# How do we get to know the total mass of an atmosphere?

Since atmospheres don't end abruptly but gradually get thinner the higher you go, I wonder how we can get the total mass of an atmosphere if we don't know where exactly it ends. E.g. the Earth's atmosphere's mass is defined as 5.1480 × 1018 kg. Does this value include the exosphere (which doesn't have an abrupt end either)? Or is it up to the exobase only? Or is it even the significant part only, up to the mesopause or to the Kármán line or something?

Also, if we mention the mass of a celestial body, does this value (e.g. in case of Venus 4.867 × 1024 kg) include its atmosphere's mass or not?

• If you'd like a particular number explained, please cite the source where it comes from. It could be that different sources give different numbers. Thanks!
– uhoh
Nov 26, 2020 at 11:08
• @uhoh I'm just wondering how we limit the atmosphere to conclude a particular number for its mass, and whether the mass of a planet includes its atmosphere (if it's not a gas giant that consists mostly of atmosphere of course). The provided numbers are examples. Nov 26, 2020 at 11:13
• Theoretically, a planet's atmosphere is close to infinite since interplanetary space isn't a perfect vaccuum either. Nov 26, 2020 at 11:14
• Sure, interplanetary space isn't a perfect vacuum, but that stuff isn't gravitationally bound to the planet, so it shouldn't be considered part of the planet's atmosphere. OTOH, there's no simple cut-off line, since the interaction between the planet's outer atmosphere & the stellar wind is rather complex. See en.wikipedia.org/wiki/Earth%27s_magnetic_field#Magnetosphere Nov 26, 2020 at 11:53
• I guess it's valid to consider everything inside the magnetopause, where the solar wind pressure is balanced by the Earth's atmospheric pressure, to be truly part of our atmosphere. Nov 26, 2020 at 12:34

There is a simple$$^*$$ way to know the total mass of the atmosphere: measuring the pressure it exerts on the surface, which necessarily integrate all of the atmosphere above ground level.

If you take an atmospheric pressure of $$1\cdot10^5$$ Pa, it is equivalent to a force of $$1\cdot10^5$$ newton over one square meter. Multiply by the area of the planet in square meters, you get the total weight of the atmosphere: $$1\cdot10^5 \times 5.1\cdot10^{14} = 5.1\cdot10^{19}$$ N. Divide by the acceleration of gravity to convert this weight to a mass: $$\frac{5.1\cdot10^{19}}{9.8} = 5.2\cdot10^{18}$$ kg. There you go!

$$^*$$Well, I guess it is simple on Earth, but could be more challenging on other planets...

• "Divide by the acceleration of gravity to convert this weight to a mass" To get a more precise value, we need to compensate for the fact that the acceleration of gravity isn't constant, either over the planet's surface, or as the altitude increases. But just using the mean g is certainly a reasonable first approximation. I suppose we also need to take temperature into account as well... Nov 26, 2020 at 12:00
• @PM2Ring But if you measure the pressure at ground level, you get the weight at ground level, so using an average value of $g$ at ground level should be OK to convert to mass. At least that's my understanding, I could be wrong. Nov 26, 2020 at 12:13
• Yes, but if you start with atmospheric pressure at ground level, and work backwards toward mass, you already account for all potential mass changes driven by $g$ changes. They already are present in the pressure you measure. Nov 26, 2020 at 13:22
• This gives you the weight of the atmosphere, not the mass. But it is a very good approximation since $g$ doesn;t change much over several scale heights. And actually, the increase in area compensates for the decrease in gravity, so I think it will be a very, very good approximation. Nov 26, 2020 at 13:37
• As a result, this calculation will not give the mass of the atmosphere to better than how well some sort of area-weighted average of the surface pressure can be calculated. Nov 27, 2020 at 10:00

Suppose the atmosphere has a density that decays exponentially with height. e.g. $$\rho = \rho_0 \exp[-h/h_0]\ ,$$ where $$\rho_0$$ is the density at some surface and $$h_0$$ is a characteristic height scale on which the density decreases.

If we integrate this funcion from $$h=0$$ to $$h = \infty$$, then this gives a finite result. $$\int^{\infty}_0 \rho_0 \exp[-h/h_0]\ dh = \rho_0 h_0$$

In practice when modelling an atmosphere there will be an upper limit defined which is less than $$\infty$$, but as long as that upper limit is $$\gg h_0$$ (where $$h_0$$ would be around 10 km for the Earth), then exactly where it is won't make much difference because the vast majority of the atmospheric mass is within the first few $$h_0$$.

The mass of planets, moons, etc. would include the mass of any atmosphere since it is estimated from their gravitational effects. The mass of the atmosphere (barring gas giant planets, where you would have to define what you meant) is totally negligible compared with the mass of the "solid" part of a planet/moon.

• FWIW, Wikipedia has a reasonably good article on scale height. Nov 26, 2020 at 11:40
• Your answer is good, but e.g. at 30 km (100 kft) it is said that 99% of the atmosphere's mass is below you. So it's about the total mass even though there is very little atmosphere above 16 km (where pressure is 0.1 atm). You use such argument also with the atmosphere compared to the rest of the body: while it may be negligible, I still wonder whether it is included. This is especially important in case of Venus' thick atmosphere. Nov 26, 2020 at 12:14
• @uhoh You misinterpreted what I said - see edit. Nov 26, 2020 at 13:33
• @Greenhorn To quote my own answer: "The mass of planets, moons, etc. would include the mass of any atmosphere since it is estimated from their gravitational effects.". Nov 26, 2020 at 13:34
• @RobJeffries Thank you for the edit. Still, Monsieur Prival explains the Earth's entire atm mass value, so I'll let his answer accepted. If the atmosphere's masses are included, do we have to correct the gravity for orbit a bit since the mass of the atmosphere is below the spacecraft in orbit? (Earth is very little, but in case of Venus and Titan I think the gravity must be corrected as per the atmosphere's mass) Nov 26, 2020 at 13:59

If the mass of the atmosphere is given as 5.1480 × 10^18 kg, then according to the rules of significant figures, the uncertainty in that need not be smaller than 10^14 kg (and depending on how one interprets significant digits, it can be as high as 10^15 kg). According to this site:

And [the exosphere's] mass is only 0.002% of the total mass of the atmosphere because gas molecules are far apart in the exosphere.

That would make it 10^14 kg, within the error bounds allowed by the significant digits, and any difference based on where the exosphere is considered to end would be much smaller.

Also, if we mention the mass of a celestial body, does this value (e.g. in case of Venus 4.867 × 10^24 kg) include its atmosphere's mass or not?

The main way we estimate a planet's mass (and the main reason we care) is its gravitational effects, and apart from probes that have entered the atmosphere, the atmosphere has just as much gravitational effect (per kg) as any other part of the planet.

However, if the mass of Venus is given as 4.867 × 10^24 kg, that implies a error bar no smaller than 10^21 kg. Wikipedia gives the mass of Venus's atmosphere as 4.8 x 10^20 kg. It also says this is nearly 100 times the mass of Earth's, so Earth's atmosphere would be an even smaller percentage of its total mass.

• The planet's mass is also used to define the surface gravity. When we calculate the surface gravity on a solid or ocean planet, we have to exclude the atmosphere's mass from that of the planet. Otherwise, the radius provided would have to end at the atmosphere's upper boundary (let's say at the mesopause in case of Venus and Earth) and not at its surface, so that the result isn't falsified. Nov 27, 2020 at 7:01
• @Greenhorn That much is correct. The surface gravity calculation should exclude the atmosphere. A correction in the 6th significant figure for the Earth and the 4th for Venus. These are far smaller than the variations caused by local geography, density inhomogeneity, non-sphericity and in the Earth's case, centrifugal acceleration. Nov 27, 2020 at 9:19
• @Greenhorn As I will say for the FINAL time. That mass includes the mass of the atmosphere. If someone has used that for calculating the surface gravity to 6 significant figures, then the 6th significant figure will be in error. However the local gravity on the Earth's surface varies in the third significant figure due to other things I mention - so the inclusion or not of the atmosphere is totally unimportant. Nov 27, 2020 at 9:59

If I'm geoengineering on the back of a napkin, I use C^2 / pi * 10^4 kg

(14.696 psi ~= 10,000 kg/m^2) so,

(4x10^7)^2 / 3.14 * 10^4 kg ~= 5.1 x 10^18 kg, which is about 99% of the generally accepted value of 5.1480 x 10^18 kg.

For reference, pi = C/D, so 4 pi r^2 = pi D^2 = C^2 / pi, and we have an easy value for the earth's circumference (~40,000 km).

• Are you sure this is meant as answer to this question? If so, be more elaborate. If not, delete and repost in the right place. Even then explain the physical meaning of C and D May 30 at 7:07
• Welcome to Stack Exchange! What is C? What is D? Where does C^2 / pi * 10^4 kg come from? Stack Exchange answers need to be well-supported, right now this is "well my guess is" which is not an answer to the actual question "How do we get to know...?" Please add some further explanations. Thanks!
– uhoh
May 30 at 7:11
• I think it's fairly clear that C= circumference and D= Diameter of the Earth, and this answer is, as above, "Area of Earth * pressure" as in the accepted answer. May 31 at 9:00
• By knowing the surface area of the Earth and the mean pressure at the surface, the calculation of an estimation is straightforward, not a guess. Surface area of a sphere with a circumference of ~40,000 km is (4x10^7 m)^2 times ~3.14 = 5.1 x 10^8 km^2, which when is multiplied by sea level pressure of 14.696 psi = ~10^4 kg gives the accurate to within 2 significant digits of the accepted answer 5.1 x 10^18 kg. There is no "guess" here. While the answer in rectilinear terms is complicated, it doesn't have to be. May 31 at 11:25
• edit,: 14.696 psi is = ~10^4 kg/m^2, not ~10^4 kg. May 31 at 11:38