GRADES 9–12
NGSS: Earth's Systems:
Use a model to describe how variations in the flow of energy into and out of Earth's systems result in changes in climate.
CCSS: Functions: Linear, Quadratic, and Exponential Models:
Distinguish between situations that can be modeled with linear functions and with exponential functions.
In addition to hurricanes, typhoons, and other tropical cyclones, students may want to look at regular warm and cold fronts on a current weather map and compare them to what the animation shows. During the winter in either hemisphere, one can see a rather strong (and cold) polar vortex
circulating at high altitude above the wintering pole.
The animation also shows sea surface data, including waves and currents. One can find the Gulf Stream as well as several other ocean currents. These currents are not simple, smoothlyflowing bodies of water; there are eddies that pinch off of their
edges and meander away.
Advanced students with some knowledge of calculus may want to derive how to calculate the atmospheric pressure as a function of altitude with an eye to converting “pressure altitude” to geometric altitude. One can draw a thin horizontal box at some altitude “z” and
calculate the forces on the air in the box. If the horizontal crosssection area of the box is “A” and the (vertical) thickness of the box is “D”, then the mass of the air in the box is “r A D", where “r” is the density of the air. The
force of gravity “F_{g}” acting on the air is given by the formula (with a minus sign indicating that it is pulling down)
In this equation, “g” is the gravitational acceleration. It actually varies slowly with altitude, but we will neglect that variation for lower altitudes. We will refer to the pressure at the altitude “z” at the bottom of the box as the
function “P(z)” and the pressure at the altitude “z + D” at the top of the box as “P(z+D)”. Because the pressure is the force per unit area, the force “F_{PB}” that the air exerts pushing up on the bottom of the box is given by the formula
Similarly, the force “F_{PT}” of the air pushing down on the top of the box is given by the formula
Since our block of air is sitting still, the forces on it must balance:
Substituting in gives
The “A” in each term cancels out, meaning that our derivation does not depend on the horizontal size of the box. This is an excellent commonsense check.
We now need to use the perfect gas equation of state. This is a generalization of Boyle’s Law (that the pressure on a given amount of gas, multiplied by the volume of that amount of gas, at constant temperature, is a constant) and Charles’ Law (that the volume of a given amount of gas, divided
by the temperature of that amount of gas, at constant pressure, is a constant). It is given by the formula
where “R” is our “gas constant.” We solve for the density “p” and substitute this into our balance of forces equation:
Astute readers will note that our “P” in the first term does not have an argument. Strictly speaking, we took our density at the center of our box and so our pressure should be taken at that altitude (which would be “z+ (1/2) D”) as well. In fact, when we specify “D” to be small enough, we can specify
that the pressure in the first term can be taken from anywhere inside the box.
Now the knowledge of calculus comes in. The definition of the derivative of a function at a point “h” is the slope of the curve of the function:
In this formula, “lim” means “find the limit as a small quantity called ‘dh’ approaches zero.” Our balance of forces equation has for its second and third terms a function (the pressure “P”) evaluated at two points (“z” and “z+D”) that are very close
together. We rearrange the terms in our balance of forces equation to show this more clearly:
We match this equation with the equation that defines the derivative of a function. Our pressure “P” is our function “f”; our altitude “z” is our variable “h”; and our box thickness “D” is our small change “dh” to our variable. If we let our box thickness “D” get arbitrarily small (and this is
why it does not matter where in the box we take our pressure in the first term of the force balance equation), our force balance equation becomes
If we wish to be
pedantic, we should specify that “D”
becomes arbitrarily small but stays large enough that the size and spacing of
the molecules in the air does not become an issue.
Now let us consider the exponential function:
In this equation, “C” and “k” are arbitrary constants.
If we find the derivative of this function, our formula gives the
following:
The next step is to
evaluate the limit. If we let “dh” go to zero, then our constant “k” multiplied by “dh” will also go to zero. We
multiply and divide our limit by “k”:
This limit is the
“rise over the run” of the exponential function at “h = 0”, which is one. Our
formula for the derivative is therefore
Now we look at the last
version of our force balance equation.
Our constant “k” is “–g/RT”
and our function “f(h)” is “P(z)”. Our other constant “C” is the pressure at zero altitude, “P_{0}”, which has a value of 1,013.25 hPa. We find that our pressure as a function of
altitude is given by an exponential function:
It should be noted that
this derivation assumes that the temperature does not change with altitude;
while this is not exactly true, it is close enough for rough calculations. I suggest that an “average” sort of
temperature be used, such as 250 K; the temperature in theatmosphere below
about 25 kilometers (82,000 feet) ranges between 288 K and 216 K. Gravity at sea level is 9.81 m/s^{2}. The gas constant “R” for air is 287 m^{2}/s^{2}K. The students can now calculate the pressure
altitude in terms of the geometric altitude.
Depending on the choice for the temperature, they should get results
that are within about ten percent of the given values.
Sixty Years Ago in the
Space Race:
February 13, 1957:
The first Australian
Skylark rocket was launched at Woomera, Australia to a height of 50,000 feet
and landed 20 miles downrange.
