Expert: Philip A. Stahl - 11/5/2009

Question
Okay, I think I have worked this out, but can you tell me if it’s right?

From the link you gave in the previous answer, I can write:

m – M = 5 log (d/10)

where m = +8

M = +4.8 (absolute magnitude for a solar mass star)  and d is the distance to the binary

So:  

(8 – 4.8) / 5   =   log (d/10)  =  0.64

Doing anti-logs:

4.36 = d/ 10 so    d = 4.36 x 10 = 43.6 or call it about 44 pc.

The basic relation between distance and parallax angle p is p = 1/d

So in this case:  p = 1/44 =  0.”0227

The angular separation is given as a” = 1.”00

So the semi-major axis in AU is  A = a”/p  =  1.”00/(0.”0227) = 44 AU

Kepler’s 3rd law gives the form (for two separate masses):

(M1 + M2) = A^3/ P^2

Where we are given P = 100 yrs.

We know M1 is a solar mass star so M1 = 1 (solar mass) and need to get M2

So then:  (M1 + M2) =   (44)^3/ (100)^2   =  8.5 solar masses

So that M2 = 8.5 – 1 = 7.5 solar masses

Is that correct?


Thanks!

Answer
Congrats! You have it correct!  The key to the solution was in seeing that, after you obtain the distance (d) to the system via the distance modulus, then you need to obtain the parallax angle applicable, p.  We know the basic definition for parallax implies a distance of 1 parsec for an angle of 1" subtended at the Earth. Since the distance obtained via the distance modulus is  ~ 44 parsecs, that means the applicable value for p is 44 times *less*, or:

p = 1"/44 =  0."0227

Which is then employed to obtain A, the actual semi-major axis in A.U.. The only remaining chore was to get the unknown mass of the binary system (M2) using Kepler's law as you did.

Good job!