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I was having one of those obnoxious conversations with a friend where he was arguing that a fat skydiver would reach the ground faster than a skinny skydiver. To me it seemed as obvious that the world is round (that is: not very obvious unless you have some education).

We weren't getting anywhere so I punched in the data in Wolfram Alpha and I was surprised that indeed a fat person falls faster. I don't quite grasp why a heavier person with the same surface area has less slowdown factor. Can anybody explain this to me?

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There's a difference between fat skydiver & heavy skydiver.. – Evil Angel Apr 4 '14 at 19:44
up vote 19 down vote accepted

Ah, this gives me a chance to give a proper home to an analysis I first posted on Reddit. (I would much rather have first posted it here :-P)

Mathematical derivation

It all starts with a blog post I've written that comes very close to addressing the exact question you're asking. In the post, I calculated how fast an object would be moving after falling a given distance, assuming quadratic drag. But one of the formulas I used to get to that result is the time it takes for an object to fall a certain distance.

Here's the argument from my post. If you write out Newton's second law for an object falling through the air, you get

$$\frac{1}{2}CA\rho \dot{y}^2 - mg = m\ddot{y}$$

i.e. drag force minus gravitational force equals mass times acceleration. In this equation, $m$ is the object's mass, $A$ is the cross-sectional area it presents, $C$ is the object's drag coefficient, $\rho$ is the density of the fluid it's falling through, $g$ is the acceleration of gravity, and $y$ is its height at any given moment. Solving this equation for $\dot{y}$ gives

$$\dot{y} = -\sqrt{\frac{2mg}{CA\rho}}\tanh\biggl(\sqrt{\frac{gCA\rho}{2m}}t\biggr)$$

You can then integrate this with respect to time and solve it for $t$ to get

$$t = \sqrt{\frac{2m}{gCA\rho}}\cosh^{-1}\exp\biggl[\frac{CA\rho}{2m}(h - y)\biggr]$$

A couple more steps are shown in my blog post, but they're not really important. The point is that this formula gives the time $t$ it takes for an object to fall a distance $h - y$.

You'll notice that the properties of the falling object occur in this formula only as part of the particular combination $\frac{CA\rho}{2m}$. So the behavior of a falling object can be entirely characterized by that ratio. If you call this ratio $r$, then the formula becomes

$$t = \sqrt{\frac{1}{rg}}\cosh^{-1}\exp[r(h-y)]$$

For a few sample values of $h - y$, here's what this looks like as a function of $r$:

Plot of time taken to fall a given distance as a function of r

You'll notice that the time to fall a given distance always increases with increasing values of $r$. So the larger an object's value of $\frac{CA\rho}{2m}$, the longer it takes to fall. Conversely, an object with a smaller ratio of cross-sectional area to mass (i.e. smaller $r$, assuming the same shape) will fall faster.

Now, roughly speaking, a fat person tends to be larger than a skinny person in all three dimensions. So their mass will roughly be bigger by a factor of $k^3$ for some $k$, whereas their cross-sectional area will only be bigger by $k^2$. (This is a huge approximation, of course, but it should still work for the question of "faster" vs. "slower.") Accordingly, $r$ for a fat person will be smaller (by a factor of $k$), which means they take less time to fall.

Physical interpretation

That's all well and good, but just going through the math doesn't necessarily make it clear why (physically) fat people fall faster. The crux of the explanation is in that last paragraph: a fat person has a larger mass in proportion to their surface area. Since the drag force is proportional to area, but weight is proportional to mass, as a person gets fatter, the weight (going down) increases more than the drag force (going up), which means the person accelerates more.

More math: slowdown factor

Now what about this "slowdown factor" that Wolfram Alpha is coming up with? If you look down toward the bottom of the results, it tells you that the slowdown factor is just the ratio of the time actually taken to fall, which I showed you how to calculate above, to the time that would be taken without air resistance. You can get the latter time by setting $C$, $A$, or $\rho$ to zero, or taking the limit as $m\to\infty$. (Does it make sense why all of these assignments correspond to making the effect of the air insignificant?) Or, of course, you could just as well take $r\to 0$. Now, before you start wondering how you're going to get away taking the square root of $\frac{1}{0}$, you actually need to take a limit, and the limit of $t$ as $r\to 0$ is well defined:

$$t_0 = \lim_{r\to 0}\sqrt{\frac{1}{rg}}\cosh^{-1}\exp[r(h-y)] = \sqrt{\frac{2(h - y)}{g}}$$

The formula for the slowdown factor is then

$$\text{slowdown factor} = \frac{t}{t_0} = \sqrt{\frac{1}{2r(h - y)}}\cosh^{-1}\exp[r(h-y)]$$

This depends only on the product of the "drag ratio" $r$ and the height fallen $h - y$. Essentially it's a way of characterizing how much the presence of air resistance affects the time of flight.

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The drag force is not strongly dependent on the mass of the falling object.

So if we take two objects with mass $m$ and $M$ where $M > m$, and we assume that $F_{m,\,drag} \approx F_{M, \, drag}$ then we see that the acceleration on $M$ is less than the acceleration on $m$.

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Assuming the same surface area is probably incorrect, but even so the area/volume ratio suggests that fat guy will come out a little ahead.

It's just that the area goes by roughly $(\mathrm{volume})^{\frac{2}{3}} \propto (\mathrm{mass})^{\frac{2}{3}}$ (with an assumption of roughly equal density) while but the downward force is proportional to the mass, so the terminal velocity can be expected to be a little higher.

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Both cases have "projected surface area 1.28 m^2" so isn't Wolfram Alpha assuming same volume? – RedGrittyBrick Oct 13 '12 at 11:54

In the simplest situation, the answer is very straightforward. For the sake of simplicity let's just consider two equal size spheres, one made of iron and the other made of plastic. In the presence of air friction, when releasing the two spheres from rest at a given height, the iron sphere will touch the ground first. There are two reasons for that : 1) the acceleration of the iron sphere is greater than the acceleration of the plastic sphere. 2) the terminal speed of the iron sphere is greater than the terminal speed of the plastic sphere.

To understand the last statements, just draw a force diagram, with gravitational force pulling downwards and air friction pulling upwards and apply Newton's second law. Because the strength of air friction does not depend on the mass of the object falling, it happens that acceleration and terminal speed depend on the mass of the object. This is not true of course on the Moon where there is no air friction.

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Ignoring fat bodies, and treating this as different size spheres with same density:

Using the empirical equation for drag force, the larger the sphere, the greater the

terminal velocity.

Also, at any given velocity, the larger sphere will have a greater acceleration rate.

So : the larger sphere has greater acceleration and terminal velocity, so gets ahead and stays ahead.

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