 Solution

ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2021/6/d/solution

Use the positive rotation-rate convention implicit in the northward planetary vorticity gradient. One Pluto day is

T_{P} = 6.4 \times 86400 = 552960 s

. At the equator,

r = T_{P}^{- 1} = 1.8084 \times 1 0^{- 6} s^{- 1}, β = \frac{2Ω}{a _{p}} = \frac{4 π}{T _{P} a _{p}} = 1.9129 \times 1 0^{- 11} m^{- 1} s^{- 1} .

The width of the western boundary current is therefore

δ_{W} = \frac{r}{β} = \frac{a _{p}}{4 π} ≃ 9.45 \times 1 0^{4} m ≃ 95 km .

Here

b = a_{p} π /3 ≃ 1244 km

is the full meridional scale, and

L = a_{p} (50 π /180) ≃ 1037 km

. For these data the exact exponents give

λ^{- 1} ≃ 89.7 km

λ L ≃ 11.56

and

γ L ≃ 0.593

. The western layer is narrow, but

γ L ≪ 1

is only marginal. The leading estimates below should therefore be distinguished from evaluation of the exact linear model.

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