 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s2/3/b/solution

Treat the gravitational field as approximately spherical. The circular orbit condition gives

M (< 8 kpc) = \frac{v _{c}^{2} r}{G} = 9.0 \times 1 0^{10} M_{⊙}, Υ = \frac{M / M _{⊙}}{L / L _{⊙}} = 4.50.

If every star had a representative mass

m

and obeyed

L / L_{⊙} = (m / M_{⊙})^{3}

(m / M_{⊙})^{- 2}

m ≃ 0.47 M_{⊙}

. This is a one-mass estimate, not an actual average inferred uniquely from the data: a galactic stellar mass function, stellar remnants, dust and nonstellar mass alter the relation. The flattening of the galactic disk also changes the spherical conversion by an order-one factor.

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