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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?

Glorfindel
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Greenhorn
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    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 '20 at 11:08
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    @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. – Greenhorn Nov 26 '20 at 11:13
  • Theoretically, a planet's atmosphere is close to infinite since interplanetary space isn't a perfect vaccuum either. – Greenhorn Nov 26 '20 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 https://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Magnetosphere – PM 2Ring Nov 26 '20 at 11:53
  • @PM2Ring So you'd consider the boundary either halfway to the Moon (120,000 mi respectively 190,000 km) where Earth's gravity ceases to hold the atmosphere to it, or at the magnetosphere's outer border at 1 million mi (1.5 million km)? – Greenhorn Nov 26 '20 at 12:19
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    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. – PM 2Ring Nov 26 '20 at 12:34
  • Can the downvoter explain their vote? It's not like the answer was obvious, the atmosphere doesn't end abruptly (unless you consider the boundary where the homopause ends or at the exobase, but the question is whether it's one of them perhaps). – Greenhorn Nov 28 '20 at 15:20

4 Answers4

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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...

WarpPrime
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Jean-Marie Prival
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    "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... – PM 2Ring Nov 26 '20 at 12:00
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    @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. – Jean-Marie Prival Nov 26 '20 at 12:13
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    Hey, I could be wrong too. :) But I think you do need to account for the fact that 1kg of air at 50 km altitude has slightly less weight than 1 kg of air at sea level. – PM 2Ring Nov 26 '20 at 12:24
  • At 50 km (31 mi) altitude you should be about 1.5% lighter. – Greenhorn Nov 26 '20 at 13:10
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    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. – Jean-Marie Prival Nov 26 '20 at 13:22
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    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. – ProfRob Nov 26 '20 at 13:37
  • @Greenhorn Yes, at 50 km, g is about 98.45% of its value at sea level. And using a scale height of 8.5 km, only 0.28% of the atmosphere (by mass) is above 50 km. So we can safely ignore the changes to g unless we're trying to get very accurate values. – PM 2Ring Nov 26 '20 at 15:18
  • @PM2Ring For gravity it is actually quite a difference. You have to be more precise with gravity. But, as Mr. Jeffries states, it is a very good approximation anyway. – Greenhorn Nov 26 '20 at 15:26
  • @Greenhorn Yes, it is a very good approximation. FWIW, I was going to post an answer similar to this one, but Jean-Marie beat me to it. :) Rob's point that the increase in area compensates for gravity is a good one: gravity is proportional to $1/r^2$, where $r$ is the distance from the centre of the Earth, but area is proportional to $r^2$. However, sea level atmospheric pressure isn't a constant, it varies with temperature and humidity (water vapour has lower density than dry air). – PM 2Ring Nov 26 '20 at 15:59
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    @PM2Ring Sorry! :) If you feel you can contribute, please feel free to edit my answer! I'm sure there is room for improvement. We could add some of the discussion we had here about $g$, I put some thoughts into it and I think you're right. – Jean-Marie Prival Nov 26 '20 at 16:26
  • Yeah especially how the percentage of the atmosphere's masses is in contrast to the percentage of the solid/liquid celestial bodies; Venus and Titan in particular where the atmosphere's percentage might be quite a lot. – Greenhorn Nov 26 '20 at 16:31
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    @Jean-Marie Well, Rob has explained why we can ignore the change of g with altitude. We're calculating the weight of a tapered column of air. At ground level, the cross-section area of the column is 1 m². At higher altitudes, the area is larger, but gravity is smaller, so the effects of increased area & decreased gravity balance exactly. – PM 2Ring Nov 26 '20 at 16:42
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    But the effects of temperature and humidity are more complicated. If only temperature varied, we'd still get pressure variation (in accordance with the universal gas law) even though the mass of the atmosphere stays constant. But water makes it complicated because water moves back & forth between the atmosphere and the oceans (and lakes, rivers & the ground), changing the mass of the atmosphere. – PM 2Ring Nov 26 '20 at 16:47
  • @PM 2Ring: And another complication is that atmospheric pressure varies due to weather. Still, it's a pretty good approximation. – jamesqf Nov 26 '20 at 22:04
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    By the way this implicitly assume that the atmosphere is in equilibrium. But it isn't and that's why the surface pressure can vary by a few per cent. – ProfRob Nov 27 '20 at 09:13
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    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. – ProfRob Nov 27 '20 at 10:00
  • Assumption: the atmosphere's own gravity is negligible. Definitely true on Earth, probably true on Venus, but don't try it on Neptune (which probably has a solid surface). – Joshua Nov 28 '20 at 18:30
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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.

ProfRob
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  • FWIW, Wikipedia has a reasonably good article on scale height. – PM 2Ring Nov 26 '20 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. – Greenhorn Nov 26 '20 at 12:14
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    @uhoh You misinterpreted what I said - see edit. – ProfRob Nov 26 '20 at 13:33
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    @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.". – ProfRob Nov 26 '20 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) – Greenhorn Nov 26 '20 at 13:59
  • yep I did, looks great – uhoh Nov 26 '20 at 14:04
  • @Greenhorn there are lots of refinements required for calculating orbits of low-Earth orbit satellites, though I would have thought atmospheric drag much more important than perhaps missing a hundred millionth (or probably much less) of the Eath's mass that lies outside the orbit. – ProfRob Nov 26 '20 at 14:55
  • @RobJeffries The amount of mass by the atmosphere is about 1 millionth, isn't it? But, as I said, I rather meant Venus and Titan whose atm's mass' percentage is much higher compared to the mass of the solid/liquid body. – Greenhorn Nov 26 '20 at 15:01
  • @Greenhorn Not above a low-Earth orbit satelite it isn't. Indeed my hundred millionth is probably a vast over-estimate. If the scale height is 10 km, then something at say 200 km has only a faction of $10^{-9}$ of the atmosphere above it. i.e. $10^{-15}$ of the Earth's mass. You can scale numbers up by a factor of 100 or so for Venus and the scale height will be bit bigger, but it is still not an important fraction. You are not being clear as to what you want to know, The quoted mass of a palnet includes the atmosphere and cetrainly doesn't affect th 4th significant figure. – ProfRob Nov 26 '20 at 15:13
  • @RobJeffries The mass of (the entire) Venus is a bit smaller than Earth's while its atmosphere's mass is almost a 100 times that of Earth as you state. Titan's atmosphere's mass is a bit more than that of Earth's atm while Titan's overall mass is much, much lower than Earth's (about 1/44). This makes the percentage of the atm's mass much higher in contrast to the body's mass to the surface. Gravity decreases with the square of its distance from the surface on, however with such massive atmospheres I wonder whether the gravity due to these atm's masses should be considered in gravity calcs. – Greenhorn Nov 26 '20 at 15:22
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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.

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  • 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. – Greenhorn Nov 27 '20 at 07:01
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    @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. – ProfRob Nov 27 '20 at 09:19
  • @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. – ProfRob Nov 27 '20 at 09:59
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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).

James J Youlton
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    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 – planetmaker May 30 '23 at 07:07
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    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 '23 at 07:11
  • 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. – James K May 31 '23 at 09:00
  • 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. – James J Youlton May 31 '23 at 11:25
  • edit,: 14.696 psi is = ~10^4 kg/m^2, not ~10^4 kg. – James J Youlton May 31 '23 at 11:38