Calculating Mass of Equal-Mass Binary Star System

In summary, the binary star system consists of two stars with equal mass, separated by 360 million km and taking 5.0 Earth years to orbit around each other. After calculations, the mass of each star is determined to be 1.11x10^30 kg, which is twice the answer given in the book. However, this could be due to a missprint in the book or a different calculation method used. The equations used include GmM/R^2 = mR[omega]^2 and M = R^3[omega]^2/G, where G is the gravitational constant, m is the mass of one star, M is the mass of the other star, R is the center-to-center distance of the
  • #1
liljediboi
28
0
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?
 
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  • #2
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= 2π/(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?
 
  • #3
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.
 
  • #4
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= 2π/(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.
 
  • #5
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)
 
  • #6
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:
  • #7
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. )
 

1. How do you calculate the mass of an equal-mass binary star system?

The mass of an equal-mass binary star system can be calculated using the formula M = 4π²a³/GT², where M is the combined mass of the two stars, a is the distance between them, G is the gravitational constant, and T is the period of their orbit.

2. What is the gravitational constant used for in calculating the mass of an equal-mass binary star system?

The gravitational constant, denoted as G, is a universal constant used in the calculation of gravitational force between two objects. In the context of calculating the mass of an equal-mass binary star system, it is used to determine the strength of the gravitational force between the two stars.

3. Can the mass of an equal-mass binary star system be calculated using different methods?

Yes, there are alternative methods for calculating the mass of an equal-mass binary star system, such as using the radial velocity method or the astrometric method. However, the formula M = 4π²a³/GT² is the most commonly used and accurate method.

4. How does the distance between the stars affect the calculated mass of an equal-mass binary star system?

The distance between the two stars, denoted as a in the formula, has a direct impact on the calculated mass. The closer the stars are to each other, the stronger the gravitational force and the higher the calculated mass will be. On the other hand, a larger distance will result in a lower calculated mass.

5. Can the mass of an equal-mass binary star system change over time?

Yes, the mass of an equal-mass binary star system can change over time due to various factors such as mass transfer between the two stars or the loss of mass through stellar winds. These changes in mass can also affect the orbital period and distance between the stars.

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