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 $T=2 \pi\left(R_{1}+R_{2}\right)^{3 / 2} / \sqrt{G\left(M_{1}+M_{2}\right)} .(\mathrm{c})$ The two stars in a certain bi- nary star system move in circular orbits. The first star, Alpha, has an orbital speed of $36.0 \mathrm{~km} / \mathrm{s}$. The second star, Beta, has an orbital speed of $12.0 \mathrm{~km} / \mathrm{s}$. The orbital period is $137 \mathrm{~d}$. 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 $\mathrm{A} 0620-0090$. 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 $\mathrm{A} 0620-0090$ is 7.75 hours, the mass of $\mathrm{V} 616$ 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 $T=2 \pi\left(R_{1}+R_{2}\right)^{3 / 2} / \sqrt{G\left(M_{1}+M_{2}\right)} .(\mathrm{c})$ The two stars in a certain bi-
nary star system move in circular orbits. The first star, Alpha, has an orbital speed of $36.0 \mathrm{~km} / \mathrm{s}$. The second star, Beta, has an orbital speed of $12.0 \mathrm{~km} / \mathrm{s}$. The orbital period is $137 \mathrm{~d}$. 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 $\mathrm{A} 0620-0090$. 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 $\mathrm{A} 0620-0090$ is 7.75 hours, the mass of $\mathrm{V} 616$ 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

Kepler’s Third Law

Kepler’s Third Law of planetary motion states that the square of the orbital period of two bodies is proportional to the cube of the sum of the distances from the center of mass (or the semi-major axis in more general orbits) divided by the total mass of the system. This law, when combined with Newton’s law of gravitation, provides a quantitative relationship that allows determination of orbital periods and distances in binary star systems and other gravitationally bound systems.

Gravitational Force and Circular Motion

Gravitational attraction provides the necessary centripetal force to keep bodies in circular orbits. By equating the gravitational force between two objects to the centripetal force required for circular motion, one can derive relationships among the masses, orbital radii, and orbital speeds. This fundamental approach is key to analyzing and solving problems in orbital mechanics for systems with circular orbits.

Center of Mass

The center of mass is the point in a system of masses where the weighted relative position of the distributed mass sums to zero. In binary systems, each object orbits about this common point, and the distances of the objects from the center of mass are inversely proportional to their masses. This concept simplifies analysis of orbital mechanics by reducing a two-body problem to motion about a single reference point.

Orbital Dynamics in Binary Systems

In a binary system, two bodies in mutual gravitational attraction revolve in orbits around their common center of mass. The dynamics of these systems involve understanding how the masses and distances relate to each other, ensuring that both bodies have the same orbital period despite differing orbital radii and speeds. This equal-period motion is a consequence of the conservation of momentum and the mutual gravitational interaction between the bodies.

Transcript

00:01 So starting off with part a for question 64, the radii r1 and r2 are measured with respect to the center of mass.

00:08 And so m1r1 is equal to m2 r2 and r1 divided by r2 is equal to m2 divided by m1.

00:20 So for part b, part b asks us to explain why the two stars have the same orbital period.

00:35 So, excuse me.

00:37 We have f g is equal to g m1 m2 over r1 plus r2 squared so that equivalent expressions for the period would be m2 t squared is equal to four pi squared r1 times r1 plus r2 squared divided by g and m1 t squared equal to four pi squared is equal to four pi squared r2 times r1 plus r2 squared divided by g.

01:17 And so adding these expressions gives you m1 plus m2 times t squared is equal to 4 pi squared times r1 plus r2 cubed over g.

01:33 Or t equals 2 pi r1 plus r2 to the 3 half.

01:42 Has power divided by the square root of g times m1 plus m2.

01:49 So for part c, first we must find the radii of each orbit given the speed and period data.

02:00 So we have v equals 2 pi r over t, which goes to r equals vt over 2 pi.

02:10 So r a is equal to 36 times 10 to the third meters per second times 1 .5.

02:17 37 days times 86 ,400 seconds per day, divided by 2 pi.

02:27 And that gives you 6 .78 times 10 to the 10 meters.

02:32 And then r of beta equals 12 times 10 to the third times 137 days times 86 400 seconds per day divided by 2 pi...