### Dropping Pebbles down a Mine-shaft

Trajectory of a body falling at the Equator during a period of 10 seconds.

If you drop a pebble down a mine-shaft, it will not fall vertically, but will be deflected slightly to the East by the Coriolis force, an effect of the Earth’s rotation. We can solve the equations to calculate the amount of deflection; for a ten-second drop, the pebble falls about 500 metres (air resistance is neglected) and is deflected eastward by about 25 cm. The figure on the left shows the trajectory in the vertical xz-plane (scales are not the same).

We derive the equations after making some simplifying assumptions. We assume the mine-shaft is at the Equator; we assume the meridional or north-south motion is zero; we neglect variations in the gravitational force; we neglect the sphericity of the Earth; we neglect air resistance. We can still get accurate estimates provided the elapsed time is short. However, carrying the analysis to the extreme, we obtain results that are completely unrealistic. The equations predict that the pebble will reach a minimum altitude and then rise up again to its initial height a great distance east of its initial position. Then this up-and-down motion will recur indefinitely.

The Equations of Motion

The two forces acting on the falling pebble are gravity and the Coriolis force due to the Earth’s rotation. The equation of motion in a reference frame rotating with the Earth is

$\displaystyle \frac{\mathrm{d}\mathbf{V}}{\mathrm{d} t} = -2\mathbf{\Omega \times V - g} \,.$

We will use local cartesian coordinates ${\mathbf{x} = (x,y,z)}$ pointing east, north and upwards with unit orthogonal triad ${(\mathbf{i, j, k})}$, so that ${\mathbf{V} = u\mathbf{i} + v\mathbf{j} + w\mathbf{k}}$, ${\mathbf{\Omega} = \Omega\mathbf{j}}$ and ${\mathbf{g} = -g\mathbf{k}}$. Then, the equations in component form are are

$\displaystyle \begin{array}{rcl} \dot u &=& -2\Omega w \\ \dot v &=& 0 \\ \dot w &=& 2\Omega u - g \,. \end{array}$

If we set ${v(0)=0}$ then ${v}$ remains zero throughout the motion. We combine the remaining two equations to get a pair of decoupled equations for ${u}$ and ${w}$:

$\displaystyle \begin{array}{rcl} \ddot u + (2\Omega)^2 u &=& 2\Omega g \\ \ddot w + (2\Omega)^2 w &=& 0 \,. \end{array}$

These are easily solved so we can write

$\displaystyle \begin{array}{rcl} u &=& (u_0 - g/2\Omega) \cos 2\Omega t - w_0 \sin 2\Omega t + g/2\Omega \\ w &=& w_0 \cos 2\Omega t + (u_0- g/2\Omega)\sin 2\Omega t \,. \end{array}$

Let us rearrange the terms:

$\displaystyle \begin{array}{rcl} u &=& u_0 \cos 2\Omega t - w_0 \sin 2\Omega t + (1-\cos 2\Omega t ) g/2\Omega \\ w &=& w_0 \cos 2\Omega t + u_0 \sin 2\Omega t - (\sin 2\Omega t) g/2\Omega \,. \end{array}$

Suppose the particle starts from rest, so that ${u_0 = w_0 = 0}$. Then

$\displaystyle \begin{array}{rcl} u &=& (1-\cos 2\Omega t ) g/2\Omega \\ w &=& - (\sin 2\Omega t) g/2\Omega \,. \end{array}$

For ${2\Omega t}$ small, this means that

$\displaystyle u \approx g\Omega t^2 \qquad\mbox{and}\qquad w \approx - g t \,.$

These equations imply that, for small time, the pebble moves eastward (${u>0}$) and the downward motion increases linearly with time.

Since the Earth revolves through ${2\pi}$ radians per day, ${\Omega = 7.272\times 10^{-5}\textrm{s}^{-1}}$. After a fall of ten seconds we have ${2\Omega t = 8.73\times 10^{-4}}$ which is certainly small. At that time, ${w = -100\,\textrm{m\,s}^{-1}}$ and ${u = 10\,\textrm{m\,s}^{-1}}$.

Dropping Pebbles down a Mine-shaft

Let us consider the problem of dropping pebbles down a mine-shaft. The initial conditions are ${x = 0, z = 0 , u = 0, w = 0}$. We can integrate the velocities) using ${u = {\mathrm{d}x}/{\mathrm{d}t}}$ and ${w = {\mathrm{d}z}/{\mathrm{d}t}}$ to get the position as a function of time:

$\displaystyle \begin{array}{rcl} x &=& \frac{g}{(2\Omega)^2}(2\Omega t - \sin 2\Omega t) \\ z &=& \frac{g}{(2\Omega)^2}(\cos 2\Omega t - 1) \,. \end{array}$

For small time (${2\Omega t \ll 1}$), these imply

$\displaystyle x \approx \frac{1}{6}(2\Omega g ) t^3 \qquad\mbox{and}\qquad z \approx -\frac{1}{2}\, g \, t^2$

so the eastward deflection is small and the vertical drop does not depend on ${\Omega}$. These equations were used to plot the trajectory of the falling pebble in the Figure at the head of this post.

The Cycloid: the Pebble Rises Again

For larger times, the solution is very surprising. The solution corresponds to an inverted cycloid. We plot the solution for 60,000 seconds (16.6 hours) in the Figure below.

Trajectory of a body falling at the Equator during a period of 60,000 seconds (about 16.67 hours).

Of course, the cycloid motion is completely unphysical. The prediction is that the pebble will fall more than 800 megametres and then rise again, reaching its initial altitude 3000 megametres to the east. This is utterly unphysical. The pebble must hit the ground long before this. We have pushed the analysis far and far beyond its limits.

We remark that, in the atmosphere and ocean, cycloidal patterns due to the Coriolis force are often seen in the horizontal motion; but that is another story.