October 21, 2020. The earth spins and the equator bulges, turning from a sphere into a flattened ellipsoid. To estimate just how flattened, we assume that the potential energy is the same at the equator and the poles, and find an equatorial bulge only a few hundred metres off the correct result!

Energy in a spinning frame

Suppose you’re in a spinning reference frame with angular velocity $\omega$, and a distance $r$ from the axis of rotation. You will be subject to an acceleration

\[a = \frac{v^2}{r} = \omega^2 r.\]

This looks like a centrifugal force $F = ma$, arising from a quadratic potential

\[F = -\frac{dU_\text{cent}}{dr} \quad \Longrightarrow \quad U_\text{cent} = -\frac{1}{2}m \omega^2 r^2.\]

If your spinning reference frame is the earth, then there is a gravitational potential $U_\text{grav} = -GMm/R$, where $M$ is the earth’s mass and $R$ is the distance to the centre.

We will try to estimate the earth’s bulge by assuming that the potential is the same at the poles and the equator. If the earth was a fluid (and the surface was molten at some point), it would redistribute until there were no differences in potential across the surface, the same reason the surface of a fluid in a cup is level. At the poles, there is no spinning, so $\omega = 0$, and

\[U_{\text{pole}} = U_{\text{grav}}(R) = -\frac{GMm}{R},\]

where $R$ to be the distance from the centre of the earth to the poles. Technically, if the earth is not a sphere, then the sphere theorem (that objects are attracted to the centre) no longer applies, but the corrections will be small enough we can ignore them.

An equatorial expansion

Now, suppose that the radius at the equator is slightly larger, $R + r$. Then, including the centrifugal potential,

\[U_{\text{eq}} = U_{\text{cent}}(R+r) + U_{\text{grav}}(R+r) = -\frac{1}{2}m \omega^2 (R+r)^2 -\frac{GMm}{R+r}.\]

We want to equate this with $U_\text{pole}$ and solve for $r$, but this will lead to a horrible cubic equation to solve. To proceed, we use a typical physicist’s trick: if $\epsilon \ll 1$, then

\[\frac{1}{1+\epsilon} \approx 1 - \epsilon,\]

since $(1+\epsilon)(1-\epsilon) = 1 - \epsilon^2 \approx 1$. In our case, this gives

\[\frac{1}{R+r} = \frac{1}{R(1+r/R)} \approx \frac{1}{R} - \frac{r}{R^2},\]

and hence, after some algebra, an equation for the equatorial radius:

\[(R+r)^2 = \frac{2GMr}{\omega^2 R^2}.\]

We can expand the LHS and solve for $r$ using the quadratic formula. But since we’ve approximated away $r^2$ once, we should be consistent and do it again. In other words, we expand the LHS and keep only the linear term, to get

\[(R+r)^2 \approx R^2 + 2Rr \quad \Longrightarrow \quad r \approx R\left(\frac{GM}{\omega^2 R^3} - 1\right)^{-1}.\]

This is reasonably simple!

Comparing to reality

The mass and polar radius of the earth are approximately

\[M = 5.972 \times 10^{24} \, \text{kg}, \quad R = 6.357 \times 10^6 \, \text{m},\]

while Newton’s constant is $G = 6.674 \times 10^{-11}$ in SI units I am too lazy to typeset. The angular frequency is just 1 revolution per day, or

\[\omega = \frac{2\pi}{T} = \frac{2\pi}{24 \text{ hours}} = 7.272 \times 10^{-5} \, \text{s}^{-1}.\]

Plugging these numbers in, we get an estimate for the equatorial bulge,

\[r = 21700 \, \text{m}.\]

Now, for the real result. Drum roll please…

\[r = 21385 \, \text{m},\]

so we are only off by $300$ metres or so!