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PC1672 Advanced dynamics

2.3 Motion on the Earth

Although we don't usually notice it, the Earth provides an example of a rotating frame of reference. It spins about its axis at a rate of $\omega=2\pi\ \hbox{\rm rad day}^{-1}\simeq 7\times 10^{-5}\ \hbox{\rm rad s}^{-1}$. We shall also need the radius of the Earth, $R=6400\ \hbox{\rm km}$. We choose our $x_1$-axis to point eastwards, our $x_2$-axis northwards and our $x_3$-axis vertically upwards. If we are at latitude $\lambda$ then the vector angular velocity of the Earth is

\begin{displaymath}\hbox{\boldmath $\omega$}=\omega(0,\cos\lambda,\sin\lambda)\end{displaymath}

The centrifugal force on a body at the Earth's surface is

\begin{displaymath}{1\over m}{\bf F}_{\rm cent}=R\cos\lambda\omega^2(0,-\sin\lambda, \cos\lambda)\end{displaymath}

where $R\cos\lambda$ is our distance from the Earth's axis and ${\bf e}_r=(0,-\sin\lambda,\cos\lambda)$ is the unit vector pointing outwards from the axis. The typical magnitude of the acceleration produced by this inertial force in our rotating frame is $R\omega^2=0.03\ \hbox{\rm ms}^{-2}$. It contributes to small variations in the apparent strength and direction of the gravitational field at the Earth's surface.

The Coriolis force on a body moving with velocity ${\bf v}_r=(v_1,v_2,v_3)$in the frame rotating with the Earth is

$\displaystyle {1\over m}{\bf F}_{\rm Cor}$ $\textstyle =$ $\displaystyle 2\omega(v_2\sin\lambda-v_3\cos\lambda)\, {\bf e}_1$  
    $\displaystyle -2\omega v_1\sin\lambda\, {\bf e}_2$  
    $\displaystyle +2\omega v_1\cos\lambda\, {\bf e}_3$  

The horizontal components are the ones of most interest because the much larger force of gravity makes the vertical one hard to detect. For an object moving at speed $v=10\ \hbox{\rm ms}^{-1}$, the typical magnitude of the Coriolis acceleration is $v\omega\simeq 5\times 10^{-4}\ \hbox{\rm ms}^{-2}$. (In order to estimate the size of these effects, we can assume that all interesting places are at roughly $45^\circ$ N or S of the equator and so $\cos\lambda\simeq 1/\sqrt 2$. Manchester is actually close to $53^\circ$ N.)

For an object moving horizontally ($v_3=0$), the Coriolis force is

\begin{displaymath}{1\over m}{\bf F}_{\rm Cor}=2\omega\sin\lambda (v_2{\bf e}_1-v_1{\bf e}_2) +(\cdots){\bf e}_3\end{displaymath}

The horizontal piece of this has magnitude $2\omega v\sin\lambda$ and so its effect is largest at the poles and vanishes at the equator. Its direction is perpendicular to ${\bf v}_r$, pulling the motion to the right in the northern hemisphere and to the left in the southern hemisphere.

For an object moving vertically ($v_1=v_2=0$), the Coriolis force is

\begin{displaymath}{1\over m}{\bf F}_{\rm Cor}=-2\omega v_3\cos\lambda\, {\bf e}_1\end{displaymath}

The effect of this is a maximum at the equator. For an object falling downwards ($v_3<0$), this force points in an easterly direction.

Popular mythology maintains that the Coriolis force determines which way your bath water twists as it goes down the plug. This is not true; the distances and timescales are far too small for the Coriolis effect to become larger than other influences (the shape of the tub, how you sloshed the water when you got out of it) which could make the water twist either way. For a thorough debunking of this myth, see the web pages on bad meteorology.

Textbook references

Home: PC 1672 home page | Up: 2 Noninertial frames . . . | Weekly plan | Help: Guide to using this document |
Next: 2.4 The weather | Previous: 2.2 Rotating frames |

Mike Birse
17th May 2000