Informed Guessing!

Dimensional Analysis, Escape Speed, and the Size of Atoms...

Physicists (and astronomers!) are lazy. Super lazy. Sometimes this surprises people who might imagine all scientists as being hyper-persnickety about the 9th decimal place in our answers — maybe that view comes from typical middle- and high-school science classes where lots of effort (misplaced, in my view) is expended on things like significant figures. But really, when not producing experimental results in a research paper, many of us are pretty stoked when we get a rough understanding of what’s going on and how some physical system depends on changing some parameter or other.

And in the spirit of gaining a rough understanding of how things work, it turns out that there’s a fairly simple game we can play called dimensional analysis — literally just guessing at what might be involved and then playing around with the basic units to match them up. “Showing” is more fun than “explaining”, so here are some examples of how to play the game:

---

Tension in Circular Motion

Swing a heavy thing around you, and it will be obvious that you need to exert some force (tension) to keep the motion circular (imagine just letting it go — in the absence of such a “centripetal force” the circular motion stops and it just flies off in a straight line).

The way we play is to write down the units of force and guess at all the things that the force might reasonably depend on. If you’re whirling a rock around on a string, for example, you can imagine that the tension in the string probably depends on the size of the circle (the length r of the string) as well as how fast (v) the rock is moving — faster motion would require more force. Also, you probably figure that there would be more tension in the string if the rock (mass m) were heavier. So it sounds reasonable that the force F depends somehow on m, v, and r. “Somehow”, that is, in that we don’t really know what the dependence is. For example, is the mass squared? Maybe it’s the square root of speed v?

¯\_(ツ)_/¯

We can find out! Replace those variables with their units: Force has units of “Newtons”, which we’d write in the base form involving kilograms, meters, and seconds. Then on the right-hand-side, raise each of our variables to some unknown power:

\(\begin{align} F_T = & m^i v^j r^k\\ \left[\frac{\text{kg m}}{\text{s}^2}\right] = & \left[\text{kg}^i\right] \left[\frac{\text{m}}{\text{s}}\right]^j \left[\text{m}\right]^k \end{align}\)

Now, the units on both sides of the equation have to match, so that’s how we’ll figure it out. We don’t even have to break out the big algebra guns! Since kg only appears to the first power on the left-hand side (LHS) and only once on the right-hand side (RHS), i has to be 1. How about seconds? It’s squared in the denominator on the LHS, and only appears in one term on the RHS, so that tells us that j=2. Ok, now for the meters unit: we already know j=2, but on the LHS it should only appear to the first power. That means that k has to be -1 (exponents just add when the terms are multiplied).

So if it’s true that the tension only depends on m, v, and r, then the expression looks like

\(F_T = m^1v^2r^{-1}=\frac{mv^2}{r}\)

Which is the same thing that any beginning physics textbook will tell you, but probably only after a far more complex derivation. A physicist might look at that equation and say “Cool, so if we double the length of the string r (if the mass travels at the same speed) then there’s less tension. If you keep the same length of string but want the mass to circle twice as fast, then there’s 4 times as much tension in the string!” You can gain a lot of physical insight into a system just by looking at the equations and playing around with increasing or decreasing some of the variables.

---

Escape Speed!

If you throw something, it’ll fall down to the ground. If you throw it faster, it’ll cover a longer distance before hitting the ground.

If you keep running this in your head with faster and faster speeds, it gets interesting because it falls to the ground all right, but at some point the ground starts falling away from it too (because the Earth is spherical)! There’s a special speed where these things balance out — the object keeps falling to the ground, but the ground is curving away at exactly the same rate, so it stays the same distance above the ground all the way around — that’s an orbit! (C in the diagram)

A cannon on top of a very high mountain shoots a cannonball horizontally. If the speed is low, the cannonball quickly falls back to Earth (A, B). At intermediate speeds, it will revolve around Earth along an elliptical orbit (C, D). Beyond the escape velocity, it will leave the Earth without returning (E). (Wiki Commons)

Ok, but keep going faster: there’s another special speed when the object is no longer bound by gravity and can escape, never to return. This is called, for obvious reasons, the escape speed (E, in the diagram).

Deriving the escape speed equation is a standard exercise in Physics 1, but let’s see how we might guess at its form just by using dimensional analysis:

We’re looking for a speed, so we need to wind up with units of m/s. What might it depend on? Well, in any gravitational interaction, Newton’s gravitational constant G always pops up, so that’s probably in there (it has units, it’s not just a plain number). It seems reasonable that it’s harder to escape from a more massive planet, so M is probably in there too. And the size of the planet is likely important too — Jupiter is pretty massive, but also large (not very dense), so escaping from a gas giant’s “surface” is easier than the equivalent-mass rocky planet. Ok, same game: we have to multiply G, M, and R (raised to some powers) in just such a way to get units of m/s at the end:

\(\begin{align} v = & G^i M^j R^k\\ \left[\frac{\text{m}}{\text{s}}\right] = & \left[\frac{\text{m}^3}{\text{kg s}^2}\right]^i \left[\text{kg}\right]^j \left[\text{m}\right]^k \end{align}\)

Let’s look for the easy thing first: units of seconds only appears once on the LHS, but it’s squared in the denominator. That tells us right away that i=1/2.

Cool — now notice that units of kg don’t appear at all on the LHS, so they need to cancel on the RHS. That means that j has to be 1/2 as well! Almost done! Now look at the meters unit: we already know that i is 1/2, so that term has meters to the 3/2 power. If we want the power to be 1, to match the LHS, then k has to be -1/2. Done! Arranging G, M, and R in just this way looks like

\(v = G^{\frac{1}{2}}M^{\frac{1}{2}}R^{-\frac{1}{2}}=\sqrt{\frac{GM}{R}}\)

This is a pretty instructive example — if you look up the values for G, mass of the Earth M and radius of the Earth R and plug them in, you get an escape speed of about 7.9 km/s (~17,000 mph). That’s actually not right! Not exactly, anyway. That standard derivation in Physics 1 will tell you that the correct expression is

\(v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \approx 11\;\text{km/s}\)

That’s the price we pay for such a comparatively quick and easy way to generate physical formulae — we lose unitless numbers like π, √2, and so on. The cool thing is that, often, those unitless numbers are just between 1 and 10, so we won’t be crazy wrong. And really, physicists are mostly interested in the relationships between the variables. It’s not as important for us to get an exact answer using dimensional analysis as it is to understand that, say, if the mass quadruples then the escape speed only doubles.

---

The Size of Atoms!

Now for something really nifty! Can we, just guessing at the expression, find out how big atoms are? Yes indeed! Here’s how we might do it:

What are we after? Size! So whatever we put into the overall expression, we need to wind up with units of meters. Now the hard part, trying to decide what might be included. Well, what might determine the size of, say, a hydrogen atom? It’s the simplest one, just a proton and electron, where the force of attraction is the electric force (Coulomb’s Law). Here’s what that looks like:

\(F_e = \frac{ke^2}{r^2}\)

Now, the force has units of Newtons, k is the constant we use in electric interactions (just a number with weird units, sort of like G for gravitational stuff), e is the amount of charge of the proton or electron in units of Coulombs, and r is the distance between the two. So we might expect k and e to be involved, and we’re looking for r.

The electron is whirling around in a circle, and as we saw above, the mass of the circling thing should probably be there too, so we’ll throw in m.

It turns out that many important interactions involve quantum mechanics when you work with the physics of atoms and particles, so that’s probably important here too. I might not have to work hard to convince you that invoking quantum mechanics sort of levels up our discussion quite a bit, but it seems like most quantum mechanical formulae include a special constant ℏ (Planck’s Constant), which has units of Joule-seconds. We’ll convert that into base units and throw that in too.

It’s starting to sound like soup.

Here’s what we’ve got:

\(\begin{align} r = & e^i k^j m^k \hbar^l\\ \left[\text{m}\right] = & \left[\text{C}\right]^i \left[\frac{\text{kg m}^3}{\text{s}^2 \text{C}^2}\right]^j \left[\text{kg}\right]^k \left[\frac{\text{kg m}^2}{\text{s}}\right]^l \end{align}\)

What a mess. Ok, let’s see what we can figure out! The LHS has no units of Coulombs, so they’ve got to cancel out on the RHS. That has to mean that whatever j is, i has to be twice as much: i=2j

Now look at the seconds unit — it also has to cancel out, so whatever j is, l has to be twice as much, and negative: l=-2j

Kilograms! There aren’t any on the LHS, so for sure j+k+l=0

And finally for meters: there’s a power of 1 on the LHS, so 3j+2l=1

But we said that l=-2j, so the last equation is 3j-4j=1, or j=-1

That’s the key! Then i=-2, l=2, k=-1 from our other equations. Whew! So apparently

\( r = e^i k^j m^k \hbar^l=e^{-2}k^{-1}m^{-1}\hbar^2 = \frac{\hbar^2}{mke^2} \)

And the most amazing thing of the month is: it’s bang-on right! We’re not even off by some small factor! When you go through the (far!) more lengthy quantum-mechanical calculation you get the same expression. Ok, we had to fuss with all those unknown powers, but that’s far easier to do by comparison.

If you plug in those constants, you get that the size of a hydrogen atom is about 0.053 nanometers (billionths of a meter) — this is a famous number called the Bohr radius. Even way more complicated atoms won’t have a size bigger than about 10 times this, so it’s a good stand-in for an average atomic size. And we got it by guessing! Now, granted, I sort of already knew what to guess, and it’s certainly possible to lead yourself far astray if your guesses about what’s in the final expression aren’t close. But because the process is relatively simple, this kind of dimensional analysis is a crazy useful tool in many theorists’ toolboxes.

---

On Deck:

For next time: I’ve been taking some pictures of Venus over the last few months. When I teach beginning astronomy, I love it when we get to the topic of competing models of the Universe at the time of Galileo, and how his observations of the changing phases of Venus forever fractured the then-1,500-year accepted model. I’ll revisit those old ideas with these new images, and try to use them to show the actual scale of the Solar System!

---

If you’re a student/teacher and want to see lots of worked examples/derivations that I like to include in my classes when I teach the “standard” University Physics 1 and 2 courses, feel free to browse the (growing) collection of 150+ videos at

Physics By Example

---

This publication is free to all; if you’d like to support it, though, or if something is especially cool and you’re inclined to leave a “tip”, I don’t consider myself above taking coffee or pizza donations (the very fuel of ideas!): In fact, following Douglas Adams, I’m not above accepting coffee or pizza in the same way that the sea is not above the clouds.

Buy Me a Coffee

---

Thanks for reading First Excited State! Subscribe for free to receive new posts automatically!