 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/2015/s2/1/3/solution

L / L_{⊙} ≃ (M / M_{⊙})^{4}

gives

M_{S} ≃ 0.946 M_{⊙}

. Then

M_{P} = V_{S} (\frac{P M _{S}^{2}}{2 π G})^{1/3}, a = (\frac{G M _{S} P ^{2}}{4 π ^{2}})^{1/3} .

For

P = 11.6

days and

V_{S} = 10 m s^{- 1}

, these give approximately

M_{P} ≃ 0.107 M_{J}, a ≃ 0.0985 AU .

These estimates retain the circular, edge-on and low-companion-mass assumptions. Adopting a mass-luminosity exponent

3.5

instead changes the quoted values by about one percent; luminosity alone does not determine a unique stellar mass without such a calibration.

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