Falling Bodies [1]: Sky-diving

Aristotle was clear: heavy bodies fall faster than light ones. He arrived at this conclusion by pure reasoning, without experiment. Today we insist on a physical demonstration before such a conclusion is accepted. Galileo tested Aristotle’s theory: he dropped bodies of different weights simultaneously from the Leaning Tower of Pisa and found that, to a good approximation, they hit the ground at the same time.

Aristotle and Galileo.

A Thought Experiment

Galileo contrived a thought experiment, or Gedankenexperiment, to show that heavy and light weights must fall at the same rate. For suppose the case is otherwise. Then, if the two weights are linked by a string, the heavy one will tug the light one after it and the light one will retard the heavy one. But the two coupled weights can be regarded as a single one which, being heavier than either, should fall even faster. This contradiction leads us to conclude that the initial supposition is false.

But Galileo was not content with such reasoning: he sought to demonstrate his conclusion physically. His pupil Vincenzo Viviani wrote that in 1589 Galileo dropped balls of the same material but of different sizes from Pisa’s Leaning Tower to show “to the dismay of the philosophers” that they fell at the same speed, contrary to Aristotle. As no written record by the master himself is extant, spoil-sport historians cast doubt on this story. However, a recent authoritative biography by Heilbron (2010) points out that the tilt of the tower made it a perfect platform for the experiment.

Left: Galileo. Right: The Leaning Tower of Pisa

Air Resistance & Sky-diving

In fact, the drag of the air causes a small difference in fall-rate, so wooden and iron balls of the same size will hit the ground at slightly different times. Galileo was aware of this effect, which is minute compared to Aristotle’s idea of fall-rate proportional to weight. We all know that a hammer and a feather dropped together will not fall at the same rate. But if there is no air resistance, they will hit the ground together, as was demonstrated on the Moon by astronaut David Scott on the Apollo 15 mission.

A sky-diver plummets towards the earth, accelerated by gravity but impeded by air resistance. His terminal velocity ${V_{\rm T}}$ is that rate of fall such that the upward drag force due to air resistance exactly balances the downward pull of gravity (for simplicity, we neglect buoyancy).

At the terminal velocity, there is no nett force, so the sky-diver continues to fall at a constant speed. If he is moving slower than ${V_{\rm T}}$ , he accelerates. If he is moving faster, the increased drag slows him down. Thus, he approaches the terminal speed asymptotically. Typically, we may assume that he has reached his terminal value in about ten seconds.

The drag force depends on several factors, one being the cross-section area of the body. This is how a parachute works. A sky-diver faced down with arms stretched out can reach over 200 km/h. Speed sky-divers take a head-down position and reach much higher speeds. Felix Baumgartner jumping from 39 km, broke the speed record, reaching over 1340 km/h on 14 October 2012. This was at high altitude, where the air density is very small. In October 2014, Alan Eustace broke Baumgartner’s record when he jumped from a higher altitude of over 41 km.

Calculating the Terminal Velocity

We can write the equation of motion as

$\displaystyle m\frac{d V}{dt}= mg - \frac{1}{2}\rho C_\mathrm{D} A V^2$

where ${V}$ is the downward speed, ${C_\mathrm{D}}$ is the drag coefficient, ${\rho}$ the density, ${A}$ the cross-section area and other notation is conventional. Putting the acceleration (left hand side) to zero we immediately obtain an expression for the terminal velocity:

$\displaystyle V_\mathrm{T} =\sqrt{\frac{2mg}{\rho C_\mathrm{D} A}}$

Let’s evaluate ${V_\mathrm{T}}$ for typical parameter values for a sky-diver. The drag coefficient is approximately equal to unity, so we take ${C_\mathrm{D}=1}$ . Let $m=80 kg$ , $g=10 m/s^2$ , $A=1 m{^2}$ and $\rho=\rho_0\approx 1 kg/m^3$ the air density at the Earth’s surface. Then

$\displaystyle V_\mathrm{T} = \sqrt{\frac{2\times 80\times 10}{1\times 1\times 1}} = 40\,\mathrm{m/s}$

which is about 145 km/h.

Question: As he falls through a great height, will the sky-diver’s fall-speed increase, level out or reach a peak and then decline?

When the distance fallen is great, we must allow for the variation in density. This causes the air resistance to change with altitude. Air pressure and density decrease exponentially with height. If we ignore temperature variations, the air density varies as

$\displaystyle \rho = \rho_0 \exp (-z/H)$

where ${z}$ is the height, ${\rho_0}$ is the density at the Earth’s surface ${z=0}$ and ${H \approx 8\,}$ km is the scale-height. Putting this in the above equation we get

$\displaystyle V_\mathrm{T} = \sqrt{\frac{2mg}{\rho_0 C_\mathrm{D} A}}\exp(z/2H) = V_\mathrm{T}(0)\exp(z/2H)$

so the terminal speed varies exponentially with height. At ${z=12\,}$ km the exponential factor is ${\exp(3/4)\approx 2}$ . Thus, a sky-diver jumping from the top of the troposphere, around 12km, could reach a speed of almost 300km/h (belly down). If his parachute fails to open, he may be briefly relieved that his speed upon hitting the ground will be only half this value!

Sources

Heilbron, J L, 2010: Galileo. Oxford Univ. Press., 508pp.

Wikipedia articles: Free Fall and Terminal velocity [http://www.wikipedia.org/ ]

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Peter Lynch’s book about walking around the coastal counties of Ireland is now available as an ebook (at a very low price!). For more information and photographs go to http://www.ramblingroundireland.com/

ThatsMaths