/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Many of the stars in the sky are... [FREE SOLUTION] | 91Ó°ÊÓ

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Many of the stars in the sky are actually binary 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 spectruscopic binary stars. Figure 37.30 (next page) 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 \(4.568110 \times 10^{14} \mathrm{Hz}\) . In the light received from the stars by a tele- scope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{Hz}\) and \(4.568910 \times 10^{14} \mathrm{Hz}\) . 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, \(1.99 \times 10^{30} \mathrm{kg}\) . Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11} \mathrm{m}\) . (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 binary star system is moving towards Earth at 39.3 m/s. Each star's speed is 26.2 km/s, the orbital radius is \(3.96 \times 10^{10}\) m, and each star's mass is \(3.2 \times 10^{30}\) kg.

Step by step solution

01

Calculate the Difference in Frequency

The given frequency of hydrogen light in the laboratory is \(f_0 = 4.568110 \times 10^{14} \text{Hz}\). The observed frequencies from the stars vary between \(f_{min} = 4.567710 \times 10^{14} \text{Hz}\) and \(f_{max} = 4.568910 \times 10^{14} \text{Hz}\). The change in frequency \(\Delta f\) is given by \(\Delta f = f_{max} - f_{min} = 4.568910 \times 10^{14} - 4.567710 \times 10^{14} = 0.001200 \times 10^{14} \text{Hz}\).
02

Determine the Direction of the Binary System

Since the maximum observed frequency \(f_{max}\) is greater than \(f_0\), the binary system is moving towards the Earth. Doppler shift results in increased frequency when the source is approaching. Hence, the system moves towards Earth.
03

Calculate the System's Radial Velocity

Using \(\Delta f / f = u / c\), solve for the speed of the system \(u\):\[\frac{0.000600 \times 10^{14}}{4.568110 \times 10^{14}} = \frac{u}{3 \times 10^8}\]This gives:\[u = \frac{0.000600 \times 3 \times 10^8}{4.568110} = 39.3 \text{m/s}\]
04

Calculate the Orbital Speed of Each Star

Since the stars are identical and orbit around their center of mass, we use the maximum frequency change to find the orbital speed \(v\):\[\frac{\Delta f'}{f} = \frac{v}{c}\], where \(\Delta f' = \frac{4.568910 - 4.568110}{2} \times 10^{14}\).The equation becomes:\[v = \frac{0.000400 \times 10^{14} \times 3 \times 10^8}{4.568110} = 26.2 \text{km/s}\]
05

Determine the Orbital Radius \(R\)

The period \(T\) is given as 11 days = 950400 seconds. Use the formula for circular motion:\[ R = \frac{vT}{2\pi}\]This gives:\[R = \frac{26.2 \times 10^3 \times 950400}{2 \pi} = 3.96 \times 10^{10} \text{m}\], which shows the radius size relative to Earth's distance to the Sun.
06

Calculate Mass \(m\) of Each Star

The centripetal force is given by \(F = \frac{Gm^2}{4R^2}\), equal to gravitational attraction. Using \(v^2 = \frac{GM}{2R}\), solve for mass \(m\):\[ m = \frac{4\pi^2R^3}{GT^2}\]Insert values to find:\[ m = 3.2 \times 10^{30} \text{kg}\], which is about 1.6 times the mass of the Sun.

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Key Concepts

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

Spectroscopic Binary Stars
In the vastness of the night sky, many stars we see as individual points of light are actually part of what is known as binary star systems. When two stars orbit around a common center of mass, they are referred to as binary stars. Among these, spectroscopic binary stars are particularly interesting. They are called "spectroscopic" because their presence is detected through the observation of spectral lines rather than direct visual clues. This typically happens when the two stars are too close together to be distinguished with a telescope.

In these systems, the stars' orbital motion causes shifts in the wavelength of light they emit due to the Doppler effect, which we will discuss further. These shifts provide astronomers with valuable insights into the characteristics of the binary system, such as the speed at which each star is moving in its orbit.
  • Spectroscopic binaries are often identified by changes in the spectral lines.
  • The motion of the stars causes these lines to periodically shift towards the red or blue end of the spectrum.
Studying these spectral lines allows astronomers to deduce the presence of a binary system and calculate important information about the stars involved.
Doppler Shift
The Doppler shift is a change in the frequency of a wave in relation to an observer moving relative to the source of the wave. In the context of spectroscopic binary stars, this effect is crucial for discovering and studying these systems. The Doppler shift is responsible for the observed changes in the frequency of light that reaches us from binaries.

When a star in a binary system moves towards Earth, its light waves are compressed, leading to a higher frequency or blue shift. Conversely, when the star moves away, the waves elongate, resulting in a lower frequency or red shift. This periodic change in the frequencies can reveal the motion of stars within binary systems.
  • As stars orbit each other, the Doppler effect causes alternating shifts in the light's frequency.
  • A blue shift indicates the star is moving closer, while a red shift means it is receding.
The Doppler shift in these systems allows astronomers to measure how fast stars are moving in their orbits and, in some cases, whether the system as a whole is moving relative to Earth.
Orbital Speed
Orbital speed is a key characteristic in understanding the dynamics of binary star systems. It refers to the speed at which a star moves around the center of mass in its orbit. For spectroscopic binary stars, determining the orbital speed is largely made possible thanks to the Doppler shift.

By measuring the change in frequency of the starlight over time, astronomers can calculate the speed at which each of the two stars in a binary system orbits their common center of mass. This is crucial for understanding the system's dynamics and the gravitational forces at play.
  • The orbital speed is derived from the maximum change in frequency observed.
  • Knowing the orbital speed helps determine the orbital radius, providing a clearer picture of the system's scale.
Understanding the orbital speed of the stars lets astronomers estimate other factors, such as the gravitational influence each star has on its companion, ultimately aiding in revealing the mass of each star in the system.
Mass of Stars
Determining the mass of stars in a spectroscopic binary system can be quite intriguing. This is another benefit provided by observing the Doppler-shifted light they emit. Once the orbital speed and period are known, one can calculate the stars' mass using several well-established formulas in celestial mechanics.

The mass calculation hinges on the gravitational force that keeps the stars in their orbit, which is fundamentally related to their orbital speed and the radius of their orbit. By rearranging the formula for gravitational force and considering the mechanics of circular motion, astronomers can estimate each star's mass in a binary system.
  • Mass is inferred from the orbital speed, radius, and period of the orbit.
  • Utilizing Kepler's laws and Newton's law of gravitation is essential for accurate calculations.
Understanding the masses of stars in binary systems is vital for gaining insights into their evolution, lifespan, and broader implications in the structure of the galaxy. Knowing their mass is not only crucial for stellar physics but also for modeling phenomena like supernovae, neutron stars, and black holes derived from stellar evolution.

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Most popular questions from this chapter

Two events observed in a frame of reference Shave positions and times given by \(\left(x_{1}, t_{1}\right)\) and \(\left(x_{2}, t_{2}\right),\) respectively. (a) Frame \(S^{\prime}\) moves along the \(x\) -axis just fast enough that the two events occur at the same position in \(S^{\prime} .\) Show that in \(S^{\prime},\) the time interval \(\Delta t^{\prime}\) between the two events is given by $$\Delta t^{\prime}=\sqrt{(\Delta t)^{2}-\left(\frac{\Delta x}{c}\right)^{2}}$$ where \(\Delta x=x_{2}-x_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Hence show that if \(\Delta x>c \Delta t,\) there is \(n o\) frame \(S^{\prime}\) in which the two events occur at the same point. The interval \(\Delta t^{\prime}\) is sometimes called the proper time interval for the events. Is this term appropriate? (b) Show that if \(\Delta x>c \Delta t,\) there is a different frame of reference \(S\) in which the two events occur simultaneously. Find the distance between the two events in \(S^{\prime} ;\) express your answer in terms of \(\Delta x, \Delta t,\) and \(c\). This distance is sometimes called a proper length. Is this term appropriate? (c) Two events are observed in a frame of reference \(S^{\prime}\) to occur simultancously at points separated by a distance of 2.50 \(\mathrm{m}\) . In a second frame \(S\) moving relative to \(S^{\prime}\) along the line joining the two points in \(S^{\prime},\) the two events appear to be separated by 5.00 \(\mathrm{m}\) . What is the time interval between the events as measured in \(S ?[\text { Hint: Apply the result obtained in part (b).1 }\)

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

In high-energy physics, new particles can be created by collisions of fast- moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive \(\mathrm{kaon}\left(\mathrm{K}^{+}\right) :\) $$p+p \rightarrow p+p+\mathbf{K}^{-}+\mathbf{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is zero. See Problem 8.100 , but note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kimetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

A cube of metal with sides of length \(a\) sits at rest in a frame \(S\) with one edge parallel to the \(x\) -axis. Therefore, in \(S\) the cube has volume \(a^{3} .\) Frame \(S^{\prime}\) moves along the \(x\) -axis with a speed \(u\) . As measured by an observer in frame \(S^{\prime},\) what is the volume of the metal cube?

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts \((a)\) and \((b)\) , find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

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