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ID: past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2021/section-b/5/a/solution

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Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2021 / Section B / 5 / a / Solution by

Codex  0 2026-10-18

Let

M (r)

be the enclosed mass. For spherical symmetry, hydrostatic equilibrium and mass conservation give

\frac{d p}{d r} = - \frac{GM ( r ) ρ ( r )}{r ^{2}}, \frac{d M}{d r} = 4 π r^{2} ρ .

The isothermal equation of state

p = K ρ

therefore implies

K \frac{d l n ρ}{d r} = - \frac{GM ( r )}{r ^{2}} .

Multiply by

r^{2}

, differentiate and eliminate

M

:

\frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d ln ρ}{d r}) = - \frac{4 π G}{K} ρ,

or equivalently

r^{- 2} d [r^{2} ρ^{- 1} ρ^{'}] / d r = - 4 π Gρ / K

. This is the density equation for an isothermal sphere. Here

K

has units

m^{2} s^{- 2}

and equals the squared isothermal sound speed.

A regular isothermal sphere does not reach zero density at a finite radius. The finite body discussed later must therefore be truncated by a nonzero external pressure, as in a pressure-confined isothermal sphere. A finite vacuum boundary would not be compatible with this equation of state.

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