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There is a common conception that physics is a business of writing and solving exact equations.  This idea is not untrue, in the sense that physicists generally prefer to produce exact solutions when they can.  But precise equations can be slow: they are often cumbersome to work with and can obscure important concepts with a tedium of error-checking and term-collecting.  For these reasons, physicists often figure things out (at least in the initial stages of problem solving) using a kind of semiquantitative reasoning that doesn’t make use of exact equalities.

In this kind of reasoning, all (or most) equations are downgraded from having an equals sign, $A = B$, which means “A is equal to B”, to having a “squiggle” sign, $A \sim B$, which means “A is equal to B up to some numeric factor that I don’t particularly care about.”

This may seem kind of dumb to you. Why reason with squiggles when you can write exact equations instead?  But the truth is that “squiggle reasoning” often allows you to figure things out much more quickly and easily than you would ever be able to if you insisted on writing only exact equations.  And as long as you are willing to live with some ignorance about exact numerical values, you sacrifice very little in terms of conceptual clarity.

As it happens, I designed and taught a short course last year for high school students that introduces basic ideas in quantum mechanics using squiggle reasoning. (I am teaching the course again this year.)  As an introduction, I gave the students the following problem:

If a bunch of animals of different sizes all jump out of an airplane together, how fast do they each fall? In this post I’ll take you through the answer to this problem, which can perhaps serve as a gentle introduction to quantitative reasoning in situations where you don’t know how to (or don’t want to) write down exact equations.

### Gravitational Force

The starting point in solving this problem is to forget that animals have particular shapes.  That is, simplify the geometry of a given animal down to a single number: its “size” $L$.  Now, obviously for any real animal you will get a different number for the “size” depending on which direction you choose for the measurement.  For example, I personally am something like 1.8 meters tall, 0.6 meters wide, and 0.3 meters thick.  But if you just want a number that is in the right ballpark, it is fair to say that I am ~1 meter in size, as opposed to 1 centimeter or 1 kilometer.

To connect to an old trope, this kind of thinking isn’t really “assuming a cow is a sphere” so much as it is “not caring about the difference between a cow and a sphere”. Now you can ask: what is the force of gravity acting on an animal of size $L$?  Well, the force of gravity is $F_g = m g = \rho V g$, where $g \approx 10 \textrm{ m/s}^2$ is the acceleration due to gravity, $m = \rho V$ is the animal’s mass, $\rho$ is the density of the animal, and $V$ is its volume.

Since we have decided to forget about all specifics of the animal’s shape, making an estimate for the animal’s volume is actually very easy: $V \sim L^3$.

In fact, in squiggle reasoning, every three-dimensional shape has volume $\sim (\textrm{size})^3$, unless you have decided to look at some shape that is especially long and skinny.  This means that we can easily write an approximate equation for the force of gravity acting on the animal: $F_g \sim \rho g L^3$.

### Drag force

Immediately after jumping out of the airplane, the $L$-sized animal in question is in freefall, and accelerates downward at a rate $\sim g$.  However, after falling for a little while its acceleration is halted by the force of all the air rushing back against it.  The animal will eventually reach a steady downward velocity determined by the two forces being in balance: So how big is the drag force $F_D$?

Of course, the exact answer to this question depends on the shape of the animal.  If you really wanted to know, with numeric accuracy, the value of the drag force, then you would need to understand the air flow pattern around the animal.  This would presumably require you to stick the animal in a wind tunnel and make careful measurements. (And you would get different answers depending on which way the animal was facing).

But at the level of squiggle reasoning, we can figure out the drag force using a simple thought exercise.  Imagine the process of throwing a big block of air at the animal: This block is taken to have the same cross-sectional size as the animal (area $L^2$) , and some length $w$.  The mass of the air block is therefore something like $m_\textrm{air} \sim \rho_\textrm{air} L^2 w$.  If the block is thrown with a speed $v$, then it has a kinetic energy $KE \sim m_\textrm{air} v^2 \sim \rho_\textrm{air} v^2 L^2 w$.  (I’m sure you learned that first equation as $KE = \frac{1}{2} mv^2$, but when you’re doing squiggle reasoning there’s no reason to fuss about $\frac{1}{2}$’s.)

In order to stop the block of air, the animal applies a force that does work on the block equal to $KE$.  The work is equal to the drag force of the air multiplied by the distance over which the force is applied.  That distance is $\sim w$; you can think that the force is applied continuously as the air block smooshes into the animal’s side.  Thus, we have $F_D w \sim KE$, and therefore $F_D \sim \rho_\textrm{air} v^2 L^2$.

Of course, when the animal is falling through the air, this drag force is applied continuously, as the animal finds itself continuously colliding with “blocks of air” that move toward it with speed $v$.

### Final Answer: never skydive in the rain

Now we are ready to get an answer: equating $F_g$ with $F_D$ and solving for $v$ gives us $v \sim \sqrt{\rho g L / \rho_\textrm{air}}$.

Thus we arrive very quickly at an important semi-quantitative conclusion: larger animals fall faster, with a terminal velocity that grows as the square root of the animal’s size.

In fact, you can use this equation to get a pretty good order-of-magnitude estimate for the terminal velocity $v$, using the fact that pretty much all animals have the same density as water, $\rho \sim 1 \textrm{ g/cm}^3$, while air is about 1000 times less dense.

In particular, the squiggle equation for $v$ suggests that a meter-sized human has a terminal velocity on the order of $\sim 100 \textrm{ m/s}$.  (For reference, one m/s is about 2 mph — within the accuracy of our squiggle reasoning you can take a meter-per-second and mile-per-hour to be roughly the same thing.)  A centimeter-sized cockroach has a terminal velocity of $\sim 10 \textrm{ m/s}$, and a 10-meter-sized whale falls at about $300 \textrm{ m/s}$; three times faster than you do.

Thus, you can see pretty quickly why falling off a building is deadly for you (hitting the ground at ~ 100 mph is worse than just about any car accident) but not deadly for insects (hitting the ground at a couple mph is no big deal).

In fact, there is a pretty practical implication of this result (besides “don’t fall off a building”): You should never go skydiving in the rain.  You might think (as I initially did) that it would be a sort of magical and pleasant experience, wherein you fall together with the raindrops like an astronaut playing with zero-gravity water droplets.  But the truth is much more unpleasant: the meter-sized you will be falling at ~1o0 mph, while the millimeter-sized raindrops fall at a slow ~3 mph.  So, from your perspective, you’ll be getting stabbed by raindrops that blast you in the face at ~97 mph.

Highly unpleasant, and just a small amount of squiggle reasoning before you jump can save you the trouble.

12 Comments leave one →
1. October 20, 2016 10:03 pm

The rain thing never occurred to me. That would be painful!

2. October 21, 2016 12:14 am

Excellent! (I must say, parenthetically, that the ‘delta T’ between your delightful posts did feel like free-falling, bored, and wishing for something inspiring to read during the long-leap toward oblivion, something that, though I won’t survive, at least restoreth my faith in well-grounded and clear-thinking younger-and-luckier-than-I physicists.)

I was waiting to read: ‘to a first approximation’, or ‘first-order’ solution’. as it relates to squiggles..
Here in Israel we have soft-ball-sized porcupines abundant, who, when frightened, (understandably, upon being ejected from business-class in a doomed 747 at 30 thousand feet by sudden decompression, ) retract even their life-saving spines into a resultant ‘sphere’. And so, counting on you guys, I’ll prepare myself to land in a pile of ’em. Oops, no, they’ll ‘rain’ on me as I belatedly write my Will. (With due historical respect for the Leaning Tower of Pizza ‘lessons’)
Seriously, bravo on your post, and for your contribution to the the next generation of thinkers. May they have the tools to vote for politicians with at least a grade-school respect for science. Amen

• Brian permalink*
October 21, 2016 9:35 am

Ha! Thanks. Actually, this post was written largely in order to set up the next one. So you should have another post to read in the very near future.

3. Noel permalink
October 21, 2016 1:22 pm

Nice post, but I was hoping you’d go a little deeper. I have always been puzzled by how molecules of different molecular weight and electrostatic properties are air born and others grounded. Why chlorofluorocarbons rise up to the ozone layer while alkali metals that are light are grounded? Perhaps the same principles of the macro can be applied to micro world, but I suspect there are other factors to be considered. I look forward to reading about it.

• Brian permalink*
October 21, 2016 1:43 pm

I’m sure you’re right that there are some electrostatic/chemical properties at work here. I’m also sure that I don’t know enough to explain them!

• Noel permalink
October 21, 2016 2:35 pm

I have read you enough to disagree with you that you do not know enough. I am sure you would have great insights if you are interested and do a little research. It is easy to calculate the effect of earth’s electric field (~150 V/m) on an electron, but to calculate this for a noble gas, a molecule, or even a micro organism is another story. The implication of how much this electrostatic component contributes to the earth’s gravity or levity is beyond my skill level, and I have gone as far as to question the number of significant digits used for some fundamental constants, and I have done classical simulations. I hope I gave you an idea for a post.

• Anonymous permalink
December 19, 2016 11:04 pm

Brian, please read this article. I would like to know if the ‘squiggle reasoning’ applies to free falling atoms as well. http://physicsworld.com/cws/article/news/2016/dec/13/quantum-free-fall-at-8500-m
I have been considering the design of an experiment to test this idea that individual atoms, and small molecules, do fall with different accelerations, and evidently even negative acceleration (unless I am missing something in my equations or reasoning, and that is why i need someone like you to collaborate). I am open to collaborate with someone, and share credit, to test this idea. I hope you are interested. Please email me. Thank you

• Noel permalink
December 19, 2016 11:07 pm

I forgat to add my name. The previous anonymous post is from the same Noel as this post and that posted above.

4. October 24, 2016 9:20 am

I’ve never jumped with any animals (let alone in the rain, which isn’t allowed, anyway), but the 100 MPH figure is accurate for humans. As you say, depends on orientation. Falling head-down, giving the smallest cross-section possible, allows speeds of up to 200 MPH. Falling as spread-out as possible (“grabbing the air”) can slow one down to the 80-90 MPH range, depending on the person’s size and clothing, but the rule of thumb is that free-fall terminal velocity is roughly 120 MPH.

Which is a blast. Not often a human can go that fast all by their self! 🙂

5. October 25, 2016 9:36 am

Great stuff. Is this course available somewhere?

• Brian permalink*
October 25, 2016 9:43 am

No, it isn’t; sorry. Maybe someday I’ll put my notes together into a nice form and publish them online.

• November 4, 2016 2:58 pm

Of course you can always go through the archives here. Every post is a treat, pashut kef, ad c’dei smarmoret.