Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 2 / b / i

Table of contents
- Solution i

Solution

The linear term

- (3/2) Ω_{K} x e_{y}

is the local Keplerian shear in the shearing sheet. The remaining constant

V = r_{0} [Ω (r_{0}) - Ω_{K} (r_{0})]

is the gas's azimuthal velocity relative to the local Keplerian frame. The outwardly decreasing midplane pressure found in part (a)(iii) makes the gas Sub-Keplerian, so

V < 0, ∣ V ∣ \sim \frac{c _{s}^{2}}{r _{0} Ω _{K}} \sim Ω_{K} r_{0} (\frac{H}{r _{0}})^{2} .

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