Problem 72 - University Physics with Modern Physics

Question

Binary Star-Different Masses. 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 \(T=2 \pi\left(R_{1}+R_{2}\right)^{3 / 2} / \sqrt{G\left(M_{1}+M_{2}\right)}\) . (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 \(\mathrm{km} / \mathrm{s}\) . 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 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. 13.27\() .\) The orbital period of \(\mathrm{A} 0620-0090\) 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.

Step-by-Step Solution

Answer

For parts (a-b), derived ratios and period formula. For part (c-d), calculated masses and orbits using velocity and radius formulas for circular motion.

Step 1: Understand the System

First, identify that both stars revolve around a common center of mass. The distance from each star to this center of mass is different, and their masses and radii play a crucial role in the analysis.

Step 2: Apply the Center of Mass Concept

For two masses in a binary system, the center of mass equation is: \(M_1 R_1 = M_2 R_2\), where \(R_1\) and \(R_2\) are the distances from the center of mass to masses \(M_1\) and \(M_2\), respectively. This ensures their moments about the center of mass are equal.

Step 3: Show the Ratio of Radii Equals Reciprocal Mass Ratio

From the equation \(M_1 R_1 = M_2 R_2\), divide both sides by \(M_1 M_2\) to get \(\frac{R_1}{R_2} = \frac{M_2}{M_1}\). This demonstrates part (a) of the problem.

Step 4: Explain Equal Orbital Periods

Both stars share a common center of rotation, sharing the same angular velocity \(\omega\). This ensures they have the same orbital period \(T\), given by \(T = \frac{2\pi}{\omega}\).

Step 5: Derive the Orbital Period Formula

For circular motion, use \(\omega^2 = \frac{G(M_1 + M_2)}{(R_1 + R_2)^3}\). The formula for the period then is \(T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{(R_1 + R_2)^3}{G(M_1 + M_2)}}\), confirming part (b).

Step 6: Use Known Speed to Find Masses (Part c)

For circular orbits, the centripetal force \(F_c = M v^2/R\) must equal gravitational force \(F_g = G\frac{M_1 M_2}{(R_1 + R_2)^2}\). Given speeds, apply \(v = \frac{2\pi R}{T}\) to find the stars' masses using their given speeds and orbital period.

Step 7: Calculate System "A0620-0090" (Part d)

Using the given mass ratios \(M_1 = 0.67M_\odot\), \(M_2 = 3.8M_\odot\), apply the period formula \(T = 2\pi \sqrt{\frac{(R_1 + R_2)^3}{G(M_1 + M_2)}}\) to find \(R_1\) and \(R_2\). Use \(v = \omega R\) to find the speeds and compare to Earth’s properties.

Key Concepts

Center of Mass

In a binary star system, the two stars orbit around a shared center of mass. This is the point where the masses balance each other out, and it plays a crucial role in understanding their motions. The positions of the two stars relative to this center are not just a visual spectacle but are mathematically governed.
The relationship between their masses and how far each star is from the center of mass is given by the equation:

• \(M_1 R_1 = M_2 R_2\)

This ensures that their "moments," or the product of mass and radius, are equal. It's like a seesaw in balance, where one heavy end counteracts a lighter end farther away. Solving for the distances gives us their positional ratio, which helps us understand their orbits better.
This idea is mirrored in how planets orbit stars, but uniquely applied to systems with two orbiting bodies.

Orbital Period

The orbital period is the time it takes for a star or any orbiting body to make one full rotation around its partner. In a binary star system, something fascinating occurs - both stars complete their orbits in the same period. This is because they share the same center of rotation and thus have identical angular velocities. Their periods, therefore, remain synchronized despite the potential differences in mass and radii.
The expression for the period, derived from their identical angular velocity, is:

• \( T = 2\pi \sqrt{\frac{(R_1 + R_2)^3}{G(M_1 + M_2)}}\)

This formula shows us how the period is dependent not just on the orbital paths of the stars, but also crucially on their combined mass and the gravitational influence between them. It's a dance dictated by math and physics, not unlike the rhythms of celestial music.

Gravitational Force

Gravitational force, the glue holding stars in their cosmic orbits, operates with an unseen yet profound strength. In a binary star system, this force is calculated through Newton's Law of Universal Gravitation:

• \(F_g = G\frac{M_1 M_2}{(R_1 + R_2)^2}\)

where \(G\) is the gravitational constant. This law tells us that the gravitational pull between the stars increases with larger masses and decreases with greater distances.
Gravitational force doesn't only keep stars revolving around each other but also maintains the stability of their orbital paths. It's the tug-of-war zero-sum that ensures they stay bound in their intricate journey together across the cosmos. Without this force, the stars would drift apart, losing their celestial partnership.

Centripetal Force

As stars whip around the center of mass in a binary system, an inward force called centripetal force acts on them, keeping them in their circular orbits. This is not a separate force but a role played by the gravitational pull between the stars. The expression for this force in circular motion is:

• \(F_c = M v^2 / R\)

Centripetal force ensures that as stars gain velocity, they don't just fly off into space. The balance between this inward force and the outward inertia is what creates stable orbits.
Without this force, there would be no circular paths, causing chaos for celestial mechanics. Thus, centripetal force and gravitational force together maintain the graceful arcs of binary star orbits in harmonious equilibrium.