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?


3 Answers 3


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.

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

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

  • 2
    $\begingroup$ 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. $\endgroup$
    – NadAlaba
    Sep 7, 2020 at 18:09
  • 1
    $\begingroup$ May I ask what tool you used to make the diagram? $\endgroup$
    – saulspatz
    Sep 7, 2020 at 18:16
  • $\begingroup$ @saulspatz Mypaint. It's what I've been using as a blackboard for screenshared remote teaching. $\endgroup$ 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 $R$

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

enter image description here

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 $\Big<R_E\cos(\theta),0,R_E\sin(\theta)\Big>$ to be modeled by the parametric curve $$\vec{r}_{\theta}(t)=\Big<R_E \cos(\theta)\cos\Big(\frac{2\pi t}{24}\Big),R_E\cos(\theta)\sin\Big(\frac{2\pi t}{24}\Big),R_E\sin(\theta)\Big>$$ Here $t$ is measured in hours. Then $$\|r_{\theta}'(t)\|=\frac{2\pi R_E}{24}\cos(\theta)$$


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