# Carousel Earth

| math

We sit on a rotational ellipsoid spinning around its axis roughly every 23 h 56 min. What is the speed at which we travel on the surface at a given latitude?

Let's center our planet on a three-dimensional coordinate system. The axis of rotation is the z-axis, x- and y-axis follow from the right hand rule. If we make a cut at y = 0, we get an ellipse with a semi-major axis a = 6378 km and a semi-minor axis b = 6357 km. $$\frac{x^2}{a^2} + \frac{z^2}{b^2} = 1 .$$

Given a latitude $$\phi$$, we can set up a line through the origin with a slope $$m = \tan (\phi \cdot \frac{180}{\pi}).$$ $$z = m x$$

This line intersects the ellipse centered at $$O = (0, 0)$$ in point $$P.$$ By substituting the linear equation into the equation of the ellipse, we can compute the x-component of point $$P$$, which is the radius $$r$$ of the circular arc on which somebody sitting on the surface at latitude $$\phi$$ is spinning around the axis. $$x = \sqrt{\frac{a^2 b^2}{b^2 + m a^2}} = r$$ Assuming constant circular speed, we get the velocity by computing the circumference and dividing it by the time for a full rotation. Take, for example, 45° latitude, somewhere north of Bordeaux, France. $$v = \frac{2 \pi r}{\Delta t} \approx \frac{2 \pi \cdot 4502 \text{km}}{23.93 h} \approx 1180 \frac{\text{km}}{\text{h}}$$

Hold on tight! It's a rough ride.