Many of the stars in the sky are actually binarystars, in which two stars orbit about their comm

Name: Problem 68, Chapter 37: University Physics with Modern Physics
Uploaded: 2019-05-03T23:09:38-08:00
Duration: 16 min
Channel: Ceren Uzun

step 1 This is chapter 37, problem number 68. We have a system of binary stars, and we're given the fre

Question

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.)

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.)

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 37, Problem 68

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Solved by Expert Ceren Uzun

05/03/2019

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568110 \times 10^{14}$ Hz. The frequency of the light received from the stars varies between $f_{+} = 4.568910 \times 10^{14}$ Hz and $f_{-} = 4.567710 \times 10^{14}$ Hz. We can use the Doppler shift formula $\Delta f/f = u/c$ to find the velocity of the binary Show more…

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Key Concepts

Newtonian Gravitation and Kepler's Laws

Newton's law of universal gravitation, along with Kepler's laws of planetary motion, underpins the analysis of binary star systems. These principles are applied to relate the orbital period and radius of the stars' orbits to the masses of the stars, enabling astronomers to deduce stellar masses from observed orbital parameters.

Orbital Mechanics in Binary Systems

This refers to the study of the motions of binary stars as they orbit their common center of mass. In systems where the orbits are circular, the stars exhibit symmetric Doppler shifts that provide information about their orbital speeds and, in conjunction with the orbital period, allow determination of orbital radii.

Radial Velocity Measurements

Radial velocity is the component of an object's velocity that is directed along the line of sight of the observer. By measuring the shifts in the spectral lines due to the Doppler effect, astronomers can determine the speed and direction of a star’s motion relative to the Earth.

Spectroscopic Binary Stars

Spectroscopic binary stars are systems in which two stars orbit a common center of mass in such a way that their orbital motions are revealed by periodic Doppler shifts in their spectral lines. This method is crucial in astronomical studies when the stars are too close together to be resolved by a telescope as separate objects.

Doppler Effect

This concept refers to the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the source of the wave. In the context of astronomy, the Doppler effect is used to detect changes in the spectral lines of light emitted by stars, revealing their motion toward or away from the observer.

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Transcript

00:01 This is chapter 37, problem number 68.

00:03 We have a system of binary stars, and we're given the frequency of a heated hydrogen gas measured on earth.

00:12 It's, i'm going to call it f not, 4 .568, 1 .11 times 10 to power of 14 hertz.

00:20 This will be 14.

00:24 And the frequency of the hydrogen gas that is received from the stars, that that, that, it varies in frequency.

00:33 So the higher frequency, i'm going to call it f plus.

00:37 It's 4 .56891 times centroboral 14 hertz.

00:43 And the lower one that is measured is 4 .56771 times 10 to barrel 14 hertz.

00:50 Now let's remember the figure.

00:53 We have two stars with the orbital velocities.

00:57 So they're rotating around their central mass, and this portion, this shows the direction of the earth.

01:07 Now, let's write down what b is.

01:10 This is the orbital velocity.

01:14 Let's also define as you as the velocity of the central mass, velocity of center of mass.

01:26 And now we're going to make an assumption.

01:28 Assume that it is already approaching earth.

01:33 And if it did we find it out to be negative, then it's going to be the other way around moving away from earth.

01:39 But if it comes out to be positive, then it's going towards earth, and we're right from the beginning.

01:45 And now assuming that, if you use the relativistic doppler frequency equation, so the plus is going to be t plus u plus v.

01:56 I'm writing this for the upper star here.

01:59 That's why i'm adding the velocities together, because the velocity of central masses in the same direction as the orbital velocity, and divided by c minus u plus v, f0, right? and this would be the frequency that we're measuring is higher, because as you know, from the conceptual portion of this topic, that if the object is approaching the observer, we're actually detecting a higher frequency emanating.

02:29 From it.

02:30 Now, if we do the same for the second star, the only difference is going to be now the direction of these two velocities u and v are opposite.

02:38 So it's still, though, we're assuming that it's approaching the earth.

02:41 So what we're going to have here is going to be u minus v this time, c minus u minus v, f0, right? now, what we need to do, well, what's asking in part a is that we need to determine the speed of this motion, so the orbital velocity and velocity of central mass of the stars.

03:06 So what i'm going to try and do is i'm going to try and get u plus v term out of the first equation and u minus v term out of the second equation so that i can add these two equations and finally find what u and v are.

03:21 Now, in order to do that, first things first, we're going to have to take the square of this equation, right? so f plus square is going to be c plus u plus b over c minus u plus b, f not squared.

03:37 Now let's take multiply both sides by the denominator.

03:42 So what we have then, well, let me put it here underneath.

03:48 What we have is f0, f plus squared times c minus u.

03:54 Plus v equals c plus u plus v f not squared right and now then let's actually redistribute these same thing for the right hand side so c times f plus squared minus u plus v f plus squared is going to be equal to c f not squared plus plus u plus v f not square now these two terms have the u plus v as a multiplier now let's group them right in order to do this we're going to have to take this term to the left -hand side so we're going to have c f plus squared minus c f -not square equals u plus v for both then we have f -not squared plus f plus f squared right now let's go on the next page then we have c parentheses f plus squared minus f not squared equals to u plus v f not squared plus f plus square now if we divide both sides by f not square plus f plus squared these are going to cancel if not square plus f plus now, u plus v is going to be c, f plus squared, minus f not squared, over f not squared plus f plus squared, this is our u plus b.

05:44 Now, if we do the same algebra for the second equation, same path, please do you soles, then what we're going to find for u minus v this time is going to be predomize.

06:00 The same.

06:01 The only thing that changes is instead of f plus we have f minus.

06:05 F minus squared, f not squared over f not squared plus f minus squared.

06:14 Then again in this equation we're given everything right.

06:19 We know what f plus is.

06:23 It is this value 4 .5, 4 .56891 and and we're given f minus, and we know f not.

06:36 So we know all these terms, and we know the c.

06:40 C is also three times 10 to the power of 8 meters per second, right? so if we build the algebra and plug everything in there from here, our u plus v is going to turn up to be 5 .25 times 10 to the power of 4 meters per second.

06:57 And if you plug in all the values here in the second equation, then u minus v is going to be found out to be negative 2 .63 times 10 power 4 meters per second.

07:11 Now, what we're going to do, we're going to add these two equations, equation 1 and equation 2 to each other.

07:17 So when we do this, we're adding the left -hand sides.

07:20 So u plus v plus u minus v is going to be equal to 5 .25 times center per of 4.

07:27 Minus 2 .63 times center of over 4.

07:31 This is going to be equal to, let's look at the left -hand side...

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