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Chapter 5: Problem 15

We observe a binary system in which the two stars are 1 and 2 arc sec, respectively, from the center of mass. The system is 10 pc from us. The period is 33 yr. What are the masses of the two stars, assuming that \(i=90^{\circ} ?\)

Short Answer

Expert verified
The masses of the two stars are \(M_1 = 3.04\ M_{\odot}\) and \(M_2 = 1.52\ M_{\odot}\).

Step by step solution

01

Calculate the total separation

The total separation between the two stars is the sum of their angular distances from the center of mass. Given that the two stars are 1 and 2 arc seconds apart respectively, the total angular separation is 3 arc seconds.
02

Convert angular separation to linear separation

The linear separation can be calculated using the formula for small angles: \(D = \theta \cdot d\), where \(\theta\) is the angular separation in radians and \(d\) is the distance to the binary system. First, convert arc seconds to radians: \(3\ arcsec = 3 \times \frac{\pi}{648000} = 1.4544 \times 10^{-5}\ radians\). Then, \(D = 1.4544 \times 10^{-5}\ \cdot 10\ pc = 1.4544 \times 10^{-4}\ pc\).
03

Use Kepler's Third Law

Kepler's Third Law states that \(P^2 = a^3 / (M_1 + M_2)\), where \(P\) is the period in years, \(a\) is the semi-major axis in AU, and \(M_1 + M_2\) is the sum of the masses of the two stars in solar masses. First, convert the distance from parsecs to AU: \(1.4544 \times 10^{-4}\ pc \times 206265 \ AU/pc = 30.0\ AU\). Thus, in the case of a binary system \(P^2 = (30/2)^3 / (M_1 + M_2)\).
04

Calculate the sum of masses

Re-arranging the formula to solve for \(M_1 + M_2\) yields: \(M_1 + M_2 = (30/2)^3/33^2 = 4.56\ M_{\odot}\), where \(M_{\odot}\) is solar masses.
05

Determine individual masses

Assuming the distance of 1 arc second corresponds to \(M_2\) and 2 arc seconds to \(M_1\), we have \(M_1 \cdot 2 = M_2 \cdot 1\). Therefore, \(M_1 = 2M_2\). Solving the two equations \(M_1 + M_2 = 4.56\ M_{\odot}\) and \(M_1 = 2M_2\), we get \(2M_2 + M_2 = 4.56\ M_{\odot}\), then \(M_2 = 1.52\ M_{\odot}\) and \(M_1 = 3.04\ M_{\odot}\).

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Key Concepts

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

Kepler's Third Law
Kepler's Third Law is a fundamental rule in astronomy that relates the period of orbit of two bodies to the distance separating them. The formula is expressed as: \[P^2 = \frac{a^3}{M_1 + M_2}\], where:
  • \(P\) is the orbital period in years.
  • \(a\) is the semi-major axis in astronomical units (AU).
  • \(M_1 + M_2\) represents the sum of the masses of the two objects in solar masses (\(M_\odot\)).
In this exercise, it helped us to determine the combined mass of the two stars, given their period (33 years) and separation (30 AU). Understanding this law is crucial for analyzing the dynamics of binary star systems.
Angular Separation
Angular separation is the angle between two celestial objects as seen from Earth. It's measured in arc seconds, where an arc second is 1/3600 of a degree.
In this problem, the stars in the binary system are 1 and 2 arcseconds from the center of mass. Therefore, the total angular separation is 3 arc seconds. To calculate the linear separation, we need to convert this angular measurement into a linear distance using the distance to the stars (10 parsecs in this case). This step is crucial to finding the real spatial separation, which is used in calculating the semi-major axis.
Mass Calculation
Solving for the masses of the stars in a binary system involves a series of steps. First, using Kepler's Third Law, we determined the sum of their masses.
  • The total mass \(M_1 + M_2\) was found to be 4.56 solar masses.
Given the proportion of their angular separations (2:1), we assigned \(M_1\) to the star further from the center and \(M_2\) to the nearer star.
We have a linear relationship: \(M_1 = 2M_2\). This lets us solve for individual masses using simple algebra, resulting in \(M_2 = 1.52 M_{\odot}\) and \(M_1 = 3.04 M_{\odot}\). This process illustrates how observations can be translated into physical properties like mass.
Parsecs to AU Conversion
Astronomical distances are often expressed in parsecs (pc). One parsec is equivalent to 206,265 astronomical units (AU). To convert parsecs to AU: \[ \text{Distance in AU} = \text{Distance in pc} \times 206,265 \] This conversion factor played a key role in this problem.
  • The linear separation in parsecs: \[ 1.4544 \times 10^{-4} \text{ pc} \]
  • Converted to AU: \[ 1.4544 \times 10^{-4} \times 206,265 = 30 \text{ AU} \]
This conversion ensured accurate usage of Kepler's Third Law to find the star masses. Understanding these basic conversions is vital in astronomy.

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Most popular questions from this chapter

For a binary system with stars of masses \(m_{1}\) and \(m_{2},\) in circular orbits, with a total separation \(r,\) find an expression for the ratios of the kinetic energies of the two stars.

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