Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 317 / 2 / i

Table of contents
- Solution i

Solution

For the centrally condensed point-source model,

M_{r} = M

,

L_{r} = L

, and gas pressure dominates. The equations are

\frac{d P}{d r} = - \frac{GMρ}{r ^{2}}, \frac{d T}{d r} = - \frac{3 κ _{0} ρ ^{2} L}{16 π a _{rad} c r ^{2} T ^{13/2}}, P = R ρT,

where

R = k_{B} / (μ H)

. Dividing the first equation by the second and eliminating

ρ = P / (R T)

gives

P \frac{d P}{d T} = \frac{16 π a _{rad} c GM R}{3 κ _{0} L} T^{15/2} .

The surface conditions

P = T = 0

then yield

P = (\frac{64 π a _{rad} c GM R}{51 κ _{0} L})^{1/2} T^{17/4},

ρ = \frac{1}{R} (\frac{64 π a _{rad} c GM R}{51 κ _{0} L})^{1/2} T^{13/4} .

Substitution into hydrostatic equilibrium cancels the common power of

T

and gives

\frac{d T}{d r} = - \frac{4 GM}{17 R r ^{2}} .

Using

T (R) = 0

,

T (r) = \frac{4 GM}{17 R} (\frac{1}{r} - \frac{1}{R}) .

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