Objectives

• Apply basic concepts of pressure and density to an interesting real-world situation
• Use basic physics to set up and solve a differential equation
• Gain intuition for the properties of exponential functions

Introduction

When asked how air pressure changes with altitude, most will be aware that it decreases as you get higher and the air gets “thinner”, but how can we properly understand this phenomenon? The first question to ask is why there’s any atmospheric pressure at all? We can understand air pressure very simply as just the weight of the atmosphere. Just as a box sitting on a table will exert a pressure equal to its weight divided by the contact surface area, the pressure at sea level is the weight of the air above a given area divided by that area. So how heavy is the atmosphere? Well, we know that the pressure at sea level is 101325 Pascals which is 101325 Newtons per square meter. Using \latex@g \approx 9.8N/kg@ we can express the mass as \latex@\mbox{weight}/g \approx 10,000kg@ — every square meter at sea level has 10 tonnes of air pressing down in it!

Derivation

To understand how this changes as we ascend we need to be able to figure out the weight of air that remains above any given altitude.

Simple case: Incompressible fluids (constant density)

In a liquid like water this would be very easy to do because the density is constant. \latex@1 m³@ of water weighs about \latex@1000kg@ \latex@(\rho = 1000kg/m³),@ so if we consider a column of cross sectional area \latex@A@ and height \latex@h@, at its base we will have

so, as you go up the height of water above you decreases so that at a height \latex@x@ the pressure will be

and above \latex@h@, \latex@P = 0@, because we are above the surface and there is no more water pressing down above us. This “linear” pressure change with altitude is characteristic of incompressible fluids whose density does not change appreciably when pressure is applied.

Setup: Compressible fluids

Gasses, however, are compressible. In fact, the ideal gas law states that density is linearly proportional to pressure when temperature is kept constant. For simplicity, let’s make that assumption of constant temperature and write out a completely generic linear relationship between density and pressure

or, calling the sea-level pressure

and the sea-level density

It will be more useful to work in terms of altitude than pressure so we need to change our expression a bit to have \latex@h@ above sea level as the variable. Fortunately, this is easy because pressure and density will both depend only on altitude (because we have omitted temperature as a variable for simplicity and there are simply no other variables in the problem on which pressure and density could depend). Thus, we write:

We now have most of what we need to solve the problem, but there’s one very important step left to do.

Setup: Differential equations

We already know how to solve the problem of pressure vs altitude when the density is constant. That might not seem very helpful here because density varies with altitude but we can actually still use the same method to set up an equation that we can then solve. Suppose that we are at an altitude h and go up slightly to \latex@h+\Delta h@. For a really small \latex@\Delta h@ the pressure and density will both change by only a tiny amount. In order to figure out how big that tiny change in pressure is, we want to use the same method we used before and just say

\latex@\Delta P = -\rho(h) \times g \times \Delta h@

(note the minus sign because as we go up the pressure decreases)

But wait, didn’t we say that density changes too? How can we simply use \latex@\rho(h)@ for the entire interval from \latex@h@ to \latex@h+\Delta h@? The answer touches on some core concepts of calculus but, to put it simply, it’s all a question of proportionality. As you can see from the equation, the tiny change \latex@\Delta P@ that we are interested in is proportional to the altitude step that we took (\latex@\Delta h@). If we now say that at \latex@h@ the density is \latex@\rho(h)@ and at \latex@h + \Delta h@ the density is \latex@\rho(h) + \Delta \rho@ it’s safe to say that in ignoring the \latex@\Delta \rho@ our error will be no more than about

\latex@\mbox{Error in }\Delta P = -\Delta \rho \times g \times \Delta h@

And since \latex@\Delta \rho@ is, itself, proportional to \latex@\Delta h@ the error will be proportional to \latex@\Delta h²@ which becomes increasingly irrelevant the smaller we make \latex@\Delta h@.

So, going back to the expression for \latex@\Delta P@ we can put everything in terms of pressure as

\latex@\frac{\Delta P}{\Delta h}= -\frac{\rho_0}{P_0} \times g \times P(h)@

In the limit of \latex@\Delta h \rightarrow 0@ the approximation of constant density for each step becomes exact and we obtain an ordinary differential equation

\latex@\frac{dP}{dh}= -\frac{\rho_0}{P_0} \times g \times P(h)@

Solving the ODE

This equation of the form \latex@\frac{dy}{dx} = -B \mbox{ }y@ is one of the most common differential equations in physics and you can easily verify that the solution is

\latex@y = C \mbox{ } e^{-B \mbox{ } x}@ i.e., exponential decay

Exercise: show that this expression solves the equation

Just take the derivative with respect to x and note that the C cancels out

In our case, we use what we know about the conditions at seal level (the initial conditions) to set the constants \latex@B@ and \latex@C@

\latex@P(0) = P_0 \rightarrow C = P_0@

\latex@B = \frac{\rho_0}{P_0} g@

Which we can plug back in to get

\latex@P(h) = P_0 \mbox{ } e^{-\frac{\rho_0}{P_0} g \times h}@

That’s it, we now have an expression that we can solve to calculate pressure at any altitude.

Exercise: what is the pressure at the summit of Mt. Everest (8848m elevation)?

Simply plug in numbers (1.225kg/m³, 101325Pa, 9.8N/kg, 8848m) to the above and we get roughly 0.35atm or 35kPa

Exercise: what percentage of the atmosphere, by weight, lies below a balloon flying at 30km?

We solve first for the pressure at 30km directly from the equation to get 0.03atm. As we initially explained, the pressure is simply the weight of the atmosphere above this altitude so 97% of the atmosphere is below the balloon’s altitude.

Some more intuition

Although we have derived a solution, we can often learn a lot more by re-writing it in a more intuitive manner

\latex@P(h) = P_0 \mbox{ } e^{-\frac{h}{\frac{P_0}{\rho_0 \times g}}}@

where \latex@\frac{P_0}{\rho_0 \times g}@ has units of \latex@\frac{\frac{[Force]}{[Length²]}}{\frac{[Mass]}{[Length³]} \times \frac{[Force]}{[Mass]}} = [Length]@

so this is some characteristic distance; let’s call it

\latex@h_c = \frac{P_0}{\rho_0 \times g} \approx 8km@

which we can re-write as

\latex@P_0 = \rho_0 \times g \times h_c@

This should look familiar — in fact, it’s exactly what we derived for the pressure under an incompressible fluid of density \latex@\rho_0@ and depth \latex@h_c!@ If we were to somehow force the atmosphere to have the same density from sea-level all the way to the top, it would only be about 8km thick. Of course, as we just proved, atmospheric pressure is given by

\latex@P(h) = P_0 \mbox{ } e^{-\frac{h}{h_c}}@

So rather than running out of air at an altitude of \latex@h_c@ we will find the pressure there is

\latex@P(h_c) = \frac{P_0}{e}@

i.e., lower by a factor of \latex@e@. If we go up by another \latex@h_c@, it will become thinner by another factor of \latex@e@ and so on forever.

Exercise: how much pressure will a balloon feel at 24km?

\latex@\frac{24km}{h_c} \approx 3@ so we will have \latex@P(24km) \approx \frac{P_0}{e³} \approx \frac{P_0}{20} = 0.05atm@

In fact, every exponential equation of this form that you encounter in physics can be formulated similarly — with a variable divided by some value of that variable in the exponent. This is necessary by dimensional analysis because the number in the exponent much be dimensionless in order for the equation to evaluate to something with well-defined dimensions. It’s quite useful to calculate and consider the characteristic value in the denominator because it often has some intuitive physical significance in the same way as we saw above.

Discussion

1. Does this derivation for pressure give us any counter-intuitive conclusions? Ask, for example, what the air pressure would be at an altitude of 400km where the International Space Station orbits. What does it mean that the answer is not 0? (The ISS has to be regularly boosted to higher altitude to make up for the energy lost to atmospheric drag)
2. If the ideal gas laws have been introduced, what would happen if we actually factored in temperature? If the temperature were to have some really extreme fluctuations would it be possible for density to sometimes increase with altitude (yes). Would it be possible for the pressure to sometimes increase with altitude? (no)
3. Try to think of other physical processes that obey an exponential decay curve (e.g., radioactive decay, discharging of a capacitor) or exponential growth (e.g., population of a colony of well-fed bacteria) Consider the units of the decay/growth constants and what different values imply (e.g., a radioisotope with a decay given by -t/1s vs one with -t/1 million years)

Followup

• Given a balloon kit with a known weight and a balloon with a known burst diameter, ask students to use the equations given here to derive the maximum altitude. Compare a range of different balloons and note how the size-to-weight ratio of the balloon itself eventually limits the gains as the balloon size is increased
• Consider a way to derive the solution by using similarity properties rather than solving the differential equation
• Introduce ideal gas laws and revisit the derivation for a more general case
• Examine the 1976 Standard Atmosphere equations (often used in HAB projects to convert pressure to altitude) and ask students to consider why the range of altitudes is broken up in the way it is (temperature gradients play a major role)

We’re always working to improve; please send suggestions or errata to andrey@loonar.tech