Height of the atmosphere - conflicting answers

Question

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:

$P=\frac{F}{A}$

$PA=mg$

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

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

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

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

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

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

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

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?

Answer 1

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.

Answer 2

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:

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