Is the total mass of Earth's atmosphere essentially constant in time?

Question

Per a few questions and comments on this site such as Huge buildings affect Earth's rotation? I wonder, is the mass of the Earth's atmosphere roughly constant?

We're burning an incredible amount of fossil fuels which put a huge amount of water and $\textrm{CO}_2$ into the atmosphere. Global temperature increases are causing more water evaporation. A very stormy season can carry a huge amount of additional moisture over parts of the globe.

Is there some physical process such as the gravitational gradient between the surface of the Earth and the edge of space or something else that keeps the Earth's atmosphere roughly constant in mass? Related, is the pressure at sea-level roughly constant? I'd expect that as we burn more fossil fuels the amount of mass in the atmosphere would increase slowly but I wonder now if there is a counterbalancing effect.

An increase in the overall mass of the atmosphere would imply a tiny lengthening of a day. Has such a lengthening been measured?

Answer 1

The paper "The Mass of the Atmosphere: A Constraint on Global Analyses" from Trenberth & Smith (2004) contains a detailed analysis of the mass of the Earth's atmosphere. They note that the mean mass $m$ can be derived from the mean surface pressure $p_s$ with the formula $$ m = 5.22371\times 10^{15}p_s, $$ with $m$ in kg and $p_s$ in hecto-Pascals (the pressure is not measured at sea level, but at a mean altitude of 232 m). Using satellite data, they find that the mean surface pressure in the period 1979-2001 is $p_s= 985.5$ hPa, so that $$ m = 5.148\times 10^{18}\;\text{kg}. $$ The mass of the atmosphere does fluctuate. The largest fluctuation is seasonal: the mean pressure changes from $985.41$ hPa in January to $985.64$ hPa in Augst. In other words, $\Delta p_s = 0.23$ hPa, so that $$ \Delta m \approx 1.2\times 10^{15}\;\text{kg}. $$ The reason for this seasonal variation is almost entirely due to a change in water vapour. The authors find that the mean pressure of dry air is $p_d=983.05$ hPa and the mean pressure of water vapour is $p_w=2.44$ hPa, and $p_w$ varies between $2.33$ hPa in January to $2.62$ hPa in July, so $\Delta p_w=0.29$ hPa. It seems that the dry air pressure also varies slightly, but apparently it isn't clear if this effect is real or due to measurement error. If $p_d$ is actually constant, then the change in mass due to water vapour change would be $$ \Delta m \approx 1.5\times 10^{15}\;\text{kg}. $$

The reason that the water vapour pressure is higher in July-August and lower in December-January is that the Northern Hemisphere has more landmass than the Southern Hemisphere, which leads to higher temperatures in Northern summer compared to Southern summer, which in turn leads to more moisture in the atmosphere during Northern summer.

There are also long-term effects. The water vapour content is higher during El Niño events and lower during La Niña events. These effects are of the order of $0.1$ hPa, equivalent to $0.5\times 10^{15}$ kg. Apparently there was also a slight decrease after the Mount Pinatubo eruption. And there is also evidence that the amount of water vapour is also slowly increasing due to global warming.

The effect of changes in $\text{CO}_2$ is complicated but very small. The burning of fossil fuels does not just add carbon dioxide, it also removes oxygen. Taking into account the interactions with the oceans and the biosphere, the net change in surface pressure is of the order of $0.01$ hPa (and, according to the authors, most likely a net loss instead of a net gain), too small to measure. The effects of mass gain due to meteors and mass loss of hydrogen and helium are even smaller.

Edit

The length of a day does vary due to interactions between the atmosphere and the solid Earth. According to wiki, the angular velocity has an annual period with an amplitude of 0.34 milliseconds and a semi-annual period with an amplitude of 0.29 milliseconds. But it seems that these fluctuations are due to changes in the large-scale wind field, rather than variations in the atmosphere's mass.

Answer 2

We are constantly losing gas to space. According to this article in Scientific American, "the loss rate is currently tiny, only about three kilograms of hydrogen and 50 grams of helium (the two lightest gases) per second." As AlanSE rightly points out, we also gain some mass from space, but his 100 tons would not be enough to balance the ~290 tons of gas we lose per day according to Sci Am's numbers. Wikipedia's page on this provides a nice overview of various processes leading to atmospheric losses.

But that's Earth's mass loss to space. As for transfer of mass in the form of Carbon from the crust, this paper states that the net increase in atmospheric $\mathrm{CO}_2$ per year is 4.1 petagrams ($4.5\times 10^9\, \mathrm{tons}$), which accounts for the amount re-absorbed in the oceans and biomass. As @Pulsar points out, the Oxygen was mostly already in the atmosphere. About 1/4 of $\mathrm{CO}_2$'s weight is from the Carbon, so that's roughly $10^9\, \mathrm{tons}$ of Carbon per year added to the atmosphere, which dwarfs the numbers for mass loss to space.

So how does this affect the rotation rate of earth? Well, ignoring relevant external torques, we can treat the total angular momentum of Earth as constant. And since angular momentum is the moment of inertia times the angular frequency ($I\, \omega$), this product is constant. The moment of inertia (MOI) for a spherical shell in terms of inner and outer radii $R_i$ and $R_o$ is \begin{equation} I = \frac{2}{5}\, M\, \frac{R_o^5-R_i^5}{R_o^3-R_i^3} \end{equation} Let's imagine that the initial MOI is just this formula with $M = M_\oplus$ (mass of Earth), $R_o=R_\oplus$ (radius of Earth), and $R_i = 0$. Then, we take $M_\mathrm{C} = 10^9\, \mathrm{tons}$ from the crust and put it in the atmosphere. Just as an approximation, let's say that the Carbon was initially a uniform spherical shell ranging from the surface to $500\, \mathrm{m}$ below the surface (typical depth of a deep coal mine). Then, let's treat the final state as a uniform spherical shell extending from the surface to $50\, \mathrm{km}$ above (a large overestimate). So the ratio of initial MOI to final is \begin{equation} \frac{\Delta I}{I_\mathrm{initial}} = \frac {M_\mathrm{C}\, \frac{(R_\oplus+50000\, \mathrm{m})^5-R_\oplus^5}{(R_\oplus+50000\, \mathrm{m})^3-R_\oplus^3}\, - M_\mathrm{C}\, \frac{R_\oplus^5-(R_\oplus-500\, \mathrm{m})^5}{R_\oplus^3-(R_\oplus-500\, \mathrm{m})^3}\, } {M_\oplus\, R_\oplus^2} ~. \end{equation} Plugging in the numbers, I get a fractional change of a couple times $10^{-15}$, which will be the fractional change in the rotation frequency – a fraction of a nanosecond change per year.

But, this is probably a significant overestimate, because the $\mathrm{CO}_2$ is not really uniformly spread out to $50\, \mathrm{km}$. And, of course, as @anna v points out, there are other processes that involve larger masses that will probably dominate this.