SPECIAL FEATURE: E XTREME P HYSICS www.iop.org/journals/physed The physics of Colonel Kittinger’s longest lonely leap
A W Robinson1,3 and C G Patrick2
Department of Physics and Engineering Physics, University of Saskatchewan, 116 Science Place, Saskatoon, SK, S7N 5E2, Canada 2 Department of Civil Engineering, University of Saskatchewan, 57 Campus Drive, Saskatoon, SK, S7N 5A9, Canada E-mail: andrew.robinson@usask.ca
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Abstract We present a case study of the physical principles necessary to model the high altitude parachute jump made by Colonel Joseph Kittinger, USAF, in 1960, in order to determine the maximum speed attained and to calculate whether this speed was sufficient to exceed the speed of sound at that altitude. There is considerable discrepancy in the value of the maximum speed attained— 614 miles per hour (mph) or 714 mph—in material available over the internet. Using a very simple physical model we are able to conclude that the lower figure is correct and that it is likely that Colonel Kittinger did not ‘break the sound barrier’ during his descent. The wealth of audio-visual material and animations available makes this a particularly attractive case study to instruct students in elementary kinematics, approximation and physical modelling.

Introduction
On 16 August 1960, Captain (later Colonel) Joseph Kittinger, USAF, jumped from the helium filled balloon Excelsior III, at an altitude of 31 600 m (102 800 ft). His parachute jump remains the highest altitude jump ever made, and the balloon ascent is also the highest altitude ever attained by a manned balloon [1, 2]. Meanwhile, somewhat less heroically and some 45 years later, one of the authors (AWR) began to include ‘back of the envelope’ or ‘Fermi problems’ in one of his first-year university physics courses. The objectives of this are to get the students to think like a real physicist and develop a physical model of a situation, apply some simple mathematics and produce an answer
3 Author to whom any correspondence should be addressed.

which hopefully matches reality. It challenges the students to move away from the ‘find an equation and substitute numbers in as quickly as possible’ approach which is often a successful strategy for earning high marks in high school. The students in the class refer to this approach as ‘plug and chug’! The conclusions from a study of student grades and levels of attainment are that the types of skills needed to solve ‘back of the envelope’ calculations have to be taught and are not innate ones possessed by first-year undergraduate students [3]. So what is the connection between these two apparently unrelated events? One of the current authors (CGP), a former student of this physics class, who had taken the ‘back of the envelope’ skill training, saw a ‘You-Tube’ video clip of Kittinger’s jump [4] and had serious doubts as to whether some of the claims made in several
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A W Robinson and C G Patrick internet-based sites about the jump were actually correct. He therefore applied the modelling and problem solving skills taught in the physics class to this problem and was able to disprove some of the facts claimed on the internet. It turns out that Colonel Kittinger’s parachute jump is an excellent case study to demonstrate to students that information collected from the internet is not always correct, to stimulate critical thinking skills and allow the construction of rather simple physical models to disprove erroneous claims and substantiate the correct value. In this article we will explore the various aspects of physics and the level of approximation which need to be used in this particular case study, and highlight the various instructional tools and multimedia resources available to the educator. We will give many references to material which is available over the internet, easily accessible by students. air, there will be less air resistance [12]. Next, we need to account for the magnitude of acceleration due to gravity. At an altitude of 31 600 m, is g significantly smaller than it would be at the surface of the earth? One could do a direct calculation using Newton’s law of gravity to find a value for the magnitude of the acceleration due to gravity, g , at an altitude h above the earth:

g(h) =

GM (r + h)2

(1)

The jump
An inspection of various internet-based data sources gives the maximum speed attained by Kittinger as either 614 or 714 mph. The latter figure would be sufficient to break the speed of sound at high altitude, given the low air temperature, and this ‘breaking the sound barrier’ feat is often claimed by many of the data sources. Interestingly, the lower figure of 614 mph is reported in both of the print magazine articles written in 1960 [1, 2], and the higher figure is often quoted in US-based websites, particularly those associated with the US military [5, 6]. In the interests of fairness, it should be noted that some US military sources do quote 614 mph as the maximum speed [7]. These various figures have been tabulated by Elert and co-workers [8]. We can model the initial stages of Colonel Kittinger’s jump using a very simple physical model. First, we will assume that the air resistance is negligible. At an altitude of 32 km, the air pressure is only 868.02 Pa according to the International Atmosphere Model [9–11], compared to a value at sea level of 101 325 Pa. The phenomenon of air resistance is a complicated one, as it also depends on the density of the air and the compressibility and the temperature, but broadly speaking, if the falling body is surrounded by less
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where M is the mass of the earth, G is the universal gravitational constant and r is the radius of the earth. We will assume that the earth is a sphere with a radius of 6.38 × 106 m and a mass of 5.98 × 1024 kg [13]. For an altitude of 31 600 m, this allows us to calculate the acceleration due to gravity at 31 600 m as 9.70 m s−2 . An alternative, for more mathematically inclined students, would be to use equation (1) to determine the ratio of the magnitude of g(h) at an altitude h above the surface of the earth to g(0), the value at the surface of the earth at the latitude of the continental United States [13]. From a teaching perspective, this would be the preferred method of calculation, as this is a standard method in physics of factoring out unknown constants. In this case the values for the mass of the earth and the gravitational constant are now not required. We also note that many students in elementary physics courses need practice in taking ratios. Using the ratio method, we obtain

g(h) = g(0)

r r +h

2

.

(2)

This can be reduced to

g(h) h = 1+ g(0) r

−2

.

(3)

We could now use the binomial expansion of equation (3) because h is small compared to the radius of the earth. The fact that this expansion is valid demonstrates that the ratio g(h)/g(0) only varies by a small amount over the range of heights under consideration. If we take the value for g(0) to be 9.80 m s−2 , then, using equation (3), we can calculate g(31 600 m) as 9.70 m s−2 , a negligible correction (when carrying out order of magnitude
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Physics of Colonel Kittinger’s leap calculations) over the value of g at the earth’s surface. We note here that many students are unaware that the acceleration due to gravity varies with latitude. As a Canadian institution, we are at relatively northerly latitudes and the local value in Saskatchewan is 9.81 m s−2 , the value used in our physics laboratory experiments. Our textbooks, on the other hand, are sourced from the United States, where the acceleration due to gravity is taken to be 9.80 m s−2 . Although the difference is small, this does cause some confusion for our students. Now we will examine whether assuming that the earth is a perfect sphere is a reasonable approximation. The international gravity formula [14] assumes that there is flattening in the polar regions, and so the earth is an oblate spheroid. This means that the acceleration due to gravity at a latitude φ varies as

Excelsior III balloon

stabilization parachute opens maximum speed measured at 614 mph

Figure 1. Colonel Kittinger’s parachute jump, as described in reference [1].

g(φ) = 9.780 3185(1 + 0.005 278 895 sin2 φ (4) − 0.000 023 462 sin4 φ).
A comprehensive review of the various models for the shape of the earth is given by HofmannWellenhof and Moritz [15]. The latitude of New Mexico, where the jump took place, is 34◦ , and so we could calculate a value for the acceleration due to gravity at the surface as 9.80 m s−2 . The firstorder correction term in equation (4) is of the order of 0.5%, and hence is negligible. In our model of the parachute descent, we will take the simplest possible physical model, and assume a constant value for g of 9.80 m s−2 . In Kittinger’s own account, written in National Geographic in December 1960 [1], he states that he jumps at an altitude of 102 800 ft (31 600 m). We will assume that his initial velocity is zero when he jumps from the balloon. He falls freely to an altitude of 96 000 ft, where a small (1.8 m diameter) stabilization parachute opens (see figure 1). At this point we will assume that this parachute does not exert a significant retarding force, consistent with Kittinger’s own observation that he felt no effects when the parachute opened. He then falls for another 6000 ft, reaching a reported top speed of 614 mph on a wrist-mounted velocity meter. This means that he is now at an altitude of 90 000 ft (27 700 m) and has fallen through a distance of 12 800 ft (3900 m). Colonel Kittinger then fell for another 4 min, albeit at
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lower speeds, until at 18 000 ft he opened his main parachute. We use elementary kinematics, assuming constant acceleration, to model the descent from the initial jump from the gondola of the balloon to the point at which the maximum speed was recorded:
2 v 2 = v0 + 2ah,

(5)

where v is the final velocity, v0 is the initial velocity, a is the acceleration due to gravity and h is the distance fallen. Assuming that the initial velocity is zero when he jumps, √ v = 2ah. (6) Using our value for the acceleration due to gravity of 9.80 m s−2 and a height difference of 3900 m, equation (6) gives a figure for the final speed as 276 m s−1 , or about 621 mph. As we have neglected air resistance, the drag from the stabilization parachute, and used an acceleration due to gravity slightly higher than the actual figure at this altitude, the actual maximum speed must be slightly less than this figure. We therefore conclude that, as this figure is close to the value of 614 mph cited in many of the references, the value of 714 mph is not reasonable. We also have no information on the accuracy of the speed measurement, so it is a reasonable assumption that our model is a good one under the circumstances, and that our initial decision to neglect air resistance was justified. This scenario has also been investigated by Yoon [16] using a value of g(h) of 9.72 m s−2 . This value is calculated using the aerospace web atmospheric calculator [17] and is based on the
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A W Robinson and C G Patrick US Atmosphere Model [18, 19], which corrects for altitude and latitude. As we have already demonstrated, this level of approximation is unnecessary for an order of magnitude calculation. Yoon’s model gives a value of 275 m s−1 , or 619 mph (not 615 mph, as quoted in [16]). The internet site [6] which purports to prove a final speed of 714 mph (317 m s−1 ) makes the claim that this higher speed was attained in 23 s, although it is not clear where this figure for the time has come from. Let us continue our simple kinematic analysis by determining what acceleration would be required to reach a final speed of 317 m s−1 from rest in 23 s. We use another of the elementary equations of kinematics:
80 70 60 altitude (km) 50 40 30 20 10 0 200 220 230 K 240 260 temperature (K) 280 300 28 km

Figure 2. Atmospheric temperature profile calculated from the International Standard Atmosphere model [9–11].

v = v0 + at.

(7)

We can solve equation (7) for the acceleration a , given an initial velocity, v0 , of zero, to find that a = 13.8 m s−2 . As this value is much larger than the acceleration due to gravity at the surface of the earth, we can conclude that either the final speed, or the time, or both are incorrect. So where does the value of 714 mph come from? Yoon [16] has concluded that it is simply a misprint of 614 mph made in some publication at some time after the jump. The articles in National Geographic and Time [1, 2] which were published soon after the event in 1960 have the more plausible figure of 614 mph in them, as do several other internet-based information pages. We also note that internet-based supplementary material from the US television series Nova [20], which generally has excellent standards of background research, also supports the lower figure. The Nova website even has a downloadable game, which allows a student to simulate the entire jump. The physics in the simulations appears to be correct, and the fastest speed is given as 614 mph. One might speculate that an error occurred somewhere in the US Defense Department, which passed on a press release to various military museums and other institutions. It seems likely that some of the internet sources are citing secondary sources of evidence. You might want to pass this article on to colleagues in history departments as a lesson in using primary data sources. We now turn to testing the assertion that Colonel Kittinger broke the sound barrier while falling. If we assume that the earth’s atmosphere
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is an ideal gas, then the speed of sound, vsound , is given by [21] √ vsound = (γ ∗ R ∗ T /M), (8) where M is the molar mass of air, R is the ideal gas constant (8.31 J K−1 mol−1 ), and T the temperature of the air in kelvin [15]. The variable γ is the ratio of the heat capacity at constant pressure, C p , to that at constant volume, Cv . For a diatomic ideal gas, the value of γ is 7/5. The temperature of the air at different altitudes varies significantly. In the troposphere, up to 10 km above the earth, the temperature falls with altitude. In the stratosphere, from 10 km up to approximately 50 km, the temperature increases with altitude, due to the absorption of ultraviolet radiation by ozone molecules [23]. Colonel Kittinger’s freefall descent starts within the stratosphere, at an altitude of ∼32 km. Figure 2 shows the variation of temperature with altitude using the International Standard Atmosphere (ISA) model [10]. The temperature of the air at an altitude of 27 700 m, the altitude at which the fastest speed was attained, can be calculated as 230 K. We will also assume that air is composed of 80% nitrogen (N2 ) and 20% oxygen (O2 ) and hence has a molar mass of 29 × 10−3 kg mol−1 . Using these values in equation (8), we can calculate that the speed of sound at an altitude of 90 000 ft (28 km) in the stratosphere is 304 m s−1 . This is slightly greater than the calculated maximum speed of 276 m s−1 using our simple model, or using Hoon’s model (275 m s−1 ), so we conclude that he probably did
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Physics of Colonel Kittinger’s leap not break the sound barrier on the way down. This value is actually rather sensitive to the temperature of the air at a particular altitude, and we do not have data for the local temperature of the air at the altitude at which maximum speed was attained. One interesting point to note is that a figure for the minimum temperature is often cited in accounts of the jump [5, 7] as −94 ◦ F (203 K). The minimum temperature in the ISA model occurs at altitudes between 10 and 20 km, and is 222 K. Thus there is a considerable discrepancy between the atmospheric model and the temperature recorded during the jump. If this temperature occurred when the maximum speed was reached, then the speed of sound would have been 285 m s−1 . The accounts do not say at what altitude the minimum temperature was reached, but the temperature profile in figure 2 suggests that the temperature minimum is at an altitude between 10 and 20 km high, which is somewhat lower than the altitude at which the maximum speed was reached. We conclude that it is unlikely, but not impossible, that Colonel Kittinger broke the sound barrier during the jump.

Acknowledgment
The authors would like to thank Dr Caroline Mitchell for a critical read through and helpful comments during the writing of this paper.
Received 4 February 2008 doi:10.1088/0031-9120/43/5/002

References
[1] Kittinger J W Jr 1960 The long, lonely leap Natl Geogr. 118 854–73 [2] Kittinger J W Jr 1960 The twenty mile fall Time Mag. www.time.com/time/magazine/article/0, 9171,939169,00.html (Accessed 10 December 2007) [3] Robinson A W 2008 Don’t just stand there, teach Fermi problems! Phys. Educ. 43 83–7 [4] www.youtube.com/watch?v=81gn2oLeC U (Accessed 10 December 2007) Several other versions of the same footage are also posted on youtube.com [5] The official USAF website www.af.mil/history/ person.asp?dec=&pid=123006518 (Accessed 11 December 2007) [6] www.elchineroconcepts.com/Joe%20Kittinger. htm (Accessed 11 December 2007) [7] National Museum of the USAF website www. nationalmuseum.af.mil/factsheets/factsheet. asp?id=1114 (Accessed 11 December 2007) [8] http://hypertextbook.com/facts/JianHuang.shtml (Accessed 11 December 2007) [9] International Organization for Standardization 1975 Standard Atmosphere ISO 2533, p 1975 [10] www.atmosculator.com/ The%20Standard%20Atmosphere.html (Accessed 11 December 2007) [11] http://en.wikipedia.org/wiki/ International Standard Atmosphere# note-NASA GSFC (Accessed 11 December 2007) [12] Wegener P P 1991 What Makes Airplanes Fly? (New York: Springer) p 57 [13] Serway R A and Jewett J W Jr 2008 Physics for Scientists and Engineers with Modern Physics 7th edn (California: Brooks-Cole) ISBN-10: 0495112453 [14] Smith J R 1997 Introduction to Geodesy (New York: Wiley) ISBN-10 0471 16660x [15] Hofmann-Wellenhof B and Moritz H 2005 Physical Geodesy (Vienna: Springer) p 84 ISBN-10: 3211235841 [16] www.aerospaceweb.org/question/aerodynamics/ q0243.shtml (Accessed 11 December 2007) [17] www.aerospaceweb.org/design/scripts/ atmosphere/ (Accessed 11 December 2007)
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Conclusions
What conclusions can we draw from this work? First, the discrepancies in the reported top speed should not detract from the remarkably courageous feat performed by Colonel Kittinger in 1960. Second, this is an excellent case study in checking the reliability of information sources, particularly from publications which have not been peer reviewed (the majority of postings on the internet). Third, it demonstrates that a very simple physical model can be used to determine the correct value for the highest speed attained during the parachute jump. Fourth, there are excellent resources available on the internet to provide interesting pedagogical material in this case study, including video footage and interviews with Colonel Kittinger [22], and a downloadable simulation game of the jump [20]. The case study also provides a lot of practice in unit conversions between SI and non-SI unit systems. Finally, it also allows us to explore the level of approximation needed to arrive at a satisfactory physical model which can answer the questions regarding the maximum speed attained.
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[18] US Standard Atmosphere 1976 (Washington, DC: US Government Printing Office) [19] http://nssdc.gsfc.nasa.gov/space/model/atmos/ us standard.html (Accessed 11 December 2007) [20] www.pbs.org/wgbh/nova/escape/skydive.html [21] Tipler P A and Mosca G 2008 Physics for Scientists and Engineers 6th edn (New York: Freeman) pp 499–618 ISBN-13 1-4292-0132-0 [22] The Planets. BBC Documentary Series DVD 2000 (London: BBC Worldwide) www.bbcshop. com/invt/bbcdvd1003 [23] Wallace J M and Hobbs P V 2006 Atmospheric Science: An Introductory Survey 2nd edn (San Diego, CA: Academic) p 11 ISBN-10 0-12-732951-X
Andrew Robinson, PhD, is an assistant professor in the Department of Physics and Engineering Physics, University of Saskatchewan, Saskatoon, Canada. His research interests are the use of new technology in teaching, self-assembled nanostructures on the surfaces of metals and medical imaging in the near-infrared region. He specializes in teaching introductory physics to first-year undergraduate students.

Charles Patrick is a fourth-year undergraduate student in the Civil Engineering programme at the University of Saskatchewan.

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