Two stars, with masses M1 and M2, are in circular orbits...

Two stars, with masses ${M_1}$ and ${M_2}$, are in circular orbits around their center of mass. The star with mass ${M_1}$ has an orbit of radius ${R_1}$; the star with mass ${M_2}$ has an orbit of radius ${R_2}$. (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses$-$that is, ${R_1}$/${R_2}$ $=$ ${M_2}$/${M_1}$. (b) Explain why the two stars have the same orbital period, and show that the period $T$ is given by $T = 2\pi$(R1 + R2)$^{3/2}$/$\sqrt{G(M1 + M2)}$. (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 km/s. The second star, Beta, has an orbital speed of 12.0 km/s. The orbital period is 137 d. What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.28). The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

Two stars, with masses ${M_1}$ and ${M_2}$, are in circular orbits around their center of mass. The star with mass ${M_1}$ has an orbit of radius ${R_1}$; the star with mass ${M_2}$ has an orbit of radius ${R_2}$. (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses$-$that is, ${R_1}$/${R_2}$ $=$ ${M_2}$/${M_1}$. (b) Explain why the two stars have the same orbital period, and show that the period $T$ is given by $T = 2\pi$(R1 + R2)$^{3/2}$/$\sqrt{G(M1 + M2)}$. (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 km/s. The second star, Beta, has an orbital speed of 12.0 km/s. The orbital period
is 137 d. What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an
orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.28). The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be
0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the
orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

Key Concepts

Center_of_Mass

In a two-body system, the center of mass is the unique point where the mass-weighted positions of the bodies balance each other out. For two bodies, the distances from each object to the center of mass are inversely proportional to their masses. This means that the more massive object will be located closer to the center of mass, which is fundamental in understanding the dynamics of binary systems.

Circular_Orbit_Dynamics

In circular orbits, the gravitational force between two bodies acts as the centripetal force required to keep them in orbit. By setting the gravitational force equal to the centripetal force for each body, one can derive key relationships such as the dependence of orbital radius on mass and speed. This concept is essential when analyzing the motion of objects in a binary system.

Keplers_Third_Law_Generalization

Kepler’s Third Law, when generalized using Newton’s law of gravitation, relates the orbital period of two bodies to the sum of their masses and the size of the orbit. Specifically, the period is proportional to the three-halves power of the sum of the orbital radii divided by the square root of the combined mass, providing a powerful tool for determining orbital characteristics in binary systems.

Conservation_of_Momentum_in_Binary_Systems

In binary star systems, conservation of momentum guarantees that both stars must complete their orbits in the same period despite their different masses and orbital radii. This principle underpins the derivation of relationships between the masses and orbital radii, ensuring that the system’s overall momentum remains zero in the center of mass frame.

Transcript

00:01 Okay, so for this problem, so first, since the center is at their center of the mass.

00:08 So according to the definition of the center of mass, we have mass 1 times radius 1, equal to mass 2 times radius 2.

00:16 So we can know that the ratio relation is r1 divided by r2 is equal to mass 2 divided by mass 1.

00:23 So for question b, the force between the two stars is the gravitational force, which is the gravitational constant, times the mass of this two divided by the distance between them, which is the sum of this r1 and r2 and this square.

00:42 So the force is also the centropical force, so we can use the centropical equation.

00:48 So for star 1, you have m1, velocity 1 squared divided by r1.

00:56 Since the velocity is the circumference divided by the period, we can get the expression.

01:03 Of the centrifugee force by, okay, we're talking here, we have 4 pi square r1, m1, divided by the period square.

01:14 So this one is the force between the two stars.

01:20 From question one, we know that m1, r1 equal to m2 r2, so here, no matter whether it's r1, m1, or r2, m2, they are equal to each other.

01:33 Okay.

01:34 So since they have, have the same force, the numerator is the same, the dominator should be the same.

01:40 So this should have the same period.

01:45 Now since they have the same period, we name it t and we compare the equation the left side here and the right side here.

01:56 We can have, we have m1, m1 get cancelled.

02:00 We have that m2 times t squared, put you 4 pi square r1 times r1 plus r2 squared divided by the gravitation of 10g.

02:12 So similarly, we can also have m for m1, we have m1, t squared, equal to 4 pi r2, r1 plus r2 squared, divided by 2.

02:24 So we add these two equation together.

02:27 We have the sum of the mass times t square equal to 4 pi square, the sum of the radius cube, divide by g.

02:37 And we divide m1 plus m2 on both sides, take a square root.

02:43 We have the expression of t as showing in the problem.

02:51 So for questions three, we know the velocity, we know the radius, we know the velocity, we know the period, right? so we have this equation that velocity is equal to the circumference 2xar divided by the period.

03:11 So we can find the radius of the two stars...

Please give Ace some feedback