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I am trying to solve the problem below:

Calculate the mass of the Earth's atmosphere given a mean pressure at the surface of $1.013\times 10^5$ Pa and $g=9.81$ m/s/s.

I have been provided a hint in the textbook to use the hyrdostatic equation. From my understanding, the hyrdostatic equation is of the form $$\frac{\partial p}{\partial z}=-\rho g.$$Here, $\rho$ denotes density and thus $$\rho=\frac{m}{v}\implies \frac{\partial p}{\partial z}=-\frac{mg}{v}.$$I am unsure of how to proceed. Any tips would be greatly appreciated.

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The pressure at the earth's surface is the force per unit area needed to support the weight of the column of air above that area, extending from the surface to outer space. So the mass of the column per unit area is equal to the pressure divided by g.

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  • $\begingroup$ I have derived the following for the entire atmosphere $$\text{mass}=\frac{\text{pressure}\times\text{surface area}}{\text{g}}.$$ Is this correct? $\endgroup$ – JulianAngussmith Feb 22 at 0:01
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    $\begingroup$ That's what I said. $\endgroup$ – Chet Miller Feb 22 at 0:03
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    $\begingroup$ Thanks. I ended up integrating from $z$ to $\infty$ and realised that $$\int_{z}^{\infty} \rho \ dz$$ was just the mass per unit area of the atmospheric column. $\endgroup$ – JulianAngussmith Feb 22 at 0:05

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