π! I can rattle off 36 significant figures on demand, but how accurate do you really need to be? The JPL uses 15 digits for interplanetary navigation. What is needed, though?

Let’s start with a 1 m diameter circle, and add successive digits to the approximation. How many digits before you’re being absurd? I’ll give a definition of absurd, too: a difference in the circumference of less than the Planck length, approximately \(\LARGE{1.616229(38)\times10^{-35} {\rm m}}\)

- For 1 m, the answer is fairly obvious: 36 significant figures
- For d
_{⊕}, 43 significant figures are needed - For the distance from the sun to Voyager 1, ~141 AU at time of writing, just 49 significant figures
- For the Solar System, out to the distance where we believe the Oort cloud ends, ~100000 AU (i.e., the edge of the solar system), 53 significant figures
- For the size of our observable universe, roughly 93000000000 light years, 63 is enough

What if the Bohr radius (approximate size of a hydrogen atom) was our standard, at \(\LARGE{5.2917721067(12)\times10^{-11}{\rm m}}\)? For the size of the observable universe, we only need 36 figures. Handily, that’s what I know! How much would one need for a circle the size of the orbit of Neptune? 23 digits.

So how accurate *is* JPL if they were measuring the circumference of a circle at the orbit of Neptune? They could be off by as much as a centimeter! Which, assuming appropriate feedback systems, isn’t going to be anything near significant.

How many do we know? 22459157718361 digits. Why? As George Leigh-Mallory put it, “Because it’s there”.