Just what the title states please... and perhaps naive too...

Wikipedia pegs the mass of our Earth at 5.9722 × 10^24 kg.

Does this figure include the mass of the Atmosphere?

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    $\begingroup$ The difference is (just) past the last digit of accuracy in your number, so it is impossible to tell. $\endgroup$ – Ron Maimon Jan 4 '12 at 6:38
  • $\begingroup$ Then would it be fair to say the atmospheric mass is, for all practical purposes, negligible? $\endgroup$ – Everyone Jan 4 '12 at 6:44
  • $\begingroup$ If you consider this accuracy sufficient. $\endgroup$ – Ron Maimon Jan 4 '12 at 6:47

The difference is (just) past the last digit of accuracy in your number, so it is impossible to tell.

You can estimate the fraction of the mass of the atmosphere by the ratio of the atmosphere's height to the radius of the earth (which gives the order of magnitude of the fraction of the Earth's volume in the atmosphere, about 10km/6000km, or 1/600), times the ratio of density of gas to ordinary solid (which is about 1/300). This gives 1 part in $10^{-5}$, and this is an overestimate, because the Earth's core is much denser than an ordinary solid because of the immense pressure. The mass figure you quote is correct to about 1 part in $10^{-5}$, so it is not clear if it includes the atmosphere or not, because this is a negligible error.

The mass of the Earth is found by measuring g, and the 1 part in $10^{-5}$ (or less) variations due to the atmosphere as you go up will be hard to distinguish from oblateness corrections, or just experimental error.

Atmospheric pressure

Multiplying standard atmospheric pressure by the surface area of the Earth gives the mass of the atmosphere as $5 \times 10^{18}Kg$, which is almost exactly 1 part in $10^{-6}$ of the mass of the Earth that you quote.

  • $\begingroup$ you have an error in the sign of your exponent of the atmosphere's mass. $\endgroup$ – Mark Beadles Jan 4 '12 at 17:21

The mass of the earth can be and has been measured using Newton's law of gravity. For the original experiment see this wiki article. The method by construction excludes most of the atmosphere, and the value is valid below the radius where it is measured.

Measurements from space include the atmosphere and I expect that this NASA experiment exploring mass differences will also be giving the mass of the earth including the atmosphere. In any case the mass of the atmosphere with respect to the rest of the globe is

The average mass of the atmosphere is about 5 quadrillion (5×1015) tonnes or 1/1,200,000 the mass of Earth

One should worry only if one is measuring with accuracy greater than this, as in the following fascinating experiment: I found this NASA experiment, which confirms general relativity and uses mass differences.

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    $\begingroup$ Well, the mass of Earth can be and has been measured by Kepler's method as well – from the orbit of the Moon – and this method makes it completely clear that the atmosphere, like anything else linked to Earth, is included just like the bulk of the Earth. $\endgroup$ – Luboš Motl Jan 4 '12 at 9:33
  • $\begingroup$ The website anna v references does not seem to be very trustworthy. Cavendish never computed G, as they claim, G was invented about 100 years after Cavendish died. Wikipedia is very careful to make this point en.wikipedia.org/wiki/Henry_Cavendish#Density_of_the_Earth. $\endgroup$ – Zeynel Jan 5 '12 at 2:50
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    $\begingroup$ @Zeynel I found it by google , and you may be right . It is the method that is important. $\endgroup$ – anna v Jan 5 '12 at 4:36
  • $\begingroup$ @Zeynei: Wikipedia is saying nonsense. The fact that they didn't call it "G" is immaterial. The law of universal gravitation was known, and the coefficient was known to be constant. When Cavendish measured Newtonian gravitation for masses, he knew that the same "G" would apply to all stuff. So Wikipedia is stupid here, and it is not appropriate to cite it. $\endgroup$ – Ron Maimon Jan 5 '12 at 17:40
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    $\begingroup$ @RonMaimon Hmm. mv^2/R=ma , in my books. m the mass of a small satellite, so if we know its orbit and velocity then we have a to first order? correct equations could give great accuracy. $\endgroup$ – anna v Jan 8 '12 at 17:13

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