Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 8E / c

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- Solution c

Solution

The orbital plane has normal parallel to the conserved angular momentum. If its inclination to the equatorial plane is

ψ

, then

L_{z} = L cos ψ

. Therefore

p_{θ}^{2} = L^{2} - \frac{L _{z}^{2}}{s i n ^{2} θ} = \frac{L ^{2}}{s i n ^{2} θ} (sin^{2} θ - cos^{2} ψ),

so

p_{θ} (θ) = \pm \frac{L}{s i n θ} sin^{2} θ - cos^{2} ψ .

It vanishes when

sin θ = ∣ cos ψ ∣

, equivalently at the greatest northern and southern latitudes reached by the inclined orbit. These are the turning points of the polar motion.

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