 Solution

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/2012/s26/3/5/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 / 2012 / S26 / 3 / 5 / Solution by

Codex  0 2026-10-25

For a spherical mass distribution with constant circular speed

v_{c}

,

M (< r) = \frac{v _{c}^{2} r}{G}, ρ (r) = \frac{1}{4 π r ^{2}} \frac{d M}{d r} = \frac{v _{c}^{2}}{4 π G r ^{2}} .

For

v_{c} = 220 km s^{- 1}

, this is about

0.0140 (r /8 kpc)^{- 2} M_{⊙} p c^{- 3}

. It is the spherical model's total density; it is singular if extrapolated to the very centre and is intended only over the flat-rotation region.

For a sharp outer truncation at

r_{*}

, the exterior mass is fixed at

M_{*} = v_{c}^{2} r_{*} / G

. Integrating the gravitational work from

r

to the boundary and then to infinity gives

\frac{1}{2} v_{esc}^{2} = \int_{r}^{r_{*}} \frac{v _{c}^{2}}{s} d s + \int_{r_{*}}^{\infty} \frac{G M _{*}}{s ^{2}} d s = v_{c}^{2} [lo g \frac{r _{*}}{r} + 1] .

Thus

v_{esc} (r) = 2 v_{c} 1 + lo g (r_{*} / r) .

The exterior contribution cannot be dropped: at

r = r_{*}

it correctly gives the point-mass escape speed

2 v_{c}

.

If the fastest measured radial speed is assumed to approximate both the total speed and the local escape speed of a bound star, then at

r = 8 kpc

,

r_{*} ≃ r exp (\frac{43 0 ^{2}}{2 ( 220 ) ^{2}} - 1) ≃ 19.9 kpc .

Without those extra assumptions, the measurement gives only a bound: a bound star's total speed is at least its radial component, and the sample maximum need not reach escape, so

r_{*} ≳ 19.9 kpc

within this model. An unbound interloper would invalidate even that bound. This is not a determination of the real Milky Way halo edge.

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