# Another planetary motion ? sorry!

1. Nov 24, 2008

### 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?

2. Nov 25, 2008

### 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)

Last edited: Nov 25, 2008
3. Nov 25, 2008

### lindz.12

thank you!

4. Nov 25, 2008

### D H

Staff Emeritus
Both of these results are wrong.

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,

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

Last edited: Nov 25, 2008
5. Nov 25, 2008

### ak1948

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