On Circles

I've been collecting some fun facts about circles. Here they are.

The Orbital Period of any Rock is 2 hours

Max Fagin staked this claim on twitter some years back, and I found it troubling enough to commandeer the braintrust of the coworkers to do the necessary Physics 101 algebra to convince ourselves of its truth.

Wikipedia tells us that a Low Earth Orbit is about 128 minutes or less (what the hell, Max), but we must remember that the Earth isn't a rock, it has an iron core which bumps up the density to at least twice of that of a generic rock:

However, two hours remains a relatively good approximation for the low orbital periods of many celestial, household rocky objects: Terra: 1.5hrs ibid, Luna, 2 hrs, Mars 1.8hrs. This seems unintuitive, for it feels like the duration of time it takes to go around something should be depending on the size of that thing, rather than just the density, but that's not how gravity works.

Turns out, if you take the relation between a sphere's density, mass, and volume, and use these expressions in the equation for the period of a low circular orbit in terms of the radius $R$ of the celestial body instead of the more familiar equations in terms of radius and mass $M$, then all of the $R$ terms cancel out, meaning that the period is independent of the radius!

We can derive the period of low orbit around some rock as the distance traveled circumferentially over the velocity:

where $C$ is the circumference of the rock, and $v = \sqrt{GM/R}$ is the constant velocity in terms of the gravitational constant $G = 6.67 × 10^{−11} \frac{m^3}{kg\;s^2}$ which is gobbledygook outside of the kind of physics expression we're working with here, but describes the strength of the gravitational field induced by the mass of our rock. We can describe the mass of our rock $M$ in terms of the volume of a sphere $V$ multiplied by its density $\rho$:

Plugging this back into our equation for the period, we get:

And plugging in the density of a typical rock of $\rho_{rock} = 2,700 \frac{kg}{m^3}$, this expression evaluates to almost exactly two hours:

And crucially this result is a function only of density, not the size of our rock, cool!

The volume of a unit $n$-sphere has a maximum at $n=5$

As observed above, and hopefully it was not particularly mind blowing then, the volume of a 3-dimensional sphere is: $V = \frac{4 \pi R}{3}$. We might naively expect the volume to increase with dimensionality, since more dimensions seems like more space for a sphere to fill, but that's not the case¿ In fact, the volume tends towards zero in the limit of dimensionality.

The formula for the volume of an $n$-ball with radius $R$ is given by

where $\Gamma$ is the gamma function, allowing us to extend the factorial function to real and complex numbers. For the case of e.g. $n=100$ dimensions, assuming unit radius, we compute

which is gonna be a big number divided by an even biiiiiigger number, resulting in an extremely small number. Plugging into some calculator which supports the Gamma function, we get $V_{100}(1) = 1.87 \times 10^{-70}$ which is much smaller than the volume of a unit sphere in three dimensions which would be $V_3(1) = \frac{4 \pi 1}{3} = 4.19$. Hamming speaks of this☝️.

High dimensional spheres cannot be contained by equally-high-dimensional cubes

Similar to the previous section, we can observe some other oddities of platonic¹ solids which only occur in high dimensions. Specifically, for $D > 9$ dimensions, an $D$-sphere will "slip out" of the $D$-cube it was completely enclosed by in lower dimensions.

To measure this seeming contradiction, we'll first construct an inequality in two dimensions, then generalize it till the inequality inexplicably breaks.

We begin with a 2-cube (a square), and bisect it vertically and horizontally into four cells, or $2^D$ cells in $D$-dimensions. To minimize the presence of fractions in later calculations, we'll fix the side length of our square to be $4a$:

Next we'll place $2^D$ $D$-spheres (circles) into each cell such that they are circumscribed by their enclosing cell walls, implying a radius of $r = a$:

Finally, we place a $D$-sphere as large as possible at the center of the $D$-cube such that it touches all $2^D$ spheres.

In two dimensions, this circle is obviously inside the bounding square. The distance from the center of the square to the center of each of the enclosing cells is $\sqrt{2}a$, or $\sqrt{D}a$ in $D$-dimensions because the distance is a quarter of the full diagonal of the bounding $n$-cube, which is $\sqrt D\times 4a$ in general.

The distance from the center of the cube to the center of each cell, then, must be the sum of the radius of the inner sphere $r$ and the radius of the cell-filling sphere $a$. Therefore, we have

Nothing wrong here, just playing with our shapes jmo style. For $D=2$, we get the radius of the central circle to be

which is smaller than $2a$, half the size of the square, so indeed the central pink circle is indeed fully contained by the square. However, since $r(D)$ is monotonic and unbounded above by $D$, this means that as the dimension grows, so too will the central sphere, while the linear size of the cube remains fixed! Therefore, at some point, the sphere must escape the cube. And it happens rather quickly too, we can solve the inequality hiding in the above definition of the radius and observation that it's inclosed within the circle:

which happens when $D > 9$. We can struggle to visualize what this might look like in 3 dimensional space, taking a page out of Geoffrey Hinton's book:

• If you are not used to thinking about hyper-planes in high-dimensional spaces, now is the time to learn.
• To deal with hyper-planes in a 14-dimensional space, visualize a 3-D space and say "fourteen" to yourself very loudly. Everyone does it.

which is some biblically-accurate, if not eldritch geometric soup.