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Well first of all it is an adulterated pi^2 so the odds of getting something close are substantially higher.

Second, we'd be having exactly the same conversation if it happened to be g = 2pi^2, or 4pi^2, or any other reasonably artificial number.

Third, if you do the math, the mass of the earth and gravitational constant do conspire.

g = G * M_earth / R_earth^2

M_earth is approximately (4/3) * pi * R_earth^3 * Density_earth

G is approximately (2/3) * 10^-10 m^3.kg^-1.s^-2

We can eliminate our human units and rewrite it as

G = 2/3 * 10^8 / ( Density_water * Earth_orbital_period^2)

Put this all together and you get

g = 8/9 * pi * 10^8 * (Density_earth / Density_water) * ( R_earth / Earth_orbital_period^2 )

the ratio of the densities of earth and water is a dimensionless number that is independent of our units of measurement, and is approximately 5.5. With a little rearrangement we get

g = 88/9 * (pi/2) * 10^8 * R_earth / Earth_orbital_period^2

That 88/9 happens to be equal to pi^2 to within 1% error. This comes purely from nature.

1 m/s^2 is defined to be (pi/2) * 10^8 * R_earth / Earth_orbital_period^2 and thus we get the nice and neat g = pi^2 in metric units, but getting (pi^3)/2 * 10^8 in natural units is just as remarkable.




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