The Problem of the Eiffel Tower

Too busy today (again) for anything else so I’m going to resort (again) to the Cavendish Problems in Classical Physics. I think I’ll eschew the multiple-choice format for this one, but will say that there is a small hint in the fact that the question is split into two parts:

The Eiffel Tower is 300m high and is situated at a latitude 49° N. What are the magnitude and direction of the deflection caused by the Earth’s rotation to:

  1. the bob of a plumb-line hung from the top of the Tower;
  2. the point of impact of a body dropped from the top?

Please give your answers, with reasons, through the comments box below. For legal reasons I should make it clear that you are not expected to perform either experiment.


6 Responses to “The Problem of the Eiffel Tower”

  1. Second question first correctly solved by Robert Hooke in his correspondence with Isaac Newton. Important step on the way to Newton’s theory of gravity.

  2. Since no one else has, and since I’d rather do this than grade exams, …

    Working in the rotating reference frame attached to the Earth, there are two inertial “forces”: a centrifugal force and a coriolis force. For the plumb bob, the coriolis force doesn’t contribute. The centrifugal acceleration is omega^2 r cos(l), pointing in a direction straight away from the Earth’s axis. An observer on the Earth resolves this into a vertical component and a southward component. We can neglect the vertical component in comparison with g. The south component is omega^2 r cos(l) sin (l). The plumb bob hangs at an angle that deviates from the “true” vertical by omega^2 r cos(l) sin(l) / g. Multiply this by the height of the tower to get the southward displacement of the plumb bob. If I’ve done the arithmetic right (by no means guaranteed), it comes out to about 51 cm.

    The falling body has this same southward displacement, but it is also displaced by the Coriolis force. To zeroth order, the body falls straight down with velocity gt. The Coriolis force causes an acceleration -2 omega x v, or 2 omega v cos(l), in the eastward direction. So there’s an eastward acceleration of 2 omega g t cos(l). Integrate that twice to get the eastward displacement omega g T^3 cos(l) / 3. The fall time T is related to the height by the usual rule h = g T^2/2. Clean it up and plug in numbers, and unless I’ve made a mistake (again, quite possible), you get an eastward displacement of 7.5 cm, to be combined with the 51-cm southward displacement due to centrifugal acceleration.

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