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})]

(1)

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} .

(2)

ii

 0  0

Solution

 0  0

Write

Δ u = (U, W, 0)

. The background shear gives

u_{0} \cdot \nabla u_{0} = - \frac{3}{2} Ω_{K} U e_{y} .

(1)

The constant radial and azimuthal components of the dust equation are therefore

2 Ω_{K} W - \frac{U}{τ} = 0, \frac{1}{2} Ω_{K} U + \frac{W - V}{τ} = 0.

(2)

With the Stokes number

St = τ Ω_{K}

, their solution is

Δ u = \frac{2 St V}{1 + St ^{2}} e_{x} + \frac{V}{1 + St ^{2}} e_{y} .

(3)

The gas is slower than a Keplerian particle, so the particle feels an aerodynamic headwind. Drag force removes its angular momentum, and it drifts radially inward because