Convex homomorphisms and high-\(T_c\) spin flux

AbstractThis is a collection of undergrad-level order-of-magnitude problems and their solutions

*What is the radius of the Earth?*My most recent frequent-flyer statement says that there are 3000 miles from Boston (BOS) to Los Angeles (LAX). Since planes travel along great circles, this distance \(D_u\) represents a segment of the Earth’s circumference. There are four time zones (three intervals) from Boston to Los Angeles, out of 24, so \(D_u\) represents 1/8 of the Earth’s circumference.

\[\begin{aligned} 2\pi R_\oplus &= 8 D_u \nonumber \\ R_\oplus &= \frac{8 D_u}{2 \pi} \nonumber \\ &= \frac{8 (3000 {\rm miles}}{2\times3} \nonumber \\ &= 4000~{\rm miles} \nonumber \\ &= 6400~{\rm km}\end{aligned}\]

The true answer is \(R_\oplus = 6378.1\) km.

*What is the mass of the Earth?*Since we have the radius of the Earth, \(R_\oplus\), all we need is the density of the Earth to get the mass. Based on my experience with handling rocks, I estimate that a round rock that has a diameter of \(d = 8~cm\) has a mass of 1 kg\( = 1000\) g. Such a rock has a volume of \(4/3 \pi (d/2)^3 \approx 250\) cm\(^3\), and a density \(\rho_\oplus \approx 1000~{\rm g}/250~{\rm cm}^3 = 4\) g cm\(^3\). Thus

\[\begin{aligned} M_\oplus &= \rho_\oplus (4/3 \pi R_\oplus^3) \nonumber \\ &= \rho_\oplus (4/3 \pi R_\oplus^3) \nonumber \\ &= \rho_\oplus (4/3 \times 3 \times (6.4\times10^8~{\rm cm})^3) \nonumber \\ &= 4\times 10^{27}~{\rm km}\end{aligned}\]

A more precise answer is \(M_\oplus = 5.97219\times 10^{27}\) g, corresponding to a mean density \(\rho_\oplus = 5.5\) g cm\(^3\). Note that a carefully chose the diameter of that rock, since the diameter would end up cubed. Whenever you have a term raised to a large power, be extra careful with your estimate.

*How far away from the Earth is the Moon, \(a_m\)?*The Moon orbits the Earth approximately every 28 days, so \(P_m = 28 \times 24 \times 60 \times 60 = 2.4\times10^6\) seconds. I have a hard time remembering the exact form of Newton’s version of Kepler’s third law. No worries, though. The Moon is held in an approximately circular orbit by the force of its gravitational attraction to the Earth.\[F_{\rm grav} = \frac{G M_\oplus M_{\rm moon}}{a_m^2} = M_{\rm moon} \dot{V}_m\]

Where \(\dot{V}_m = V_m^2/a_m\) is the centripetal acceleration of the Moon due to its (nearly) circular orbit.

\[\begin{aligned} \frac{G M_\oplus}{a_m^2} = \frac{V_m^2}{a_m}\end{aligned}\]

The Moon travels a distance \(2\pi a_m\) every period, so \(V_m = 2\pi a_m / P_m\)

\[\begin{aligned} \frac{G M_\oplus}{a_m^2} &= \frac{4\pi^2 a_m^2}{P_m^2 a_m} \\ a_m &= \left( \frac{G M_\oplus P_m^2}{4\pi^2} \right)^{1/3} \\ &= \left( \frac{6.7\times10^{-8} \times 6 \times 10^{27} \times (2.4\times 10^6)^2}{4\times 10} \right)^{1/3} \\ &= 3.9\times10^{10}~{\rm cm}\end{aligned}\]

A more precise estimate is \(a_m = 3.844 \times 10^{10}\) cm.

*What is the Moon’s radius?*A nice rule of thumb is that the human thumb held at arm’s length subtends about 1 degree. The moon is half as wide as my thumb, so we’ll use \(\theta_m \approx 0.5\) degrees. Using the skinny angle rule, \(R_m = \theta_m a_m\). There are \(2 \pi\) radians for every 360 degrees, so \(0.5~{\rm degrees} = 8.3\times10^{-3}\) radians.

\[R_m = 3.2\times 10^8~{\rm cm}\]

Did you catch that mistake of mine? The angular size of the Moon corresponds to its

*diameter*, not it’s radius. Thus \(D_m = 2 R_m = \theta_m a_m\), and\[R_m = 1.6\times 10^8~{\rm cm}\]

A more precise estimate is \(1.7374\times10^8\) cm.

*What is the Moon’s mass?*The Moon is made out of mostly the same stuff as the Earth, so \(M_m = \rho_\oplus (4/3 \pi R_m^3) = 5.5 \times 4 \times (1.6\times 10^8)^3 = 9\times 10^{25}\) g, or about 1% the mass of the Earth.