Find how the linear velocity of a point on the Earth's…

Find how the linear velocity of a point on the Earth's surface depends on the geographic latitude. Calculate this velocity at latitude $\varphi = 45^\circ$. The Earth's mean radius is $R = 6374\ \mathrm{km}$.

Step 1: Idea

All points on the Earth turn with the same angular velocity $\omega$, one turn per day. Their linear speeds differ because they move on circles of different sizes. A point does not circle the Earth's centre; it circles the Earth's axis, and the radius of that circle is its distance from the axis.

Step 2: Radius of the circle

Latitude $\varphi$ is the angle between the radius to the point and the equatorial plane. In the right triangle formed by the centre, the point and the axis (see the diagram), the distance from the axis is the side adjacent to $\varphi$:

\[ \rho = R\cos\varphi. \]

Step 3: Linear velocity

\[ v = \omega\rho. \]

\[v(\varphi) = \omega R\cos\varphi\]

The speed is greatest on the equator ($\cos 0 = 1$) and falls to zero at the poles ($\cos 90^\circ = 0$), where the point just turns on the spot.

Step 4: At 45°

The Earth turns once in $T = 24\ \mathrm{h} = 86\,400\ \mathrm{s}$: \[\omega = \frac{2\pi}{T} = \frac{2\pi}{86\,400\ \mathrm{s}} = 7.27 \cdot 10^{-5}\ \mathrm{s^{-1}}.\]

\[ v = 7.27 \cdot 10^{-5} \cdot 6.374 \cdot 10^6 \cdot 0.7071. \]

\[v(45^\circ) \approx 328\ \mathrm{m/s}\]

That is about $1180\ \mathrm{km/h}$, roughly the speed of sound. We do not feel it because everything around us moves with us.

Check

Equator: $ \omega R = 464\ \mathrm{m/s} $; $ 464 \cdot 0.707 $ $= 328\ \mathrm{m/s}\ \checkmark $

With the true (sidereal) rotation period of 86 164 s the value is 328.6 m/s, practically the same $\checkmark$