# How fast are you moving on the earth at a given latitude?

The radius of the earth is about $$3959\,\mathrm{mi}$$, so the earth is rotating at about $$\frac{1}{24\,\mathrm{hours}}\times 2\pi\times 3959 \,\mathrm{mi} \approx 1036\,\mathrm{mph}\,$$ and so someone standing on the equator is moving that fast too. But suppose you're standing on the earth at a latitude to $$\theta^{\,\circ}$$. How fast are you moving then?

If you're standing at a latitude of $$\theta^{\,\circ}$$, then you're rotating on a circle of radius $$R\cos(\theta)$$, where $$R = 3959\,\mathrm{mi}$$ is the radius of the earth. This circle still does one revolution in (about!) $$24\,\mathrm{hrs}$$, so we can calculate the speed from there.

$$\left(\frac{1}{24 \,\mathrm{hours}}\right) \times 2\pi R\cos\theta \,\mathrm{miles} \approx 1036\cos\theta \,\mathrm{mph}$$

So you've just gotta scale the speed at the equator by $$\cos\theta$$.

• You can also use the angular velocity of all points on the surface of the earth which is $\omega=\frac{2\pi}{24}$ radians per hour. And if you want the linear velocity of any point, you just use the equation $v=\omega.r$, and just like you said: the radius $r$ at any point is $R\cos\theta$ where $\theta$ is the latitude of that point. Sep 7, 2020 at 18:09
• May I ask what tool you used to make the diagram? Sep 7, 2020 at 18:16
• @saulspatz Mypaint. It's what I've been using as a blackboard for screenshared remote teaching. Sep 8, 2020 at 15:26

Between two longitudes the arc distance

$$L= r \Delta \theta$$

When its latitude $$=\phi,$$ radius $$r$$, earth's radius $$R$$

$$r= R \cos \phi\rightarrow L = R \cos \phi \;\Delta \theta$$

In unit time the circumferential speed ( the others are constant) $$\dfrac{dL}{dt} = R \cos \phi\; \dfrac{\Delta \theta}{dt}= \omega R \cos \phi$$

where $$\omega$$ is the earth's angular velocity, $$\omega R$$ is circumferential speed $$v_{eqtr}$$ at equator. So we can write it simply as

$$v= v_{eqtr} \cos \phi= 1036 \;\cos \phi\; mph,$$

that vanishes at poles, from where you watch the moving world standing, like from a small merry-go-round.

Let's introduce a Cartesian coordinate system in which the center of the earth resides at origin $$(0,0,0)$$ and the north pole resides at $$(0,0,R_E)$$ where $$R_E$$ is the radius of the earth.

By slightly modifying the conventional definition of the azimuth angle, we determine the trajectory of an object residing at $$\theta$$ latitude with initial position $$\Big$$ to be modeled by the parametric curve $$\vec{r}_{\theta}(t)=\Big$$ Here $$t$$ is measured in hours. Then $$\|r_{\theta}'(t)\|=\frac{2\pi R_E}{24}\cos(\theta)$$