In a certain binary-star system, each star has the same mass which is 8.7 times
ID: 1264239 • Letter: I
Question
In a certain binary-star system, each star has the same mass which is 8.7 times of that of the Sun, and they revolve about their center of mass. The distance between them is the 8.8 times the distance between Earth and the Sun. What is their period of revolution in years?
In a certain binary-star system, each star has the same mass as our Sun, and they revolve about their center of mass. The distance between them is the same as the distance between Earth and the Sun. What is their period of revolution in years?
The answer to the above question is .71 years and I don't need to see how to solve this one, but I'm not sure how to do the first question.
Thank you.
Explanation / Answer
Kepler's third law:
(M1 + M2) = a^3 / P^2
where M is the mass in Solar masses, a is the distance in AU, and P is the period in years.
In your example:
M1 = M2 = Msol
a = 1 AU
so
P = (a^3 / (M1 + M2))^(1/2)
P = (1^3 / (1 + 1))^(1/2)
P= (1/2)^(1/2)
P = 0.707 years