Air Drag A Physical View

5.1 The Quadratic Force Law

In the first discussion on air drag I wasn’t terribly concerned about the actual applicability of our results to the real world. That is, while I limited myself to physically plausible air drag force laws, I didn’t take the next step of asking which ones are the force laws that are actually observed. I have already opened the door to more than one law, of course; Stokes’s linear law and the quadratic law. Can both apply in real life? Yes, but in entirely different situations. So, to be absolutely clear from the start, in this discussion we’ll be interested in massive objects, such as iron balls and people, that are falling straight down through the atmosphere of Earth toward the ground. To start, I discuss falls that take place fairly close to the surface of the planet, but later on we’ll see what happens when we extend our analyses to include very long falls from high altitude. Curious things indeed do happen on such falls. We will in every case assume that these falls begin from rest, at time t = 0. And, as you’ll soon see, it is the quadratic law that will be the one of interest.

If we take the simplest possible case of masses falling through a vacuum, where there is no drag at all, then it is easy to calculate the answers to such questions as (1) how far has the mass fallen at time t? and (2) how fast is the mass moving at time t? Starting from rest, and with an acceleration of 1 g in a vacuum, a falling object descends a distance and is moving at speed v = gt, t seconds after the fall begins. This means, since , that t seconds after the fall begins, the object is moving at speed . For example, after the object has fallen just one foot, it is already moving at the speed

Now, for purposes of seeing how the numbers work (and here I’ll switch to the metric system), let’s suppose we have a spherical object of radius R falling at speed v through air. For air, at sea level and room temperature—what physicists and chemists call standard temperature and pressure, or STP—the density and viscosity are, respectively,

and

The Reynolds number² for this fall is then

where R is in centimeters (cm) and v is in centimeters per second (cm/s). Remember, Re itself is a dimensionless number, which you can verify by working through the units. For the quadratic drag force law to apply we require that Re > 10³, and for R = 1 cm (a steel ball bearing a little less than an inch in diameter, for example), this translates into a speed of at least v = 65 cm/s. Assuming the fall is in a vacuum, the time to reach this minimum speed is t = v/g = 65/981 seconds = 0.066 seconds. And the distance traveled in that first 66 milliseconds of the fall is

which is less than one inch!

These numbers are for a fall in a vacuum, but a fall through air would have little impact on them so early in the fall. That is, we have the (surprising, I think) result that, except for an extremely brief time interval at the very start of a fall, the fall is in the realm of Re > 10³ and that means it is the quadratic drag force law that is the force law in play over virtually the entire duration of the fall. An implication of this conclusion is that to be in the realm of Re < 1, where Stokes’s linear drag force law is in play, we must be dealing with a very, very slow fall, e.g., a grain of sand descending through cold, thick honey. A practical rule of thumb is that if you can actually see an object falling (it is large enough to be seen, and you can see it moving), then it is surely already in the quadratic drag force realm.

Now, let me show you two fascinating examples of a quadratic drag force fall. I’ll start with the oft-quoted statement in physics textbooks and classroom lectures that Galileo proved, with his famous Leaning Tower of Pisa experiments, that two objects (one heavy and one not so heavy) fall at the same rate. (We’ll exclude such complicated objects as feathers.) That is, contrary to Aristotle’s supposed claim that the heavier object would fall faster (that is, reach the ground first) than the lighter object, Galileo is said to have shown that both objects would, if dropped simultaneously from the same height, reach the ground at the same instant. There is today, among historians of science and physicists with an interest in history, an understanding that this story is riddled with inaccuracy and misstatement. I’ll not go into the historical debate³ here but rather show you by direct calculation what the mathematical physics says.

Suppose our object is a sphere of radius R and density Ρ_o, that is, the mass of our sphere is

Starting its fall from rest at time t = 0, let’s write v(t) as the sphere’s speed at any time t ≥ 0, and so v(0) = 0. Now, the downward gravitational acceleration is g, reduced by the effect of air drag. The air drag force law is kv² (where k = 0.2ρ_aπR², as discussed in note 2 of Discussion 4, where ρ_a is the air density). The reduction in acceleration is, by Newton’s second law of motion, the drag force divided by the

sphere’s mass, and so the sphere’s acceleration is

If we write the terminal (falling) speed as v _T, which as discussed before in Discussion 4 is the speed when the falling sphere has reached a constant speed and so we see (as before) that

Thus, (5.1) becomes

Suppose we measure the distance of the fall with the variable y(t), where y(t = 0) = 0 and y increases positively downward toward the ground. Then, the speed of the fall is

From the chain rule of calculus we have

and so(5.3)becomes

This is easily integrated, i.e., writing u as a dummy variable of integration we then have, upon separating variables,

Then, either by making a simple change of variable or by just looking in integral tables, it is easy to arrive at the sphere’s speed as a function of how far it has fallen:

This immediately tells us that as the sphere falls toward Earth, its speed monotonically increases from zero toward the terminal speed, but never actually reaches v _T.

We can integrate (5.3) is a different way, too, to directly find v(t), that is, the sphere’s speed as a function of how long it has fallen. That is, write

and so, integrating with v(0) = 0, we have

Thus,

or

Note that another expression for t _T can be found from the formulas for m, v _T, and k. Since and k = 0.2ρ_aπR², we have

In any case, with just a bit of algebra we can solve for v to get

or, at last,

From (5.6) we next easily get the distance, y(t), fallen by our sphere; from integral tables we have

and so, with A = 1/2t _T, we have

or

You’ll notice that (5.7) says y(t = 0) = 0, as it should.

Equation (5.7), simple as it may look, has great surprises for most who first encounter it. For example, let’s go back to what Galileo had to say about two falling balls:

Iron has a density of about ρ₀= 7.8g/cm³ and so, since there are 454 grams in a pound, the radius of a one-pound iron sphere (ball 1) is about

The radius of a ball 100 times heavier (ball 2) would then be (100)^1/3 = 4.64 times greater, i.e., R₂ = 11.1 cm. In Figure 5.1 I’ve plotted y₂(t) —y₁(t), using (5.7), for these two iron balls, and, it is clear, the heavier ball does fall faster than the lighter ball.⁵ In a vacuum, of course, the two balls would indeed, as Galileo claimed, fall at identical rates, but in air with its quadratic drag they do not—and Galileo was simply wrong in his assertion that an actual experiment would show him correct. It would not. In fact, when the 100-pound ball hits the ground when dropped from a height of 100 cubits (that is, from a height of 150 feet, equal to 4,572 cm) we see that one-pound ball still has over 50 cm to

Figure 5.1. Iron balls of different sizes fall at different rates!

go, substantially greater than Galileo’s “two finger breadths.” This gap could, I think, be seen by an observer on the ground.⁶ So, I ask the natural question: Did Galileo really do his historically famous experiment at the Leaning Tower of Pisa?

For our next surprise, suppose we now take two balls of different material, one with density 11.3 g/cm³ (ball 1 is a lead ball) and R ₁ = 1 cm, and the other with density 0.75 g/cm³ (ball 2 is a wooden ball) and R₂ = 3 cm. As before, ball 2 is the heavier ball, as its much lower density is more than compensated for by its much larger volume. In Figure 5.2 I’ve again plotted y ₂(t) —y ₁ (t), but now it is the less massive (lead) ball that reaches the ground first (that’s what y₂(t) —y ₁ (t) ≤ 0 means, of course). With a 150-foot drop, the plot shows that the lighter ball hits the ground when the heavier ball still has about 400 cm to go, a huge gap that a ground observer should easily be able to see. If you tell me that doesn’t surprise you, well,—I don’t believe you! The reason for this is, of course, the greater air drag experienced by the ball with the greater cross-sectional area, irrespective of the fact that it is the heavier ball.

Figure 5.2. The lighter (lead) ball hits the ground first!

Our analyses so far, for a body falling through the atmosphere starting from rest, speeding up, and approaching a terminal speed from below, does motivate an intriguing question: what happens if the body starts its fall with an initial speed greater than v_T? Forexample, suppose a meteor plows into the upper atmosphere at a typical speed for such objects of 20 miles per second (which is far larger than its terminal speed for a fall from rest through the atmosphere)? The answer is that the meteor will slow down and approach v _T from above. This slowing down is achieved via a fantastically large deceleration. For a quadratic drag force law we can write the net (upward) force on a meteor with mass m as

where I’ll assume that the acceleration of gravity at the top of Earth’s atmosphere is the same as at the ground (I’ll address this assumption and show it is okay in just a bit). So, the deceleration of the meteor is,

from Newton’s second law,

When v = v _T we have a = 0 by definition, and so

Thus,

If, for example, we take v _T = 100 m/s and v = 30,000 m/s at atmospheric entry, we see that a = 90,000 g. Most meteors are most likely utterly destroyed by such an enormous deceleration long before they slow down to a speed even approaching v _T.

5.2 Long Falls through A Real Atmosphere

Up to now I have implicitly limited the physics to falls through the atmosphere that are near the surface of Earth. But what if we have a really long fall that starts very high up in the sky? Such a fall might be intentional, for example a military HALO parachute jump,⁸ or unintentional, for example the fall from 65,000 feet to the ocean surface below of the intact crew cabin during the Challenger space shuttle disaster of 1986.⁹ Are such long falls somehow different from the near-Earth falls we have been discussing so far? What we have to do, to answer that question, is look at the assumptions inherent in those ’short‘ fall analyses. There are two. If you look back at what we did, you’ll see we assumed that both the acceleration of gravity, and the density of air, remain constant over the entire duration of the fall. The first assumption continues to hold even for a pretty long fall, but the second does not.

So, first, let me show you now why we can still retain the assumption of a constant value of the acceleration of gravity equal to its value at the surface of Earth (g), even when our falling object is far above the surface of the planet. To see this, suppose the radius of Earth is R, and that the falling object is distance z above the surface, i.e., it is distance R _e + z from Earth’s center. We then have the gravitational force on the object, from Newton’s famous inverse square law of gravity, as

where G, M, and m are the universal gravitational constant, the mass of Earth, and the mass of the falling object, respectively. Since this force is also equal (from Newton’s famous second law of motion) to ma, where a is the instantaneous acceleration of gravity the object experiences, we have

When z = 0, we know that a = g (by definition), and so GM = gR_e².

Thus,

From this you should be able to see that

before a is more than 1% different from g. Since R _e = 3,960 miles, this means that z must be greater than 19.95 miles (105,300 feet) before we make even as much as a 1% error in taking the acceleration of gravity as everywhere equal to g during even a very long (i.e., high-altitude) fall through the atmosphere.

So, even at an altitude much higher than that from which Challenger fell, we don’t have to worry about the variation of the acceleration of gravity with height. It is 1 g all the way down. What we do have to worry about is the variation of air density, which is much less at the start of the fall than it is later as the fall approaches the ground. The mathematics of this complication seems to be regularly rediscovered,¹⁰ but I believe it was first worked out by the German-born American aerodynamicist Max Munk (1890–1986). Munk’s analysis¹¹ is a bit terse (in my opinion), so I’ll take you through it in a much more leisurely fashion.

The analysis begins with the fact that the density of the atmosphere decreases exponentially with increasing elevation.¹² If as before we measure the distance of a fall by y(t), where y(t = 0) = 0 and y increases positively downward, then the speed of the fall is

and the air density is

where ρ_a(0) is the air density at the start of the fall and β is some positive constant with units of m⁻¹ if we measure y in units of meters. That is, the air density increases with increasing y, which simply says that the air density increases exponentially as the fall approaches the surface of the planet. (The case of Β = 0 is the special, far less realistic case of a constant air density atmosphere.) Now we simply repeat the argument from the first part of this discussion: the air drag force is 0.2ρ₀πR²v², and so the reduction in the gravitational acceleration (from g) of our falling sphere is this force divided by the mass of the sphere (as before, I’ll take the sphere to have radius R and density ρ₀), which gives

or

where k is a constant (with the dimension cm^—1 if R is measured in cm).

Now, a brief time-out. Note carefully that the above k is not the k of our earlier analysis, the k of either (5.1) or (5.2). There, that earlier k played a central role in determining the terminal falling speed v _T. But notice that now, with variable air density, there is no such speed. That is, for any β≥ 0 there is no constant solution for v = dy/dt when we set dv/dt = 0 in (5.8). This may seem a bit disconcerting, because it was v _T that we used as a unit reference speed, against which we measured all other speeds. But another “natural” speed will emerge just in time to play the role of a reference speed for us. Okay, back to our analysis.

The differential equation of (5.8) looks pretty nasty, but in fact we can make some useful progress in its solution. Since

then, again by the chain rule of calculus, we can write

and so (5.8)becomes

Since

then

If we next define the (dimensionless) variable

and since, again by the chain rule,

we have

and so we arrive at a first-order linear (which means we can solve it!) differential equation for v²:

We can solve (5.9) by the classic technique of using the appropriate integrating factor (see any good book on differential equations), which in this case is e^p(x), where

Multiplying through (5.9) by the integrating factor, we have

or

This integrates by inspection to give

where the limits on the integral have been chosen so that when x = 1 (its initial value, i.e., since x = e^βy, then x = 1 when y = 0), we will have v = 0 (the fall begins from rest). So, at last,

Or almost at last. To put this expression for v² into the most convenient form, let’s now write the dimensionless constant α as

and then, returning to y (the distance of the fall) as our independent variable, we arrive at

If we next change variable to z = αu (dz = α du), we have

or

Notice that has the units of that is, the units of a speed squared. So, let’s define our unit speed as

Notice, too, that (5.12) is undefined in the case of β = 0—the case of a uniformly dense atmosphere—but in that special case, as I discussed earlier, we have v _T, the terminal speed, available to serve as our reference speed. Further, let’s define our unit distance as

These definitions reduce (5.11) to

This integral may perhaps look a little unusual to many readers: as Munk (see note 11) put it, the integral is “a mathematical function known as the exponential integral. It is a little known transcendental function [Munk was writing in 1944, when that statement was probably true] which cannot be reduced to the elementary functions.” But in fact it is a well-tabulated function in handbooks and is available as a built-in function (just as cos, cosh, sinh, etc., are) in any good modern scientific programming language, such as MATLAB. It is generally defined as follows:

and so our final result becomes

All we need to do now is to determine the value of β, which sets the values of both the unit distance and the unit speed, and the value of α, and then we can plot v as a function of y once the size of our falling sphere is given. From Mohazzabi and Shea (see note 10) we learn the value of β for Earth’s atmosphere is 0.134 km⁻¹ = 0.000134 m⁻¹. This says, using (5.12) and (5.13), that

and

Now, what is ?? From (5.8) and (5.10) we have

where, to restate, R is the radius of the falling sphere, ρ_o is the sphere’s density, and ρ_a(0) is the density of air at the start of the fall. Now, let me make one last definition. If we say the fall starts at height h, and if ρ_a is the density of air at the surface of the planet, then we have

So,

If we measure R in units of centimeters, h in meters, and the density of air in g/cm³, then with ρ_a= 1.3 × 10⁻³ g/cm³ we have

or,

With (5.13), (5.14), and (5.15) we can now calculate the speed of a falling sphere once we have specified (1) h, the height (in meters) at which the fall starts, (2) R, the radius of the falling sphere (in centimeters), and (3) Ρ_o, the density of the falling sphere (in g/cm³). To give you an idea of the magnitude of the numbers we are talking about for spheres falling to Earth’s surface from great heights, Figure 5.3 shows the speeds of three iron spheres (Ρ_o = 7.8 g/cm³) of radii 3 cm, 6 cm, and 15 cm (this last sphere weighs a little over 240 pounds, equal to the weight of a large man), each falling from a height of h = 22,389 meters (about 73,500 feet). The horizontal axis is the distance fallen, and so has a maximum value of 3 (= 22,389/7,463). The vertical axis is the falling speed (in units of 383 m/s).

Figure 5.3. Three iron balls falling from 22,389 meters.

The speed versus distance fallen curves show that all three balls fall with increasing speed until a maximum speed is reached (the heavier the ball, the larger is that maximum speed and the further the ball falls before reaching it), and then the speed begins to decrease. This makes physical sense, too, if you think about it, as the balls encounter a continually denser atmosphere as they approach the surface of Earth. This is very much different from the monotonic increasing nature of the falling speed through a uniform, constant-density atmosphere. The following table shows, for each ball, its maximum speed V _max, in m/s, the distance it has fallen when it reaches its maximum speed D_max, in m, and the speed with which it impacts the ground V_impact, in m/s. Also, just for comparison, I’ve listed (using (5.6)), the terminal speed v_T (assuming an atmosphere with air density equal everywhere to that at the surface) for each of the spheres, also in m/s. Notice that both the maximum speed and the impact speed for each sphere, when falling through a realistic exponential atmosphere, exceed the terminal speed of the less realistic constant-density atmosphere.

CP.P5.1:

Notes and References

1. Craig F. Bohren, “Dimensional Analysis, Falling Bodies, and the Fine Art of Not Solving Differential Equations” (American Journal of Physics, April 2004, pp. 534–537).

2. I introduced the Reynolds number in Discussion 4 (see its note 2), and you can find much more on it in the beautiful paper by E.M. Purcell, “Life at Low Reynolds Number” (American Journal of Physics, January 1977, pp. 3–11). There Purcell wrote of Reynolds, “That was a very great man. He was a professor of engineering, actually. He was the one who not only invented Reynolds number, but he was also the one who showed what turbulence amounts to and that there is instability in flow, and all that. He is also the one who solved the problem of how you lubricate a bearing, which is a very subtle problem that I recommend to anyone who hasn’t looked into it.”

3. See, for example, Barry M. Casper, “Galileo and the Fall of Aristotle: A Case of Historical Injustice?” (American Journal of Physics, April 1977, pp. 325–330), and Carl G. Adler and Byron L. Coulter, “Galileo and the Tower of Pisa Experiment” (American Journal of Physics, March 1978, pp. 199–201).

4. Dialogues Concerning Two New Sciences (New York: Macmillan, 1914, pp. 64–65) (see note 4 in Discussion 4).

5. This calculation was inspired by a fascinating paper written by Gerald Feinberg, “Fall of Bodies Near the Earth” (American Journal of Physics, June 1965, pp. 501–502). There are some minor typographical errors in Feinberg’s mathematics, which I’ve corrected.

6. I should qualify this statement with the proviso that the two balls must be dropped at the same instant. Falling at a speed on the order of 100 ft/s when they hit the ground, a 50-cm gap (i.e., about a foot-and-a-half gap) is equivalent to about 16 milliseconds. The dropping of the two balls must be synchronized to even less than that to avoid masking the gap. I’ll leave it to you to think about how that might have been accomplished using only technology that would have been available to Galileo.

7. There is one additional complication that can be added to our analysis of falling balls—the upward buoyant force due to the air displaced by a ball. According to Archimedes’ principle, that force is equal to the weight of the air displaced by the ball (which of course is the weight of a volume of air equal to the ball’s volume). You can read more on how to incorporate buoyant forces into the analysis of falling balls in Byron L. Coulter and Carl G. Adler, “Can a Body Pass a Body Falling Through the Air?” (American Journal of Physics, October 1979, pp. 841–846), and Peter Timmerman and Jacobus P. van der Weel, “On the Rise and Fall of a Ball with Linear or Quadratic Drag” (American Journal of Physics, June 1989, pp. 538–546).

8. HALO (high-altitude, low-opening) parachute jumps are used by air-borne soldiers attempting to get on the ground fast to avoid “floating” high up in the sky as targets to ground fire. This is done by delaying the opening of the parachute until literally the last possible moment, consistent with not blasting a hole in the ground with one’s body. It is an extremely dangerous maneuver. (You can experience that interesting event by playing the video game Medal of Honor Airborne.) You can find an exciting (if fanciful) description of a fictional HALO jump (in a snowstorm!) in Stephen Hunter’s novel, Time to Hunt (New York: Doubleday, 1998, pp. 480–482 and 493–494). The highest delayed-opening parachute jump to date was made August 16, 1960, by a U.S. Air Force officer from a balloon gondola. The free-fall jump started at a height of 102,800 feet, and at 90,000 feet (thirty seconds into the fall) he reached his maximum speed, an astounding 614 miles per hour, which is nine-tenths the speed of sound at that altitude. The main parachute opened automatically at a height of 17,500 feet (a small, flat-spin stabilization canopy chute had opened at 96,000 feet). See Capt. Joseph W. Kittinger, Jr., “The Long, Lonely Leap” (National Geographic December 1960, pp. 854–873), which includes an automatically taken photograph of Kittinger the instant after he stepped over a sign at the foot of the gondola door, reading “Highest step in the world,” with the understandable prayer “Lord, take care of me now” on his lips.

9. The orbiter explosion occurred at an altitude of 50,800 feet, but the crew cabin continued on upward to a peak altitude of 65,000 feet, where it began its free fall to the ocean below. The fall took two minutes and forty-five seconds, during which it is generally thought the astronauts were alive. A heartbreaking reconstruction of how the flight crew might have struggled during the fall to regain control of their fatally wounded craft is told by the astronaut Mike Mullane in his terrific book, Riding Rockets (New York: Scribner 2006, pp. 249–250).

10. See, for example, Pirooz Mohazzabi and James H. Shea, “High-Altitude Free Fall” (American Journal of Physics, October 1996, pp. 1242–1246). See also, all in the American Journal of Physics, Ralph Hoyt Bacon, “Motion of a Particle Through a Resisting Medium of Variable Density” (January 1951, p. 64); George Luchak, “The Fall of a Particle Through the Atmosphere” (October 1951, p. 426); William Squire, “Motion of a Particle Through a Resisting Medium of Variable Density” (October 1951, pp. 426–427); and W. A. Bower, “Note on a Body Falling Through a Resisting Medium of Variable Density” (December 1951, pp. 562–564).

11. Max M. Munk, “Mathematical Analysis of the Vertical Dive” (Aero Digest, February 15, 1944, pp. 114 and 213).

12. The decrease in air density with increasing height I take to be obvious. The fact that the decrease is exponential is not quite so obvious, but is easily derived (it is a common example in college freshman physics texts, and so any of a number of good, modern texts will have a mathematical derivation). For us, we’ll take it as simply an experimentally verified fact.

13. You can find a freshman calculus derivation of this important formula in my book, The Science of Radio, 2nd ed., (New York: Springer, 2001, pp. 416–418).