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Many of the stars in the sky are actually \(binary\space stars\), in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called \(spectroscopic\space binary\space stars. \textbf{Figure P37.68}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m\), orbiting their center of mass in a circle of radius \(R\). The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of 4.568110 \(\times\) 10\(^{14}\) Hz. In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between 4.567710 \(\times\) 10\(^{14}\) Hz and 4.568910 \(\times\) 10\(^{14}\) Hz. Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (\(Hint\): The speeds involved are much less than \(c\), so you may use the approximate result \(\Delta f/f = u/c\) given in Section 37.6.) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, 1.99 \(\times\) 10\(^{30}\) kg. Compare the value of \(R\) to the distance from the earth to the sun, 1.50 \(\times\) 10\(^{11}\) m. (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

Short Answer

Expert verified
The star system is moving toward Earth at 52.5 km/s. The orbital speed of each star is 39.4 km/s. Solving gives the radius and mass of each star.

Step by step solution

01

Understanding the Frequency Shift

We observe that the frequency of hydrogen light from the binary star system varies between 4.567710 \(\times\) 10\(^{14}\) Hz and 4.568910 \(\times\) 10\(^{14}\) Hz. The frequency in earth laboratory conditions is 4.568110 \(\times\) 10\(^{14}\) Hz. We can calculate the change in frequency to understand the motion of the stars.
02

Calculating the Change in Frequency

The observed maximum frequency change (shift) is \(\Delta f = 4.568910 \times 10^{14} - 4.568110 \times 10^{14} = 8.00 \times 10^{10}\) Hz. The minimum frequency change is \(4.568110 \times 10^{14} - 4.567710 \times 10^{14} = 4.00 \times 10^{10}\) Hz.
03

Determining if the System Moves Toward or Away

We note a higher frequency when both stars are moving towards Earth (blue shift), compared to when they move away (red shift). Since the maximum frequency observed (\(4.568910 \times 10^{14}\) Hz) is greater than the laboratory frequency, the system as a whole is moving toward Earth.
04

Calculating the System's Motion Speed

Using the approximation \(\Delta f/f = u/c\), where \(u\) is the speed of motion and \(c\) is the speed of light, we use \(\Delta f = 8.00 \times 10^{10}\) Hz and \(f = 4.568110 \times 10^{14}\) Hz. Thus, \(u = c \times \Delta f / f = 3 \times 10^8 \times 8.00 \times 10^{10} / 4.568110 \times 10^{14} \approx 52.5 \text{ km/s}\).
05

Calculating Orbital Speeds of Stars

Using the maximum variation frequency \(\Delta f = 6.00 \times 10^{10}\) Hz which is the difference from the laboratory frequency to the mean frequency of 4.568810 \(\times\) 10\(^{14}\) Hz. Using the same equation \(\Delta f/f = v/c\), where \(v\) is the orbital speed, we get \(v = c \times \Delta f / f = 3 \times 10^8 \times 6.00 \times 10^{10} / 4.568110 \times 10^{14} \approx 39.4 \text{ km/s}\).
06

Determining the Orbital Radius

Given that light varies from max to min in 11 days, this is half a period \(T\). Since \(T = 2 \times 11 \text{ days}\), the complete period is 22 days. Using \(v = 2\pi R/T\), we solve for \(R\): \(R = vT / (2\pi) = 39.4 \times 10^3 \times 22 \times 24 \times 3600 / (2\pi)\) meters.
07

Calculating the Mass of Each Star

Using Newton's law of gravitation \(F = G \times m^2 / (4R^2) = m v^2 / R\), solve for \(m\). We substitute known \(v\), \(R\), and \(G\), and solve for \(m\).
08

Comparing to Known Distances and Masses

Having calculated \(R\) and \(m\), compare \(R\) to Earth's distance from the sun \((1.50 \times 10^{11} \text{ m})\) and \(m\) to sun's mass \((1.99 \times 10^{30} \text{ kg})\).

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Key Concepts

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

Doppler effect
The Doppler effect is a fascinating phenomenon that refers to the change in frequency or wavelength of a wave as observed by someone moving relative to the wave source. When it comes to astronomy, this effect helps us study stars and galaxies. Especially in binary star systems, the Doppler effect reveals vital insights about the star's motion.

If a star or galaxy moves toward us, the light waves become compressed, leading to a higher frequency, which is called a 'blue shift'. Conversely, if the object is moving away, the waves stretch out, resulting in a lower frequency, known as a 'red shift'. This is what allows astronomers to determine whether a binary star system is approaching or receding from Earth just by measuring the shift in frequency of their light.

By calculating the frequency shift using the formula \( \Delta f / f = u / c \), where \( u \) is the star's speed and \( c \) is the speed of light, astronomers can gauge whether these celestial bodies are drawing closer or moving farther. This is why the Doppler effect is integral to understanding motions in the cosmos.
spectroscopic binary stars
Spectroscopic binary stars are pairs of stars that are so close together on the sky that they cannot be individually distinguished by a telescope. Their existence is inferred from the Doppler effect, as they orbit around a common center of mass.

As the stars move in their orbits, they alternately speed toward and away from us, causing shifts in the wavelengths of their emitted light. This change in wavelength is detected in the spectral lines, hence the term 'spectroscopic'. By observing these lines over time, astronomers can uncover the system's orbital period and even calculate the orbital speeds of the stars.

The precise measure of the frequency shift provides clues to the dynamics of the binary system. The greater the shift, the higher the velocity of the stars in their orbits. Spectroscopic binaries are essential because they often reveal individual characteristics and dynamics of the stars that may not be observed in solitary stars.
orbital mechanics
Orbital mechanics, or celestial mechanics, is the study of the motions of celestial bodies under the influence of gravitational forces. It applies Newton's laws of motion and universal gravitation to predict and explain the trajectories of planets, moons, and stars.

In the context of binary stars, orbital mechanics allows us to determine crucial orbital parameters, such as the period, radius, and velocity of the stellar bodies. Observing the orbital motion helps to construct a mathematical model of the system that reveals the path and behavior of these stars over time.
  • Use Kepler's laws to describe the motion of binary stars.
  • Measure the orbital velocity to find the radius of the orbit using \( v = 2\pi R / T \)
  • Calculate the period by observing how long it takes one object to complete its orbit.


Orbital mechanics is key in understanding not just the paths of stars, but also helps in predicting future movements and potential interactions with other celestial bodies.
stellar mass determination
Determining the mass of stars in binary systems is vital for understanding their physical characteristics and evolution. One of the most effective methods for finding stellar mass is through observations of binary stars.

The gravitational interactions in a binary star system, when combined with Kepler's laws and Newton's law of gravitation, provide the necessary information to determine each star's mass. For instance, by knowing the orbital radius \( R \) and velocity \( v \) of the stars, one can use the formula
\[ F = G \frac{m_1 m_2}{r^2} = m v^2 / R \]
to solve for mass \( m \). Here, \( G \) represents the gravitational constant.
  • Use the stars' velocities to calculate the gravitational force.
  • Apply the determined force to calculate the mass by rearranging the formula \( m = v^2R / G \).


This method provides not just the mass values but offers insights into how stars interact and affect each other's evolution. Knowing stellar masses is fundamental for constructing accurate models of how stars evolve, form planetary systems, and end their life cycles in phenomena such as supernovae.

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Most popular questions from this chapter

Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 7.50 \(\times\) 10\(^5\) eV. (a) What is the ratio of the speed \(v\) of an electron having this energy to the speed of light, \(c\)? (b) What would the speed be if it were computed from the principles of classical mechanics?

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame \(S'\) moving relative to \(S\), the second event is observed to occur 2.15 s after the first. What is the difference between the positions of the two events as measured in \(S'\)?

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of 0.890c. Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

In the earth's rest frame, two protons are moving away from each other at equal speed. In the frame of each proton, the other proton has a speed of 0.700\(c\). What does an observer in the rest frame of the earth measure for the speed of each proton?

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

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