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# Flat vs. round earth

## The two models

### Spherical Earth

The spherical Earth model holds that the Earth is roughly the shape of a ball, while the flat Earth holds that it is roughly the shape of a disk (so that it will be some kind of thin cylinder). Obviously, neither of these models are true, as any cursory glance will tell us, so we're going to have to deal with some uncertainty.

As a good approximation, we can roughly say that any spot on Earth is within ten kilometers of ground level. Furthermore, the spherical model does not assume a perfect sphere for ground level, but roughly an oblate ellipsoidal shape. The surface of an oblate ellipsoid is defined by the constraint

\begin{equation} \frac{x^2 + y^2}{a^2} + \frac{z^2}{b^2} = 1 \end{equation}

The oblateness means that $b > a$. This means that a point at its furthest from the center is at distance $b$, while the closest point will be at distance $a$. In the case of Earth, we have approximately $a = 6,350\ \mathrm{km}$, and $b = 6,380\ \mathrm{m}$.

of inverse flattening $f^{-1} \approx 300$. The flattening parameter for an ellipsoid

### Flat Earth

The standard idea of a flat Earth seems to be some cylinder, of thickness $h$ and radius $R$, up to whatever local topology may occur. Let's consider to round up that our thickness $h$ is only defined up to $\pm10\ \mathrm{km}$.

As a cylinder, we will be using standard cylindrical coordinates $(\rho, \theta, z)$. Our flat Earth will be some kind of cylinder with $\rho < \rho_\oplus$, for some unknown radius $\rho_oplus$, $\theta \in [0, 2\pi]$, and $z \in [-h, 0]$, with $h$ our unknown thickness of the Earth. It's often considered that the azimuthal equidistant projection is the proper representation of the flat Earth, which means that given the latitude and longitude $\phi, \lambda$, we have the relation

\begin{eqnarray} \cos(\frac{\rho}{R}) &=& \sin() \end{eqnarray}

## Astronomy

The exact astronomical model isn't quite important here, all we really need is the evolution of the sun's position in the sky. In standard astronomical theory, the position of the sun in the sky is quite complex, so to keep it simple for now let's consider the two following phenomenon :

• The Keplerian motion of the Earth around the sun
• The rotation of the Earth around its axis

Once this is done, we can change our coordinate system to a point on the surface of the Earth.

The motion of the Earth around the sun in heliocentric coordinates is simple enough, at least as far as the angular formula is concerned. If we consider the sun to be much more massive than the Earth, which it can generally be considered to be, the center of gravity of the system can be considered to be the center of the sun itself (It is about $450km$ off in fact, but this is barely measurable). If we fix the Earth's orbit to be of constant azimuthal angle (which we can do, by the conservation of angular momentum), then the earth will go along an ellipsis

\begin{equation} r \end{equation}

## Gravity

Earth's gravity is fairly straightforward in the spherical model, where we simply consider the center of our ellipsoid to be the center of gravity, with a strength that depends on our radius. For longitude $\phi$ and latitude $\theta$,

Last updated : 2021-08-26 16:15:32
Tags : physics , astronomy