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Chapter 13: Problem 3

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?

Short Answer

Expert verified
The stars have a period of approximately 0.707 years.

Step by step solution

01

Understand the Problem

We have two stars, each with mass equal to that of the Sun, orbiting around their common center of mass. The distance between the stars is the same as the distance between the Earth and the Sun (1 Astronomical Unit, AU). We need to find their orbital period in years.
02

Use Kepler's Third Law

According to Kepler's Third Law for systems like this, the square of the orbital period \( T \) is proportional to the cube of the semi-major axis \( a \) of the orbit. For two masses \( M_1 \) and \( M_2 \) that revolve around each other, the formula is: \[ T^2 = \frac{4\pi^2}{G(M_1 + M_2)} a^3 \]Given that \( M_1 = M_2 = M_{\text{sun}} \) and \( a = 1 \) AU.
03

Simplify with Given Values

Since the masses of the stars are equal to the mass of the Sun, the formula becomes:\[ T^2 = \frac{4\pi^2}{GM_{\text{sun}} + GM_{\text{sun}}} (1~\text{AU})^3 \]This simplifies to:\[ T^2 = \frac{4\pi^2}{2GM_{\text{sun}}} (1~\text{AU})^3 \]
04

Relation to Earth's Orbit

For Earth's orbit around the Sun, \[ T^2 = \frac{4\pi^2}{GM_{\text{sun}}} (1~\text{AU})^3 \]Here, \[ T_{\text{Earth}} = 1~\text{year} \]. Since the expression for our binary-star system is exactly half that for Earth's orbit,\[ T^2 = \frac{1}{2} T_{\text{Earth}}^2 \].
05

Solve for T

Taking the square root of both sides:\[ T = \sqrt{\frac{1}{2}} T_{\text{Earth}} \]This yields:\[ T = \frac{1}{\sqrt{2}} \text{ year} \approx 0.707~\text{year} \]
06

Verify Units and Answer

We confirmed that our units were consistent (AU, years, etc.) and the problem correctly applies Kepler's Third Law. The final period of the binary system is approximately 0.707 years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binary-Star System
A binary-star system is a fascinating astronomical structure where two stars are gravitationally bound to each other. They orbit around a common point known as the center of mass. This system is vital for understanding how gravitational forces and star dynamics work in our universe. In many cases, one star is more massive than the other, causing asymmetrical orbits. However, in a system where both stars have the same mass, such as in the given exercise, they revolve symmetrically around the center of mass.

Binary systems are significant in astronomy because they allow scientists to calculate stellar masses using orbital parameters, which is not easily achievable for single stars. The movement of these stars provides crucial insights into celestial mechanics and dynamics observed in many galaxies.
Orbital Period
The orbital period is the time it takes for an object to complete one full orbit around another object. In our case, it is about the two stars' revolution around their common center of mass. The orbital period is crucial for understanding the dynamics in the system, such as gravitational influences and stability of the orbit.

Using Kepler's Third Law, we can calculate the period if we know the semi-major axis and the masses involved. The law states that the square of the orbital period is proportional to the cube of the semi-major axis of its orbit, considering the gravitational influence of the masses in the system. For the binary-star system in our example, the formula tells us how the distances and masses correlate to determine the period, which turns out to be approximately 0.707 years.
Center of Mass
The center of mass in a binary-star system is the balance point around which the two stars revolve. It represents the average position of the mass distribution within the system. In a system with equal mass stars, the center of mass lies exactly halfway between the two stars, leading to symmetric orbits.

Understanding the center of mass is essential to grasp how the stars' orbits are organized. The position of the center of mass is determined by the relative masses of the stars and their distance from each other. This point does not move relative to the stars, and the laws of physics that govern the motions in space apply, making it a foundational concept in astrophysics.
Semi-Major Axis
The semi-major axis is a critical concept in orbital mechanics. It is the longest radius of an elliptical orbit and represents the average distance between the two orbiting bodies over time. For the given binary-star system, the semi-major axis is equivalent to 1 astronomical unit (AU), which is the average distance between Earth and the Sun.

In the context of Kepler鈥檚 Third Law, the semi-major axis plays a significant role in calculating the orbital period. When its value is plugged into the formula, it helps relate how the stars' distance impacts the duration of their orbit. Knowing the semi-major axis allows astronomers to predict the motion characteristics of celestial bodies accurately and is an integral part of solving problems related to orbital dynamics.

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Most popular questions from this chapter

What multiple of the energy needed to escape from Earth gives the energy needed to escape from (a) a moon of mass \(1.472 \times 10^{23} \mathrm{~kg}\) and radius \(3.480 \times 10^{6} \mathrm{~m}\) and (b) Jupiter?

An asteroid, whose mass is \(2.0 \times 10^{-4}\) times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is \(3.0\) times Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?

Assume a planet is a uniform sphere of radius \(R\) that (somehow) has a narrow radial tunnel through its center (Fig. 13-7). Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let \(F_{R}\) be the magnitude of the gravitational force on the apple when it is located at the planet's surface. How far from the surface is there a point where the magnitude is \(\frac{1}{3} F_{R}\) if we move the apple (a) away from the planet and (b) into the tunnel?

In 1993 the spacecraft Galileo sent an image (Fig. 13-32) of asteroid 243 Ida and a tiny orbiting moon (now known as Dactyl), the first confirmed example of an asteroid-moon system. In the image, the moon, which is \(1.5 \mathrm{~km}\) wide, is \(100 \mathrm{~km}\) from the center of the asteroid, which is \(55 \mathrm{~km}\) long. Assume the moon's orbit is circular with a period of \(27 \mathrm{~h}\). (a) What is the mass of the asteroid? (b) The volume of the asteroid, measured from the Galileo images, is \(14100 \mathrm{~km}^{3}\). What is the density (mass per unit volume) of the asteroid?

The Sun, which is \(2.2 \times 10^{20} \mathrm{~m}\) from the center of the Milky Way galaxy, revolves around that center once every \(2.5 \times 10^{8}\) years. Assuming each star in the Galaxy has a mass equal to the Sun's mass of \(2.0 \times 10^{30} \mathrm{~kg}\), the stars are distributed uniformly in a sphere about the galactic center, and the Sun is at the edge of that sphere, estimate the number of stars in the Galaxy.
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