 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s30/2/b/solution

Neglect disk self-gravity and take

∣ z ∣ ≪ r

around a central point mass

M_{⋆}

. Its vertical gravitational acceleration is

g_{z} = - \frac{G M _{⋆} z}{( r ^{2} + z ^{2} ) ^{3/2}} ≃ - \frac{G M _{⋆}}{r ^{3}} z = - Ω^{2} z .

d P / d z = ρ g_{z}

. For a vertically isothermal ideal gas,

P = ρ c_{s}^{2}

and

c_{s}

is independent of

z

c_{s}^{2} \frac{d ρ}{d z} = - Ω^{2} ρ z, ρ (r, z) = ρ_{0} (r) e^{- z^{2} / (2 h^{2})}, h = \frac{c _{s}}{Ω}, \frac{h}{r} = \frac{c _{s}}{v _{K}} .

Here

v_{K} = r Ω = G M_{⋆} / r

c_{s}^{2} = k_{B} T / (μ m_{p})

is the isothermal sound speed squared, and

h

is the Gaussian scale height. A different convention writing

e^{- z^{2} / h^{2}}

changes the quoted height by

2

but not the physics.

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