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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?

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    $\begingroup$ 1 year is $\simeq \pi \times 10^7$ seconds and see xkcd.com/1047 for other examples of numerology. $\endgroup$
    – ProfRob
    Commented Feb 26 at 12:51
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    $\begingroup$ 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. $\endgroup$
    – pela
    Commented Feb 26 at 14:18
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    $\begingroup$ @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. $\endgroup$
    – ScienceSnake
    Commented Feb 26 at 15:31
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    $\begingroup$ 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 $\endgroup$
    – James K
    Commented Feb 26 at 18:33
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    $\begingroup$ The error in a good rational approximation (eg from the continued fraction) $p/q$ is $<1/q^2$. See en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem 355/113 is ~3.14159292, so it gives you pi (~3.14159265) to 7 digits. $\endgroup$
    – PM 2Ring
    Commented Feb 26 at 19:14

3 Answers 3

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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.

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    $\begingroup$ FWIW, 1 g = 9.80665 m/s2 ~= 1.0323 ly/y^2, using Julian years of 365.25 days. $\endgroup$
    – PM 2Ring
    Commented Feb 26 at 19:41
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    $\begingroup$ I learned that last example as “π seconds in a nanocentury”. $\endgroup$
    – gidds
    Commented Feb 27 at 12:38
  • $\begingroup$ @gidds I mean, that works, but nanocentury definitely feels like a cursed unit though. Like kilo Watt hours or km per megaparsec per second. $\endgroup$
    – ScienceSnake
    Commented Feb 27 at 14:27
  • $\begingroup$ @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. $\endgroup$
    – Russell Borogove
    Commented Feb 28 at 7:03
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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.

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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.

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    $\begingroup$ This is only useful to Americans though… $\endgroup$
    – pela
    Commented Feb 27 at 11:29
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    $\begingroup$ Diameter of earth's orbit is ~1000 light seconds. $\endgroup$
    – David
    Commented Feb 27 at 13:03
  • $\begingroup$ Similarly, the diameter of the earth's orbit is $3 \times 10^{11}$ meters if you work in metric - the diameter of the orbit is $10^3$ seconds and the speed of light is $3 \times 10^8$ m/s. $\endgroup$
    – Michael Lugo
    Commented Feb 27 at 14:34
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    $\begingroup$ @pela Geocentric => Heliocentric => Americocentric $\endgroup$
    – Barmar
    Commented Feb 27 at 16:30
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    $\begingroup$ The americans should just change their basic length unit to nanolightsecond. This would please the science nerds, and wouldn't change much for people accustomed to feet. $\endgroup$
    – Paŭlo Ebermann
    Commented Feb 28 at 1:18

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