# Binary suns

liljediboi
suppose that a binary star system consists of two stars of equal mass. they are observed to be separated by 360 million km and to take 5.0 Earth years to orbit about a point midway between them. what is the mass of each?

P. 135 "Physics: Principles With Applications" Fourth Edition. Giancoli. 1995.

after much discussion with my dad, and working it many times, we get 1.11x10^30 kg as the mass. the answer in the back of the book is 5.5x10^29 kg. we noticed it is half of our answer, but the thought is that the book is missprinted.

help?

Homework Helper
An object moving in a circle of radius R, with angular speed omega, has acceleration R omega2. In this case, since the period of the orbit is 5 years, omega= 2&pi;/(5 years). If you are using the mks value for G, the gravitational constant, you will need to convert that to seconds.

Then use GmM/R2= m R omega2 so

M= R3omega2/G.

I don't have a value for G at hand right now. Since your answer is twice the answer in the book, I wonder if you didn't calculate the total mass of both stars?

liljediboi
we got velocity first, around 1400, and then plugged into equation rv^2/T for mass. 6.67x10^-11 for G. one thing i thought was that you would use 180 million km for radius, not 360 million km due to the way the stars will rotate around each other.

Staff Emeritus
Gold Member
Originally posted by HallsofIvy
An object moving in a circle of radius R, with angular speed omega, has acceleration R omega2. In this case, since the period of the orbit is 5 years, omega= 2&pi;/(5 years). If you are using the mks value for G, the gravitational constant, you will need to convert that to seconds.

Then use GmM/R2= m R omega2 so

M= R3omega2/G.

I don't have a value for G at hand right now. Since your answer is twice the answer in the book, I wonder if you didn't calculate the total mass of both stars?

One small problem here, the two "R"s in the two equations are not the same. the R in m R omega2 Is the radius vector of the sun's orbit, which in this case is 1/2 the R in GmM/R2 thus the equation should read:

GmM/R2= m2R omega2

thus:
M= R3omega2/2G.

or more generally:

M + m= R3omega2/G.

Staff Emeritus
Gold Member
Just in case anyone is interested how I got my general answer above, consider this:
Assume that r is the radius vector of m in its path around the common center of gravity of M and m, and R is the center to center distance of m to M.

Then r = RM/(M+m) and

GMm/R² = RMm[ome]²/(M+m)

Which reduces to the general answer I gave in my last post.

(And since in this particular example M=m, you get the first equation I gave)

liljediboi
is the solution

M=R^3omega^2/2G
or
M=R^3omega^2/G
or
2M=R^3omega^2/G

what value for R do you use and please explain the how you get the omega because that is not covered in the book, and can you use another method not dealing with omega as well?

Last edited:
Staff Emeritus
Gold Member
Originally posted by liljediboi
is the solution

M=R^3omega^2/2G
or
M=R^3omega^2/G
or
2M=R^3omega^2/G

what value for R do you use and please explain the how you get the omega because that is not covered in the book, and can you use another method not dealing with omega as well?

The first and third equation are the same equation just re-arranged, and are the correct ones. R is the center-to-center distance of the two suns (360 million kilometers).

If you read HallsofIvy's post, he gives how to get [ome]. It is:

[ome] = 2[pi]/t , where t is the period of the orbit in seconds.

If you substitute this in the equation you get

M = 2[pi]²R³/Gt²

(For this particular example where the masses of the two suns are equal. )