Circular Gravity
Is it a coincidence that the value of "π²" is very close to "g"?
Both pi and g are fundamental constants of our universe. First, let's define pi. Imagine we take a rope and form a circle in which the start and end of the rope meet perfectly. Let r be the distance from the center of the circle to any point on the circle. The length of the rope will be 2πr for every possible circle. "g" is easier to explain as we're all familiar with gravity. It's a constant that quantifies the acceleration of gravity that is pulling down on us towards the Earth. Here comes the exciting part. π (pi) is approximated at 3.14, so π² is about 9.85. The value for gravitational acceleration on earth is 9.81 meters/sec². The values 9.85 and 9.81 seem too close to be brushed away as a coincidence, especially considering they are intrinsic properties of our world.
What possible relationship could exist between gravity and a circle's geometry? Let's hold on to that question for a second and talk about simple harmonic motion (SHM). It's basically just a motion that repeats over time intervals. A common example would be a pendulum. As shown by Figure 1, a pendulum is a system consisting of mass attached to the end of a rope or string that swings back and forth on a pivot. For all types of SHM, we can define a value named the period. The period (T) is the amount of time it takes for an object in SHM to complete a full cycle, or for a pendulum to go to a far end twice. For pendulums, the period formula is T = 2π√(l/g) in which l is the length of the pendulum's rope and g is the value for gravity. If you're interested in the derivation of the formula for the period, refer to the next paragraph. If not, you can just skip it.
Figure 1
'To find the formula for the pendulum's period, we'll approach the problem using rotational motion. Let's establish the length of the rope as "l", the mass of the object as "m", and gravity on earth as "g". First, we need to calculate the torque on the mass by gravity. Torque can be calculated using the cross product of the rope and the gravitational force. Since we only take the component of gravity normal to the "rope vector", we obtain the torque as a function of the angle between the rope and the vertical: lmgsin(θ). A basic principle of rotational dynamics is that torque is equal to the moment of inertia multiplied by angular acceleration, 𝛕 = 𝐈ɑ. In this situation, we can imagine the rope as massless, leaving the moment of inertia as 𝐈 = ml². This gives us the following: lmgsin(θ) = ml²ɑ -> ɑ = sin(θ)*(g/l). You may notice this as a second-order differential equation because we're relating an angular displacement to angular acceleration. This is difficult to solve, but we can simplify it further by using a Taylor Polynomial. If we assume small angles, sin(θ) = θ and θ'' = (g/l)*θ. Now, we use a formula of simple harmonic motion that says the angular frequency "ω" can be written as x'' = ω²x. This implies that ω = √(g/l). Finally, period is equal to 2π/ω, so the period for a pendulum will be 2π√(l/g).
Figure 2
This is where we can start looking into the history of things. It started in the 1600s when Galileo realized that the period of a pendulum depended heavily on its length, but not its length and amplitude. In the late 1700s, when the need for a new metric system came, a method was proposed for a defined length based on time. The meter would be the length that a pendulum would need to have a period of two seconds. Now, if we plug this value back into our period formula, we get 2 = 2π√(1/g), which is simplified to g = π². This shows that though the metric value of gravity on earth should technically be equal to pi squared, the relationship is in the metric system, not a fundamental property of the universe.