Centripetal acceleration of Earth's rotation

Problem

Estimate the magnitude of the centripetal acceleration due to Earth’s rotation for a person standing still at latitude $\varphi\approx 45^\circ$. Compare the result with $g$ and comment on why we normally neglect this effect.

Solution

Model. Rotating with the Earth, the person traces out a circle parallel to the equator. The radius of this circle is not the Earth’s radius $R_T$, but its projection at latitude $\varphi$: $r = R_T\cos\varphi$ with $R_T \approx 6{,}37\cdot 10^6\;\mathrm{m}$.

Angular velocity. The Earth completes one turn in a day ($T = 86400\;\mathrm{s}$): $\omega = \frac{2\pi}{T} = \frac{2\pi}{86400\;\mathrm{s}} \approx 7{,}3\cdot 10^{-5}\;\mathrm{rad/s}$

Radius at 45°. $r = R_T\cos 45^\circ \approx 6{,}37\cdot 10^6 \cdot 0{,}707 \approx 4{,}5\cdot 10^6\;\mathrm{m}$

Centripetal acceleration. $a_c = \omega^2 r \approx (7{,}3\cdot 10^{-5})^2 \cdot 4{,}5\cdot 10^6 \approx 5{,}3\cdot 10^{-9}\cdot 4{,}5\cdot 10^6\;\mathrm{m/s^2}$

$\ev{a_c \approx 2{,}4\cdot 10^{-2}\;\mathrm{m/s^2}}$

Comparison with $g$. $\frac{a_c}{g} \approx \frac{2{,}4\cdot 10^{-2}}{9{,}8} \approx 2{,}4\cdot 10^{-3} \sim 3\cdot 10^{-3}\,g$

$\ev{a_c \sim 3\cdot 10^{-3}\,g}$

Comment. The centripetal acceleration due to Earth’s rotation is about three thousandths of $g$: barely a couple of hundredths of a $\mathrm{m/s^2}$. It is so small compared with $g$ that in ordinary measurements we neglect it: it is “hidden” within the effective value of $g$ measured at ground level (which is in fact already reduced by this effect). It is greatest at the equator, where $\cos\varphi = 1$ and $a_c \approx 3{,}4\cdot 10^{-2}\;\mathrm{m/s^2}$, and vanishes at the poles, where $r = 0$.

Links

Topics: Dynamics Concepts: Centripetal acceleration · Uniform circular motion · Angular velocity Skills: Fermi estimation · Symbolic set-up · Use of trigonometry