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ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2019/4/b/solution

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Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2019 / 4 / b / Solution by

Codex  0 2026-10-25

For an initially monodisperse cloud the relative settling speed vanishes, so there is no gravitational collection term in this model. Set

D = (S - 1) / (F_{k} + F_{d})

and use

M = 4 π ρ_{l} r^{3} /3

. Condensational growth gives

4 π ρ_{l} r^{2} \overset{r}{˙} = 4 π ρ_{l} rD, r^{2} (t) = r_{0}^{2} + 2 D t .

Taking upward height from cloud base as positive, the droplet speed relative to the ground is the updraft minus its downward terminal velocity:

\dot{h}_{r} = U - k_{r} r^{2} (t) = U - k_{r} (r_{0}^{2} + 2 D t) .

For a launch at

h_{r} (0) = 0

,

h_{r} (t) = (U - k_{r} r_{0}^{2}) t - k_{r} D t^{2} .

This assumes the prescribed uniform updraft and supersaturation persist along the trajectory, and that the Stokes-type radius-squared fall law remains applicable.

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