Why is "$\pi^2= g$" where $g$ is the gravitational constant?

Some months ago a professor of mine showed us a 'proof' of why $g\approx 9.8 ~\text{m}/\text{s}^2$ (the gravitational acceleration at the surface of the Earth) is 'equal' to $\pi^2\approx9.86\dots$ Using a differential equation that I think is used to model the movement of a pendulum of something like that.

Does anyone know the DE I'm talking about? Or, has anyone heard such story?

• That seems unlikely since $g$ depends on the planet you are and on the units you chose… Feb 10, 2016 at 17:24
• I also noticed that approximation years ago, and surely neither of us is the first one. And as this depends on where you are in the space-time continuum, it seems like a coincidence rather than some specific property of gravity. Feb 10, 2016 at 17:25
• Possibly check out en.wikipedia.org/wiki/… Feb 10, 2016 at 17:26
• Unless there's something built into the definitions of the meter or second, there's no physical reason why the acceleration near the earth's surface as expressed in $\text{m/s}^2$ should be approximately equal to the mathematical constant $\pi$. Feb 10, 2016 at 17:26
• @gniourf_gniourf and Asaf: I've changed the title to show I don't mean that $\pi^2=g$, but rather $\pi^2\approx g$ (with the appropiate units and whatever physics stuff happens in the background). I just found the story interesting, and thought somebody else would've heard it (and I was interesting in getting the DE from which this was derived). Feb 10, 2016 at 17:27

Maybe this helps: link Looks like some time ago the meter was defined to be essentially the length of the "seconds pendulum", i. e. the pendulum whose period is two seconds

The oscillation time of a pendulum is given by $$T = 2\pi\sqrt{\frac{L}{g}}$$. With $$T = 2$$ and $$L = 1$$ this gives $$g = \pi^2$$

• You have it. The units do not match, so it has to be an accident of how we defined the constants. Feb 10, 2016 at 17:35
• I wonder if there is a nice way to perform an asymptotic analysis of the nonlinear pendulum equation to show why, under these assumptions, $g$ should be about $5\%$ smaller than $\pi^2$.
– Ian
Feb 10, 2016 at 17:37
• Coincidence. The second was defined to be $\frac1{86400}$ part of a solar day long before the pendulum definition, which was first mooted in the 17th century. Later definitions using the pendulum were made so that the new definition matched the old one. And although Wilkins suggested defining a unit of length which was coincidentally very close to the definition of the modern meter, his definition, using the pendulum, wasn't adopted and in fact predated the actual definition of the meter by 130 years. (I am the author of the post about John Wilkins that is cited in the link you gave.)
– MJD
Feb 10, 2016 at 17:40
• Technically it was the other way around: according to the linked blog entry, Wilkins tried to define the meter as the length of the pendulum whose period was twice the already accepted value of the second. But the effect on the ratio of $\pi^2$ to $g$ (in meters per second per second) is the same. Feb 10, 2016 at 17:42
• Seems very interesting to me that earth is like a pendulum swinging around the sun, and mass of the sun and earth with respect to earths distance is approximately equal to sqrt(g) = pi, and that the meter unit is based on how far light travels in a period of time. Mar 15, 2023 at 3:12