When I was a mathematician, I used to look down on… well, non-mathematicians in general, but especially astronomers. Astronomers in particular got in the neck because their margins of error were so wide – if two different models disagreed by a factor of 100 (it was said), they’d be counted as ‘roughly in agreement’.

In the end, I sorta-kinda became an astronomer – or at least, I was looking at a star, and ended up (to my shame) using this kind of hand-wavy justification.

It turns out, if you’re an astronomer, there’s a pretty good way of getting big sums (strictly, multiply and divide sums) more or less right, using a simplified version of standard form.

There are only two numbers in this version of standard form: one and many. ‘One’ covers numbers that start with 5, 6, 7, 8, 9 or 1; ‘many’ (or $M$) covers 2, 3 and 4.

With that in mind, 1 × 1 is 1; one × $M$ is $M$; and $M$ × $M$ is ‘ten’ – or rather, one with an increased power on the standard form.

So, for example, if I wanted to work out the volume of the earth, roughly, I’d say: the volume of sphere is $\frac{4}{3}\pi r^3$, and $r$ is about $6.4 \times 10^6$m.

That would become $\left( \frac{M}{M}\right) M (1 \times 10^7)^3$.

The big bracket would become $10^{21}$. I still have to multiply by $\left(\frac{M}{M}\right) M$, which is obviously $M$, leaving me with about $M \times 10^{21}$ m$^3$ as my final answer. I’d translate the $M$ back into 3 as a decent estimate.

That’s not bad for something so rough - the real answer is $1.1 \times 10^{21}$, so I’m off by only a factor of 3 – which is practically bang on for an astronomy question!

(If I wanted to do it more accurately, I’d figure out rough estimates of the inaccuracies I introduced – $\frac{4}{3}$ is 33% higher than my estimate, $\pi$ is 5% higher, and 6.4 is 36% lower. adding up those numbers, 33 + 5 - 36 - 36 -36 = 70% off… giving $9 \times 10^{20}$, which is only 10% or so off – really not bad at all!)