Calculating the Mass of Earth's Atmosphere

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.

• It is much simpler than that. See Why does the air pressure at the surface of the earth exactly equal the weight of the entire air column above it? Commented Feb 21, 2020 at 9:54
• That pressure is the same as 14.7 psi, so just multiply by the surface area of the earth in in^2 to get the mass in lb. Commented Feb 21, 2020 at 16:25
• @ChetMiller Why though? I am not concerned with the numerical complexities of the question, but rather the understanding of the theory behind it. Commented Feb 21, 2020 at 23:06

• I have derived the following for the entire atmosphere $$\text{mass}=\frac{\text{pressure}\times\text{surface area}}{\text{g}}.$$ Is this correct? Commented Feb 22, 2020 at 0:01
• 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. Commented Feb 22, 2020 at 0:05