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In mathematics there are approximations to pi such as:

$\pi\approx\dfrac{355}{113}$

that have extraordinary accuracy and are useful ways of remembering or being able to calculate a value close to pi that may be used in calculations.

I noticed the the ratio of the diameter of the moon to the earth seems unreasonably close to $\frac{3}{11}$.

Is my observation correct?

And are there other useful ways of remembering approximate metrics of various heavenly bodies with surprising economy?

David
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    1 year is $\simeq \pi \times 10^7$ seconds and see https://xkcd.com/1047/ for other examples of numerology. – ProfRob Feb 26 '24 at 12:51
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    Fun fact: No rational approximation to π is really useful, since the number of digits to be remembered will never be smaller than the number of decimals in π to which it is accurate. For instance, 355/113 involves six digits, but is only correct to 5th decimal in π. – pela Feb 26 '24 at 14:18
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    @pela is that always true? It seems to hold for the first few terms in the continued fraction, but I can't think of why it would hold in general. – ScienceSnake Feb 26 '24 at 15:31
  • @ScienceSnake I can't give you the proof, unfortunately, but I learned it as an undergrad. I think it's true not only for π, but for all irrational numbers, and I seem to recall the proof involving continued fractions and the Diophantine approximation. – pela Feb 26 '24 at 18:30
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    can't be true for all irrational numbers eg consider 1 + pi/10^9, the rational number 1/1 involves two digits, but is accurate to the 8th decimal place. @pela – James K Feb 26 '24 at 18:33
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    The error in a good rational approximation (eg from the continued fraction) $p/q$ is $<1/q^2$. See https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem 355/113 is ~3.14159292, so it gives you pi (~3.14159265) to 7 digits. – PM 2Ring Feb 26 '24 at 19:14
  • 500 light-seconds is about 1/500th longer than one astronomical unit. – notovny Feb 26 '24 at 19:47
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    You can "upgrade" pi calculations that use 22/7 by subtracting 0.04%, since (22/7)×(1-0.0004) = 3.1416 exactly. – PM 2Ring Feb 26 '24 at 20:03
  • @PM2Ring that reminds me of the maths class back at uni, when things like finite Taylor series had a remainder term at the end. Applying that concept here: "1.234 + x" is an accurate representation of pi up to the nth digit for a sufficient "x" – Apfelsaft Feb 26 '24 at 23:30
  • @JamesK Yes, you're right. Maybe "all irrational numbers that are not too irrational"? Well, good thing I'm not a mathematician :) – pela Feb 27 '24 at 11:21
  • @PM2Ring Did both you and I count wrong? I said "5", you said "7", but only the 6 digits "141592" are correct, right? – pela Feb 27 '24 at 11:24
  • @Pela I'm also counting the leading "3". To 7 digits (i.e., 6 decimal places), we have "3.141593" – PM 2Ring Feb 27 '24 at 11:34
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    There's very little use in having approximations of constants (physical or mathematical) to a 2-10 significant figures anyway. Either it's a back-of-an-envelope calculation in which 1 sigfig, or even just the order of magnitude, is enough. Or it's a full blown numerical computation in which case numerical accuracy or the highest experimentally measured value ought to be used. Outside of undergrad exams, the middle case rarely comes up. – ScienceSnake Feb 27 '24 at 14:31
  • @PM2Ring Ah okay. I was thinking "number of decimals". Anyway, in astronomy π = 3. Everything after that is just pernicketiness. – pela Feb 27 '24 at 15:23
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    @pela But 355/113 is unsually easy to remember due to the repeating digits. – Russell Borogove Feb 28 '24 at 06:56
  • @RussellBorogove I mix them up all the time, until I start thinking, but when I start thinking, I realize that "π = 3" is just fine… – pela Feb 28 '24 at 13:41
  • @pela: As an approximation to pi, 355/113 is in error by less than one part in 12 million. Equivalently using seven characters (or six digits), 3.14159 is in error by more than one part in 1.2 million: an error ten times the size. – John Bentin Feb 29 '24 at 09:55
  • @JohnBentin Right, but I wasn't counting the "3.", only decimals. Including one more decimal makes the difference of the same order. I guess my point was that there really isn't any reason to learn rational approximations by heart. If you'd ever want to learn a better approximation, you'd need a whole new fraction, whereas as learning decimals, you'd just need one more. – pela Feb 29 '24 at 13:17
  • @pela How accurately can you tell me the size of the moon without having to look it up? – David Mar 01 '24 at 05:09
  • @David To within a few tens of thousands of kilometer? – pela Mar 01 '24 at 10:01
  • @David Ah sorry, you said size, not distance Probably a bit more accurately then. – pela Mar 01 '24 at 10:02

3 Answers3

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The angular size of the Moon and the Sun, as viewed from the Earth is pretty much exactly $1/1$ with slight variations due to the eccentricity of the Moons orbit. Another coincidence is that the acceleration of 1 $g$ at the Earth's surface is almost exactly equal to $1 c$$/$year where $c$ is the speed of light. In terms of masses I find the following chain of ratios useful to get a grip on the size of things: $80$ Moons in the Earth, $300$ Earths in Jupiter, and $1000$ Jupiters in the Sun. And, while not uniquely astronomical, I've remembered since high school that there are $\pi\times10^7$ seconds in a year (to within half a percent error!).

It's also worth bearing in mind that a lot of astronomical units are ratios to begin with (technically, all units are, but typically not with respect to a relevant baseline). Thus any time something is given in AUs or years it's a ratio to the radius and period of the Earth's orbit around the Sun.

ScienceSnake
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    FWIW, 1 g = 9.80665 m/s2 ~= 1.0323 ly/y^2, using Julian years of 365.25 days. – PM 2Ring Feb 26 '24 at 19:41
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    I learned that last example as “π seconds in a nanocentury”. – gidds Feb 27 '24 at 12:38
  • @gidds I mean, that works, but nanocentury definitely feels like a cursed unit though. Like kilo Watt hours or km per megaparsec per second. – ScienceSnake Feb 27 '24 at 14:27
  • @PM2Ring Rocket engine specific impulse is often represented in seconds (as a somewhat misleading shorthand for "pound-force-seconds-per-pound-mass"), which differs by a factor of g from the other commonly used unit, meters per second (really Newton-seconds per kilogram, but it's the same dimension). For back-of-the-envelope conversions, g=10 is a good approximation, and likewise 10N thrust against 1 kg of rocket gives you about 1 g of acceleration. – Russell Borogove Feb 28 '24 at 07:03
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1 km/s is roughly 1 pc per million years. This is roughly a consequence of there being $\pi \times 10^7$ seconds in a year and roughly $\pi$ light years in a parsec. Useful.

ProfRob
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Since the Sun is 8 light minutes away, and the speed of light is a foot per nanosecond, the diameter of the Earth’s orbit is a trillion feet.

Martin Kochanski
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