What is the period of a binary star system of two equal mass stars?
- 1 decade agoFavorite Answer
That will depend on the stars' mass and their distance from each other. However, given these two numbers, we can - and will - derive an equation that will give you the period.
The two stars will revolve around their common center of mass, attracting each other with a force F = G*m1*m2 / d², where F is the force, m1 and m2 are the masses of the two stars, and d is their distance from each other. If the stars' masses are the same, then m1 = m2 = m.
Furthermore, if we assume that the orbit is circular, then the radius of that circle r will be ½d. Also, because the stars are in a circular orbit, they will be rotating around their common center of mass at an angular velocity ω given by the equation
a = ω²r where
a is the gravitational acceleration of each star,
and ω = 2π / T, where T is the period of rotation.
Coupled with Newton's Second Law F = ma and some algebra, we now have enough to come up with a formula for the period of revolution for the binary star system:
F = ma = m(2π / T)²r = Gm² / (2r)²
Now it is merely a matter of solving for T, which gives
T = 4πr^(3/2) / √(Gm)
If r is in meters, G is 6.67e-11 N∙m²/kg² and m is in kg, the period T will be in seconds. If you prefer to work with distance d instead, the formula then becomes
T = π * d^(3/2) * √(2 / (G*m))
- ErebosLv 61 decade ago
If you need a number answer, I think you need to know the mass and the distance apart, right?