What is the mass of our planet's atmosphere, hydrosphere, and cryosphere combined?

How accurate can our estimates be?


1 Answer 1


Pretty darn accurate. To first order, the atmosphere's mass is the average atmospheric pressure at the Earth's surface times the total surface area divided by the acceleration due to Earth's gravity. Or if you prefer, $$ M_{\mathrm{atm}_\bigoplus} = \frac1{g_{_\bigoplus}}\int_{{\textstyle\!A}_\bigoplus} \!\!\!\!\! P_{\mathrm{surf}} = \frac1{g_{_\bigoplus}}\int_{\theta} \! \int_{\phi} \sin (\phi) R^2_{\scriptscriptstyle \bigoplus} P_{\mathrm{surf}}(\theta,\phi) $$ This is very slightly wrong because some of the atmosphere is further away where gravity is less, leading to a ~0.25% underestimate of the mass of $ M_{\mathrm{atm}_\bigoplus} $. The quick estimate would be pressure at sea level times total area of the Earth divided by gravitational acceleration, $$ M_{\mathrm{atm}_\bigoplus} \approx \frac{4\pi R^2_{\scriptscriptstyle \bigoplus} P_{\mathrm{sea\,level}}}{g_{_\bigoplus}} $$ Plugging in actual numbers, $$ M_{\mathrm{atm}_\bigoplus} \approx \frac{(12.6) \,\,(6.4 \times10^6\mathrm{m})^{\textstyle 2} \,(1.01\times10^5 \mathrm{kg}\,\,\mathrm{m}^{-1}\mathrm{s}^{-2})}{(9.8\,\mathrm{ms}^{-2})} \approx \underline{\mathbf{5.3 \times 10^{18}\,kg}} $$ or about 58 quadrillion American tons.

Water is rather incompressible, so integrating the density of water (1 g/cm$^2$) times ocean depth over the surface of the oceans will get you close. This is a data intensive calculation, but all the data is well known. Then, there is the liquid water outside the oceans, about 1.75% of total water, most of which is subsurface groundwater (fresh and saline). Free surface water, like lakes, is negligible. See Where is the Earth's Water.

Another 1.75% of water is frozen in glaciers and ice caps. This is the hardest to calculate because the depths of the ice caps over land and sea is not precisely known.

Using the table at the bottom of Where is the Earth's Water we find that there is about 1.34 billion cubic kilometers $(1.34\times10^9\,\mathrm{km}^3)$ of ocean and sea water. Add groundwater, and it becomes $1.36\times10^9\,\mathrm{km}^3$. Each cubic kilometer weighs a trillion kilograms $(10^{12}\,\mathrm{kg})$ so $M_{\mathrm{hydro}_\bigoplus}$ is about $\,$ $\underline{\mathbf{1.36\times10^{21}\,kg}}\,$, or 250 times the mass of the atmosphere.

Using the same source, the $M_{\mathrm{cryo}_\bigoplus}$ is about $\,\,\underline{\mathbf{2.4\times10^{19}\,kg}}$ .

  • $\begingroup$ Nice work! Do you have references for the 1.75%s and the 0.25% ? $\endgroup$
    – f.thorpe
    Oct 12, 2015 at 16:37
  • $\begingroup$ @farrenthorpe, water.usgs.gov/edu/earthwherewater.html for the 1.75%. ~~~ 0.25% is just the quadratic correction for a column of atmosphere because on an average the gas is 8km above the surface of the Earth. I did it in my head and it might be off by a factor of 2. The point is it is very small. $\endgroup$
    – Eubie Drew
    Oct 12, 2015 at 16:49
  • 1
    $\begingroup$ @farrenthorpe, the 2nd order (relative) correction due to reduced acceleration by gravity would be (R + H)^2 minus R^2, basically a factor of 2H/R =16/6400 or about 0.0025 $\endgroup$
    – Eubie Drew
    Oct 12, 2015 at 17:01

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