# The battle of the bulge

**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

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

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

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!