Two separate pairs of binary stars are observed. The separation of the pairs is identical, but the orbital period of Pair X is 4 times the orbital period of Pair Y. How does the total mass of the stars in Pair X compare to the total mass of the stars in Pair Y?

Question

Quizplus · Accepted Answer

According to Kepler's third law of planetary motion, the square of the orbital period of a planet (or binary star system, in this case) is directly proportional to the cube of the semi-major axis of its orbit. When the separation between the stars in both pairs is the same, but the orbital period of Pair X is 4 times that of Pair Y, it implies that the mass of Pair Y must be greater to account for a faster orbit. The relationship can be expressed as $T^2 \proportional M/R^3$, where $T$ is the orbital period, $M$ is the mass, and $R$ is the separation. Since the separation is constant, and $T_X = 4T_Y$, when you solve for mass, it turns out that the mass of Pair Y is 16 times greater than that of Pair X, because the square of the period ratio (16) is directly proportional to the mass ratio when the separation is constant.

Two Separate Pairs of Binary Stars Are Observed