Solutions to Exercise 2

1. The solution is given by the Java Center of Mass Calculator. Alternatively, let R1 be the distance of Sirius A from the center of mass and let R2 be the corresponding distance for Sirius B. The associated masses are given as M1 = 2.1 solar masses and M2 = 1.0 solar masses. Then, since

R1 + R2 = 20 AU

M1 x R1 = M2 x R2

we have two equations with two unknowns that we can solve simultaneously: From the second one,

R1 / R2 = (1.0 / 2.1) x R2 = 0.476 x R2

and inserting this in the first equation,

0.476 x R2 + R2 = 20 AU

Thus, 1.476 x R2 = 20 AU and

R2 = 20 AU / 1.476 = 13.6 AU

R1 = 20 AU - 13.6 AU = 6.4 AU

2. The solution is given by the Java Kepler's Laws Calculator. Alternatively, from Kepler's modified 3rd Law we have

P2 = a3 / (M1 + M2)

where P is the period in years, a is the average separation in AU, and (M1 and M2 are the masses of the stars. Thus,

a = ( P2 x (M1 + M2) )1/3 = ( 50.1 x 50.1 x 3.2 )1/3 = 19.96 AU.