 Solution

ID: past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2021/6/b/solution

Put

V_{0} = 200 km s^{- 1}

. A Taylor expansion near the centre gives

V_{c} (r) = V_{0} [\frac{r}{r _{0}} - \frac{1}{2} (\frac{r}{r _{0}})^{2} + O (\frac{r ^{3}}{r _{0}^{3}})] .

Therefore the circular angular frequency is

Ω (r) = \frac{V _{c} ( r )}{r} = \frac{V _{0}}{r _{0}} [1 - \frac{r}{2 r _{0}} + O (\frac{r ^{2}}{r _{0}^{2}})] ⟶ \frac{V _{0}}{r _{0}} .

Thus the central motion approaches solid-body rotation. Numerically

Ω_{0} = 1.6204 \times 1 0^{- 15} s^{- 1}

, and the central orbital period tends to

P_{0} = \frac{2 π}{Ω _{0}} = \frac{2 π r _{0}}{V _{0}} = 3.8776 \times 1 0^{15} s ≃ 1.229 \times 1 0^{8} yr .

Ω

accompanies

V_{c} \to 0

r \to 0

; it does not imply a finite central linear speed.

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