Okay. I have two ways of working out the height of the atmosphere from pressure, and they give different answers. Could someone please explain which one is wrong and why? (assuming the density is constant throughout the atmosphere)

1) $P=h \rho g$, $\frac{P}{\rho g} = h = \frac{1.01\times 10^5}{1.2\times9.81} = 8600m$

2) Pressure acts over SA of Earth. Let r be the radius of the Earth. Area of the Earth is $4 \pi r^2$

Volume of the atmosphere is the volume of a sphere with radius $(h+r)$ minus the volume of a sphere with radius $r$. $\frac{4}{3}\pi (h+r)^3 - \frac{4}{3}\pi r^3$ Pressure exerted by the mass of the atmosphere is:



$4\pi r^2 P = \rho V g$

$4\pi r^2 P = \rho g (\frac{4}{3}\pi (h+r)^3 - \frac{4}{3}\pi r^3)$

$\frac{4\pi r^2 P}{\rho g} = \frac{4}{3}\pi (h+r)^3 - \frac{4}{3}\pi r^3$

$3 \times \frac{r^2 P}{\rho g} = (h+r)^3 - r^3$

$3 \times \frac{r^2 P}{\rho g} + r^3 = (h+r)^3$

$(3 \times \frac{r^2 P}{\rho g} + r^3)^{\frac{1}{3}} - r = h$

$(3 \times \frac{(6400\times10^3)^2 \times 1.01 \times 10^5}{1.23 \times 9.81} + (6400\times10^3)^3)^{\frac{1}{3}} - (6400\times10^3) = h = 8570m$

I know that from Occams razor the first is the right one, but surely since $h\rho g$ comes from considering the weight on the fluid above say a 1m^2 square, considering the weight of the atmosphere above a sphere should give the same answer?

  • $\begingroup$ These calculation seem to assume a constant pressure in the air column, which is not the case... $\endgroup$ Commented Aug 7, 2012 at 14:58

2 Answers 2


Neither calculation is anything approaching physically realistic, but I guess you know that and you're just interested in why the two approaches give different answers.

Take your equation from your second method:

$$ 4\pi r^2 P = \rho V g $$

If the area is a flat sheet you have $V = Ah$ and $A = 4\pi r^2$, and substituting this in your equation gives:

$$ 4\pi r^2 P = \rho 4\pi r^2 h g $$

and dividing both sides by $4\pi r^2$ gives you back $P = \rho h g$ as in your first method. However in the second method you've taken the volume to be a spherical shell with inner surface area of $4 \pi r^2$ and thickness $h$, and the volume of this shell is greater than $Ah$ i.e.

$$ \frac{4}{3}\pi (r + h)^3 - \frac{4}{3}\pi r^3 \gt 4\pi r^2 h $$

The greater volume is why your second method gives you a smaller height.


The first formula is just a first order expansion in $1/r$ of the second formula which is thus the exact one.

The expansion is:

$$h = \frac{P}{\rho g} - \left( \frac{P}{\rho g} \right)^2 \frac{1}{r} + \frac{5}{3} \left( \frac{P}{\rho g} \right)^3 \frac{1}{r^2} + \ldots$$

  • $\begingroup$ +1, though I had to think a bit what formulas are you talking about (for $h$ it turned out). But it would be nice if you could provide 1-2 terms of the expansion. $\endgroup$
    – Yrogirg
    Commented Aug 7, 2012 at 14:07
  • $\begingroup$ h=(3×r^2*P/ρg+r^3)^(1/3)−r h=r*(3*P/ρgr +1)^(1/3)-r (1+x)^t=1+t*x+o(x^2) take x=3*P/ρgr And you get h=P/ρg $\endgroup$
    – Shaktyai
    Commented Aug 7, 2012 at 14:27
  • $\begingroup$ I meant you put it in your answer, since it is not only me who will want to see it. I've made the edit. $\endgroup$
    – Yrogirg
    Commented Aug 7, 2012 at 17:20
  • $\begingroup$ Thanks for the typing, you are the second one today to edit my post. I may need to take a few minutes to learn tex, math ML or whatever. Otherwise I am going to be kicked out ... $\endgroup$
    – Shaktyai
    Commented Aug 7, 2012 at 20:29
  • $\begingroup$ see meta.math.stackexchange.com/questions/1773/… and en.wikipedia.org/wiki/… $\endgroup$
    – Yrogirg
    Commented Aug 8, 2012 at 4:16

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