# What is the period of a binary star system of two equal mass stars?

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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))

• If you need a number answer, I think you need to know the mass and the distance apart, right?