 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/s2/1/2/solution

Equal radial velocity amplitudes imply equal stellar masses. Under the circular, edge-on eclipsing binary interpretation, each star moves at

K = 100 km s^{- 1}

, and the relative speed is

2 K

. Thus

a = \frac{( K _{1} + K _{2} ) P}{2 π} = 1.499 \times 1 0^{10} m, M_{1} + M_{2} = \frac{P ( K _{1} + K _{2} ) ^{3}}{2 π G} .

Using

P = 5.45

days gives

a ≃ 0.100 AU, M_{1} = M_{2} ≃ 2.26 M_{⊙} .

a sin i = P (K_{1} + K_{2}) 1 - e^{2} / (2 π)

and

(M_{1} + M_{2}) sin^{3} i = P (K_{1} + K_{2})^{3} (1 - e^{2})^{3/2} / (2 π G)

. Thus these numerical masses and separation use

e = 0

sin i = 1

; eclipses alone imply high inclination rather than its exact value.

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