$$

Air pressure is the mass of the air column times the acceleration due to gravity (for a thin shell of atmosphere like Earth's, we can assume that the acceleration due to gravity is constant in the atmosphere without much error).

$$P = g \sigma = \sqrt{2} g_e \sqrt{2} \sigma_e = \sqrt{4} P_e$$

From the ideal gas law, we know that density is proportional to pressure at constant temperature (and composition):

$$\rho_{air} = \sqrt{4} \rho_{air,e}$$

# Terminal velocity of a leaf

The density of a leaf is much higher than air, so I will ignore buoyancy effects. I'll also assume a drag coefficient of 1, which strikes me as reasonable for a leaf.

The velocity of the falling leaf is when the drag force due to air resistance balances the force of gravity:

$$\frac{1}{2}\rho_{air} v^2 A_{leaf} = m_{leaf} g$$

Assume the leaf has some thickness $$d$$ and density $$\rho_{leaf}$$ which are the same on both planets. Then:

$$\frac{1}{2}\rho_{air} v^2 A_{leaf} = A_{leaf}d\rho_{leaf} g$$ $$v^2 = 2 d \frac{\rho_{leaf}}{\rho_{air}} g$$ $$= 2 d \frac{\rho_{leaf}}{\sqrt{4} \rho_{air,e}} \sqrt{2} g_e$$ $$= \frac{1}{\sqrt{2}} v_e^2$$ $$v = \frac{1}{\sqrt{2}} v_e$$ $$\approx 0.89 v_e$$

# Different assumptions

If we make the atmospheric increase larger, which I think is more realistic, then the leaf will fall even slower. If we increase the surface temperature or the proportion of light gases (Helium, Neon), which are also realistic, then the air density will be less and the leaf would fall faster. Making the planet denser (rock is not very compressible, so this would probably mean more Iron relative to Silicon) would increase the surface gravity, but since the atmospheric increase was proportional to surface gravity, this still makes the leaf fall slower.

It is worth noting that maintaining Earth's surface temperature in a thicker atmosphere implies that the planet orbits farther from its star or has a dimmer star.

tldr; For these assumptions, the leaf will fall slower on a larger planet.

# Edit: Buoyant force

The buoyant force is given by the displacement of air by the leaf. The density of a fresh leaf, like other living tissues, is close to that of water, about $$1000 kg/m^3$$. The density of air at standard temperature and pressure is about $$1.2 kg/m^3$$. So the buoyant force is roungly $$0.1\%$$ of the force of gravity. If we double the mass of the planet and its atmosphere, then the density of the air increases to $$\sqrt{4} \times 1.2 kg/m^3$$, or $$1.9 kg/m^3$$, so the buoyant force increases to almost $$0.2\%$$ of the force of gravity. This is still too small to an effect to bother including in the calculation.

The fact that a leaf falls slower in air than in a vacuum is almost entirely due to drag (aka air resistance), not buoyancy. You can test this by crumpling a leaf into a ball. It has the same buoyant force it always did, but much reduced drag, and it falls much more quickly.

• Note: if you want to change the mass of the planet by some other factor, then just replace the "2" in the final answer; e.g., if the planet is 1.5 times the mass of Earth, then 1.5^(-1/6) Nov 11 '19 at 11:15
• I was double-checking the math here, but I got lost after M = and half the symbols aren't on my keyboard... so +1 for making your own alphabet and writing a poem with it Nov 11 '19 at 11:27
• @Kilisi The equation are set using mathjax with has similar notation to latex. The only non keyboard letters I see are the Greek letters pi=$\pi$, rho=$\rho$ and sigma=$\sigma$, each typeset with a backslash in front of the name and encased in dollar symbols. Nov 11 '19 at 11:56
• In your third line, you got to $v^2 \propto {g\over{\rho_{air}}}$. Since $\rho$ and $g$ are proportional, any change in $g$ is cancelled out by a corresponding change in $\rho$. So $v$ doesn't change. I don't get all the stuff after that line... Nov 11 '19 at 13:32
• @OscarBravo $\rho \propto g^2$ (because $\rho = g \times \sigma$ and, by assumption, $\sigma \propto g$) Nov 11 '19 at 19:40

Gravity is not the only thing to consider for atmospheric density; pressure (and by extension density but they are different properties) can also be increased by the amount of energy (read as heat) being stored in the atmosphere. Also, the relative strength of a magnetic field ( terrestrial or induced) can have an impact on how much of the atmosphere is retained as well.

This is best exemplified by Venus, with a very similar size and mass to earth but much denser atmosphere. There, bouyancy is your friend as it is going to be higher than on Earth but only needs to counteract the same gravitational force. Of course, building a bouyant airship that can survive that atmosphere without dramatically increasing weight is going to be a challenge, meaning that we're not comparing similar circumstances.

Finally, the other consideration is planetary density. Gravitational force increases proportionally to the square of your proximity to the center of mass, so a planet with the same mass as earth but (say) half the diameter is going to put you further down the gravity well yet may only have a similar atmospheric density. That would make the bouyancy have to work harder against gravity than on a planet with earth like density.

In short, there are a lot of factors to consider, gravity and atmospheric density being only two. All other things being equal however, I'd argue that if planetary density is equal, atmospheric density is proportional, etc. then the real deciding factor is whether the distance from the center of the planet is increasing at the square root of the increase in mass. If it's increasing at a lower rate, bouyancy will have a greater effect. If it's increasing at a higher rate, bouyancy will have less of an effect. But, it should be noted that this is a highly simplified way of looking at it and you would need a lot more specifics to figure it out on a case by case basis.

The question specifies "proportional". If we assume twice the gravity and air density, then: The answer is no. Terminal velocity would be neither faster, nor slower.

Note, the leaf would accelerate faster initially. But it would reach terminal velocity quickly. Terminal velocity is directly proportional to the sqrt of gravity and the sqrt of the inverse of the drag. So if "proportional" these factors cancel out.

• Only in a vacuum - objects such as leaves and sheets of paper have aerodynamics affecting how they fall. Nov 11 '19 at 10:34
• @Renan He includes "drag" in the answer, which is the usual shorthand for "whatever aerodynamic factors come into play". Nov 11 '19 at 10:58
• @toolforger you're right... I've reverted my DV into an UV. Nov 11 '19 at 12:10

Earth and Venus have comparable gravities but Venus has 90 times the atmosphere. Venus atmosphere

Other planets also differ widely. Saturn is huge but has close to Earth gravity, you can fit 1000 Earths into Jupiter but it has less than 3 times the gravity Planets gravity Neptune and Uranus are also close to Earth gravity, Uranus is a bit less than ours, yet both of these have atmospheres of perhaps hundreds of kilometres if not more.

• This doesn't actually answer the question though. "So, on a planet twice the mass of Earth with a proprtionally thickened atmosphere, would a leaf fall faster or slower?" Nov 11 '19 at 7:56
• The premise isn't incorrect, it just doesn't mention all possible variables. Larger planets do tend to have thicker atmospheres, but there are exceptions. Venus has a thick atmosphere because it has a lot of volcanic activity releasing heavy gases, but no plate tectonics to pull the heavy gases back into the mantle. Nov 11 '19 at 8:33
• @IndigoFenix how big is Neptune and Uranus? For all we know they may be smaller than Earth...... heck, Jupiter could be smaller than Earth Nov 11 '19 at 8:44
• @Kilisi Distance from the sun is also a factor. The gas planets' atmospheres are predominantly light elements like hydrogen and helium, which are blown away by the solar wind. Nov 11 '19 at 9:07
• I think the question is implicitly asking about rocky planets, because it talks about larger planets having higher air pressure. This is only true for planets with a solid or liquid surface, since the "surface" of gas giants is defined by convention as the point where the atmospheric pressure is 1 bar, i.e., slightly less than that of Earth. For Jupiter or Saturn, the speed of a leaf falling would depend on how deep you go in the planet. Nov 11 '19 at 9:15

To a first approximation, let's see what happens if we take the existing Earth and its atmosphere and dial up gravity to 2g.

• Pressure is just the weight of the column of air above your head. So if $$g$$ doubles, the pressure doubles.
• For a gas, $$P \propto {1\over{V}}$$ (Boyle's Law) $$\rho \propto {1\over{V}}$$ so near to the surface, the density doubles too.
• Drag (F) is directly proportional to density so it doubles.

But... terminal velocity occurs when drag equals weight. So we've doubled the drag but also doubled the weight - so the two cancel out and there is no difference.

• "For a gas, pressure and density are directly proportional" - not quite. Temperature would rise and volume is unrestructed - density would be less than 2 times more. It can be even less than on Earth! Nov 11 '19 at 10:24
• @TannerSwett Great point! I'd forgotten why it's falling in the first place... Nov 11 '19 at 13:23