Question

Astronomers discover a binary star system that has a period of 95.0days . The binary star system consists of two equal-mass stars, each with a mass twice that of the sun, that rotate like a dumbbell about the center of mass at the midpoint between them.

Answer 1

How far apart are the stars?

Answer 2

For binary stars, an equation commonly used for calculating specific properties is:
\[T = \sqrt{\frac{ 4\pi^2a^3 }{ GM }}\]This can be rewritten for what we are looking for here, which is a, or the mean distance between the two stars in the binary system.
\[a = \sqrt[3]{\frac{ T^2GM }{ 4\pi^2 }}\]The units for each variable should be geared towards the units of the gravitational constant, which are: \[m^3kg^{-1} s^{-2}\]This means that T should be measured in seconds, M in kilograms, and this will allow a to end up in meters.

In addition, M is the sum of m1 and m2 of the two stars, which are equal in mass and have a magnitude twice that of the Sun.

Defining our variables:
T = 0.95 days x (24 h / 1 day) x (60 min / 1 h) x (60 s / 1 min) = 82,080 s
G = 6.67 x 10^-11 m^3 kg^-1 s^-2
M = m1 + m2 = 2(1.9891 x 10^30) + 2(1.9891 x 10^30) = 7.956 x 10^30 kg
a = ? m

Now to substitute and solve it!
\[a = \sqrt[3]{\frac{ (82,080)^2(6.67x10^{-11})(7.956x10^{30}) }{ (4\pi^2) }}\]This yields a value (thanks to calculators!) of: 4.49 x 10^9 m, or 4.49 x 10^6 km.

I personally have never calculated properties of binaries, so I hope my brief research and knowledge have given adequate insight into this problem!

Answer 3

You should be able to calculate this from the fact that the centripetal force on each star, is provided by the gravitational force between the stars.

You'll need to use this:
\[\Large F _{c} = F _{g}\] where
\[\Large F _{c} = \frac{ m v^2 }{ r }\] and
\[\Large F _{g} = \frac{ G m _{1} m _{2} }{R^2}\]

The gravitational attraction acts over a distance R (the distance between the stars) and the centripetal force acts over half that distance (since the stars orbit about their common center of mass). So r = 0.5R, and m1=m2 in this case since the masses are equal.

Period is found from distance/velocity (or circumference/velocity here, as the stars have a circular orbit). For period T, (95 days which you'll probably want to convert into seconds), \[\Large T = \frac{ 2 \pi r }{ v }\] where r is again 0.5R and v is the same velocity as above.

Now you put it all together using Fc=Fg and find R.

Answer 4

Note that if you put everything together, you'll essentially get exactly the formula that @Pompeii00 posted. First rearrange the T = 2 pi r/v, for v = (2 pi r)/T, then plug that into Fc = mv^2/r (replace any r's with 0.5R) and equate with gravitational force, and solve for R.

Answer 5

Application of Kepler's third law: Law of periods