Introduction to Classical Mechanics · Problem 10.13

Accelerating frames of reference — Morin

Physics — Mechanics · coriolis-force-deflection

Problem

Southward deflection A ball is dropped from height

h

(small compared with the radius of the earth) at a polar angle

\theta

. Assume (incorrectly) that the earth is a perfect sphere. Show that a second-order Coriolis effect leads to a southward deflection (in the northern hemisphere) equal to

(2/3)(\omega^2 h^2/g)\sin\theta\cos\theta

. It turns out that the actual southward deflection is larger than this; it equals

4(\omega^2 h^2/g)\sin\theta\cos\theta

. So apparently there are other effects at work. The main task of this problem is to show how the factor of

2/3

turns into a factor of

4

. In what follows, we will keep terms up to order

\omega^2

(or technically

\omega^2 R/g

) and order

h/R

. Also, it will be easiest to calculate southward distances relative to the point on the ground directly below the dropping point (as indicated by a plumb bob).

Related theory

Rotating Frames The Coriolis Force

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