Find the speed at which a point in Belgrade moves because of…

Find the speed at which a point in Belgrade moves because of the Earth's rotation about its axis. The latitude of Belgrade is $\varphi = 44^\circ 45'$, and the radius of the Earth is taken as $R \approx 6470\ \mathrm{km}$.

Explanation of Method

How fast Belgrade moves because of the Earth's rotation.

Step 1: The circle the point moves on

A point on the surface circles around the Earth's axis, not its centre. At latitude $\varphi$ the radius of that circle (the circle of latitude) is

\[ r = R\cos\varphi, \]

\[ \varphi = 44^\circ 45' = 44.75^\circ, \quad \cos 44.75^\circ \approx 0.7102. \]

$\varphi$ $R$ $r = R\cos\varphi$ $\text{Belgrade}$ $\omega$ $\text{equator}$ $\text{one turn per day:}$ $v = \omega r = \frac{2\pi R\cos\varphi}{T}$ $v \approx 334\ \mathrm{m/s}\ \text{(eastward)}$

A point at latitude $\varphi$ does not circle around the Earth's centre but around the axis, on the red circle of latitude of radius $r = R\cos\varphi$. It goes once around that circle per day.

Step 2: Angular velocity of the Earth

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

Step 3: Speed

\[ v = \omega r = \omega R\cos\varphi = 7.27 \cdot 10^{-5} \cdot 6.47 \cdot 10^{6} \cdot 0.7102\ \mathrm{m/s}. \]

\[v \approx 334\ \mathrm{m/s} \approx 1200\ \mathrm{km/h}\]

Check

Circle of latitude: $ 2\pi \cdot 6470 \cdot 0.7102 \approx 28\,870\ \mathrm{km} $ per day, and $ \tfrac{28\,870\ \mathrm{km}}{24\ \mathrm{h}} \approx 1203\ \mathrm{km/h} \approx 334\ \mathrm{m/s}\ \checkmark $

Result

\[\boxed{\,v \approx 334\ \mathrm{m/s} \approx 1200\ \mathrm{km/h}\,}\]

Remark

Notes. The original answer, 332.6 m/s, comes from rounding $\cos 44.75^\circ$ to 0.707 (that is $\cos 45^\circ$); with the correct value the speed is about 334 m/s. The Earth's actual mean radius is about 6370 km (6470 is probably a typo), which would give about 329 m/s.