# Finding the density of air as a function of height

Find the density of air as a function of the radius $\rho(r\gt R_E$)(where $R_e$ is the radius of the earth in the general case. You can neglect the gravity that comes from the atmosphere, and assume the gravity is controlled by the mass of the Earth $M_e$.

We did a similar question in class, and following the same steps I came up with an answer that has the correct units, yet doesn't make sense. I also hope the question makes sense as it's translated from a different language.

I'm going to define $r'=R_E+r$ as our arbitrary height above earth.

Setting up an equation for the forces I get:

$$AP(r')-AP(r'+dr')-\frac{GM_Em}{r^2}=0$$

Where P(r'+dr') is the pressure at a location a tiny bit above r' and A is the area. Moving things around I get:

$$P(r'+dr')A-P(r')A=-\frac{GM_Em}{r^2}$$

Now we aren't given mass so we'll use:

$$m= \rho drA$$

Plugging that in gives me:

$$P(r'+dr')-P(r')=-\frac{GM_E\rho dr}{r^2}$$

Dividing both sides gives me:

$$\frac{d\rho}{dr'}=-\frac{GM_E\rho}{r^2}$$

Now using the ideal gas law:

$$PV=nRT$$

And I will multiply the left side by $\frac{\mu}{\mu}$ which is the molar mass (I believe it's called) and I get:

$$PV=\frac{mRT}{\mu}$$

Again we don't have mass so we'll use: $m=\rho V$

And it's given that T is constant. So I will differentiate P according to r':

$$\frac{dP}{dr'}V=\frac{d\rho}{dr'}\frac{RT}{\mu}$$

Now I will compare both terms, and end up with the following first order D.E.:

$$\frac{GM_e\mu}{RT}\frac{dr'}{r'^2}=\frac{d\rho}{\rho}$$

Solving it fully I get:

$$\rho(r')=\rho_0\left(e^\left(\frac{-GM_E\mu}{RTr'}\right)\right)$$

The units are correct, but the result doesn't make sense. if I take the limit as $r\to \infty$ I get one, which is incorrect as the density of air decreases with height.

I would appreciate any help or insight as to why my work is incorrect, and I prefer hints rather than full-blown solutions.

• Have a look at the derivation of the barometric formula on Wikipedia. Nov 17, 2016 at 21:40
• This is a check-my-work question, which is off-topic according to the homework-and-exercises policy. Nov 18, 2016 at 5:44

Your formula with $r^{'}$ is not correct. The easiest way to consider this (idealized) problem of gas density as a function of radius $r$ above the earth's surface is to assume a Maxwell-Boltzmann energy distribution of the number of gas molecules (with mass $m$) per unit volume (density) $n$ at an energy interval $E$, $E+dE$. When $n_0$ is the density at the earth surface $r=R_e$, then the density at radius $r$ is given by $$n(r)=n_0\exp{(-\frac{E(r)}{kT})}$$ where the potential energy $E(r)$ of an air molecule with mass $m$ at radius $r$ with respect to the surface radius $R_e$ is $$E(r)=-(\frac{GM_Em}{r}+\frac{GM_Em}{R_e})=-GM_Em(\frac{1}{r}-\frac{1}{R_e})$$ Thus the air mass density $\rho(r)=mn$ at a radius $r$ larger than the surface radius $R_e$ is given by $$\rho(r)=\rho_0\exp{(-\frac{E(r)}{kT})}=\rho_0\exp{[\frac{GM_Em}{kT}(\frac{1}{r}-\frac{1}{R_e})]}$$ where $\rho_0=mn_0$, $T$ is the absolute temperature, $k$ is the Boltzmann constant.
Check limits of integration of $$r$$ and $$\rho$$.
• I guess you meant rho($\rho$) by phro. And please try to add some more information in your answer as it seems too short. Once you've got enough reputation, you can comment on other people's questions and answers until then please make sure that you've enough information while posting an answer.