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ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s26/1/a/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S26 / 1 / a / Solution by

Codex  0 2026-10-25

A thin spherical shell of thickness

d r

has mass

4 π r^{2} ρ d r

. Balancing its inward gravitational force with the pressure difference across its faces gives

\frac{d P}{d r} = - \frac{G M _{r} ρ}{r ^{2}}, \frac{d M _{r}}{d r} = 4 π r^{2} ρ .

For a transparent estimate take constant density

ρ = 3 M_{⊙} / (4 π R_{⊙}^{3})

and negligible surface pressure. Then

M_{r} = M_{⊙} r^{3} / R_{⊙}^{3}

, so integration gives

P_{c} = \frac{3 G M _{⊙}^{2}}{8 π R _{⊙}^{4}} ≃ 1.3 \times 1 0^{14} Pa .

Combining with the ideal gas law and

μ ≃ 0.6

yields

T_{c} ≃ μ m_{H} G M_{⊙} / (2 k_{B} R_{⊙}) ≃ 7 \times 1 0^{6}

K. The less specific virial scaling gives

T_{c} \sim μ m_{H} G M_{⊙} / (k_{B} R_{⊙}) \sim 1 0^{7}

K.

The real Sun is strongly centrally concentrated: a uniform-density model substantially underestimates central pressure and is not a detailed solar model. Its value is a justified scale estimate, while more realistic structure produces central temperatures of order

1.5 \times 1 0^{7}

K and central pressures of order

1 0^{16}

Pa. Hydrostatic equilibrium alone does not uniquely determine both profiles without an equation of state and a density/thermal model.

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