Recently, I kayaked with two friends on the River Shannon, which flows southward through the centre of Ireland. Starting at Dowra, Co. Cavan, we found it easy paddling until we reached Lough Allen, when the going became very tough. It was an uphill struggle.

Could we really be going uphill while heading downstream?  That seems bizarre but, in a certain sense, it is possible.

Forces on the water

Clearly, the water is governed by the pull of the earth and must flow down the gravitational gradient. But the force on the water has two components, Newtonian gravity ( ${\mathbf{g^*}}$) and the centrifugal effect ( ${\Omega^2\mathbf{R}}$) of the earth’s rotation. Because of the interplay between these, the earth is not a sphere but an oblate spheroid, flattened at the poles like an orange (see Fig. 1). Fig. 1. The apparent gravity g is comprised of two parts, Newtonian gravity and the centrifugal force. The force g is perpendicular to the surface of the spheroidal earth.

If we slice the earth from pole to pole it has an elliptical form. If there were no land, the ocean surface would have this form. The polar radius — the distance from the earth’s centre to the pole — is about 20km less than the equatorial radius. We show this, greatly exaggerated, in Fig. 2. Points A and B, at different latitudes, are both at sea level. Point A, at higher latitude, is closer than point B to C, the centre of the earth. Fig. 2. The river source is denoted S and the mouth M. Despite flow down the gradient of gravitational potential, S is nearer than M to the centre C.

We can write the equation for an ellipse as $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

where ${a}$ and ${b}$ are the semi-major and semi-minor axes. The flattening factor ${f=(a-b)/a}$ is a measure of the eccentricity of the planet. For the earth, this is a small quantity ${f\approx\frac{1}{300}\ll 1}$, and we can approximate the ellipse by $\displaystyle r = a[ 1 - f \sin^2\theta ]$

where ${(r,\theta)}$ are polar coordinates corresponding to ${(x,y)}$. [To prove this, we can use ${x= r\cos\theta}$ and ${y = r\sin\theta}$ in the exact equation and find that, to ${O(f^2)}$, the approximate equation is satisfied.]

Shannon Waters Rising

The head of Lough Allen is at ${54^\circ 10^\prime\,}$N and the river estuary below Limerick is at ${52^\circ 40^\prime\,}$N, so the latitude span from the source to the river mouth is ${\Delta\theta=1.5^\circ}$. Taking the derivative of the approximate equation, the change of radius with latitude is $\displaystyle \frac{d r}{d\theta} = -af \times 2\sin\theta\cos\theta = -(a-b) \sin 2\theta$

so the change in radius over a span ${\Delta\theta}$ is $\displaystyle |\Delta r| = (a-b) \sin 2\theta \times \Delta\theta \,.$

Evaluating the sine function at Athlone (latitude ${53.5^\circ\,}$N), we have ${\sin 107^\circ\approx 0.956}$. With ${(a-b)\approx 20\,}$km and ${\Delta\theta=1.5^\circ= \pi/120\,}$radians, we find that the change over the span from Lough Allen to the Shannon estuary is ${\Delta r \approx 530\,}$m. That is, the eccentricity of the planet causes a difference in distance from the earth’s centre of about a half kilometre over that span!

You may argue that Lough Allen is above sea level. Indeed it is, but only 48 metres. The rise in level from source to mouth due to the flattening of the globe is an order of magnitude larger than the drop due to topography. In Fig. 2, points S and M represent the source and mouth of the Shannon. Despite the elevation of the source, it is closer to the earth’s centre C than is the mouth of the river.

Uphill or Down?

In a physical sense, the water flows downhill, or down the gravitational gradient but, in a geometric sense, it flows uphill: the water at the river mouth is farther from the centre of the earth than is the water at the source. No wonder the going was heavy on Lough Allen.

Sources

For data on the Mississippi and other rivers, see:

♦ Banks, Robert B (1999). Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics. Princeton Univ. Press, 286pp.