How Does Binary Star Mass Calculation Using Kepler's Law Work?

[lindz.12]

Consider a pair of binary stars with a separation of 3.60E12 m and an orbital period of 2.55E9s. Assuming the two stars are equally massive, determine the mass of each.

keplar's law...
so I rearranged the formula and set (2pi*r)/T=sqrt((GM)/r), and then I solved for M, which gave me the equation M={[4(pi)^2](r^3)}/(GT^2). Then, I solved for it, and I got 5.3E29kg.


i know the distance given to me was like the diameter, so technically, the radius would be 1.8E12; also, is M the mass of one binary star, considering this question is saying a pair of binary stars...

can someone tell me what i did wrong?

 

[ak1948]

I get M={[16(pi)^2](r^3)}/(GT^2) if you use R eq. half the distance. Because, in that case, (2pi*r)/T=sqrt((GM)/2r). Because the force due to gravity is (M^2)G/4(R^2)

 

[lindz.12]

thank you!

 

[D H]

Both of these results are wrong.

keplar's law...
so I rearranged the formula and set (2pi*r)/T=sqrt((GM)/r), and then I solved for M, which gave me the equation M={[4(pi)^2](r^3)}/(GT^2). Then, I solved for it, and I got 5.3E29kg.

I get M={[16(pi)^2](r^3)}/(GT^2) if you use R eq. half the distance. Because, in that case, (2pi*r)/T=sqrt((GM)/2r). Because the force due to gravity is (M^2)G/4(R^2)

So, what's wrong? lindz.12, your mistake was in using Kepler's laws. These laws are an approximation that implicitly assume that the mass of the orbiting body is very small compared to the mass of the central body. ak1948, your mistake was in using an invalid equation.

Kepler's third law can be extended to cover the case of a pair of masses orbiting one another such that neither mass can be deemed to be negligibly small. In this case,

\tau^2 = \frac{4\pi^2a^3}{G(M_1+M_2)}

where a is the semi-major axis of the orbit of the bodies about each other (i.e., not about their center of mass).Edit:
Solving for the masses,

M_1+M_2 = \frac{4\pi^2a^3}{G\tau^2}

 

[ak1948]

DH
I think its the same result: what you call "a" I called "2R". what you call M1+ M2 I call 2M.