# SBU PHY 521 - Binary Stars (6 pages)

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## Binary Stars

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## Binary Stars

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Pages:
6
School:
Stony Brook University
Course:
Phy 521 - Stars
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1 Binary Stars Consider a binary composed of two stars of masses M1 and M2 We define M M1 M2 and M1 M2 M If a1 and a2 are the mean distances of the stars from the center of mass then M1 a1 M2 a2 The mean separation of the stars is a a1 a2 If the orbit is elliptical with eccentricity e then the separation at periastron is a 1 e and at apastron it is a 1 e The total energy and angular momentum of the binary are q p 1 GM1 M2 2 2 E J GaM 1 e a 1 e2 2 a Kepler s Law is 2 2 GM 2 3 P a The projected orbital velocity of star 1 is v1 a1 sin i The quantity v13 M2 sin i 3 f1 M1 M2 i G M2 is known as the mass function since it depends only on observables v1 P If Doppler shifts from star 2 are measured then v23 M1 sin i 3 f2 M1 M2 i G M2 can also be found Then M2 v 1 M1 v2 independent of i If the binary is eclipsing the angle i can be determined and the masses individually determined as well Mass Transfer and Roche Lobes The total potential of a binary is GM1 GM2 1 2 2 d r1 r2 2 where r1 and r2 are the distances to stars 1 and 2 and d is the distance to the rotation axis Restricting ourselves to the orbital plane with the origin at the center of mass GM 1 2 GM2 GM1 q x y2 x y q 3 2 a 2 2 x a2 y 2 x a1 y 2 2 In dimensionless coordinates x x a y y a m1 M1 M m2 M2 M GM m1 m2 1 2 2 x y x y q q a 2 x m2 2 y 2 x m1 2 y 2 Contours of constant are shown in the figure There are deep minima at the stellar centers and maxima at five so called Lagrangian points The L1 point between the stars is significant because if a star expands and reaches the potential surface passing through it mass can be transferred to its companion The equipotential surface that passes through L1 is called the Roche lobe and its size depends upon the mass ratio of the binary Kopal 1959 gives for the radius RR with nearly the same volume as the Roche lobe M1 1 3 RR 0 46 1 a M A better fit is by Eggleton 2 3 1 3 1 M1 M1 ln 1 RR a 0 49 6 M2 M2 2 3 Assume the binary is circular Then a MJ2 GM12 M M1 2 da 2a M1 2M1 M M

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