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Problems
4.1. What is the limiting magnitude for naked eye length eyepiece. In the image you find that
viewing with a 5 m diameter telescope? the angular separation between two stars is
4.2. Estimate the angular resolution of a 5 m 10 arc sec. What is the actual angular separa-
diameter telescope in space. tion on the sky between the two stars?
4.3. Compare the collecting areas of 5 and 8 m 4.7. The focal length of the objective on your tele-
diameter reflectors. Comment on the signifi- scope is 0.5 m. (a) What focal length eyepiece
cance of this comparison. would you have to use to have the image of
4.4. The full Moon subtends an angle of approxi- the full Moon (whose actual size is 30 arc
mately 30 arc min. How large would the min) subtend an angle of 2 ? (b) If you then
image of the Moon be on your film if you used took a photograph with a 500 mm focal
a 500 mm focal length lens for your camera? length camera lens, how large would the
4.5. If we have two objects ( ) apart on the sky, image be on the film?
how far apart, x, are their images on the film 4.8. Scale the results in Example 4.1 to write an
of a camera with a focal length f. (Assume expression for the limiting magnitude of a
that we wish to express x and f in the same telescope of diameter, assuming that you will
units.) be viewing directly with your eye.
4.6. The focal length of the objective on your tele- 4.9. Suppose some star is at the limit of naked eye
scope is 0.8 m. You are using a 25 cm focal visibility (m 6). How much farther away can
80 PART I PROPERTIES OF ORDINARY STARS



while your CCD has an efficiency of 80%. How
we see the same object with a telescope of
diameter D? Evaluate your answer for D 5 m. long will the needed CCD images take relative
to the time for the photograph?
4.10. Suppose that we use a reflector in the coud©
4.18. The sodium D lines in the Sun™s spectrum are
focus, and each of the three mirrors reflects
at wavelengths of 589.594 and 588.997 nm.
95% of the light. What fraction of the light is
(a) If a grating has 104 lines/cm, how wide
lost as a result of these three reflections?
*4.11. (a) Using the fact that the limiting magnitude must the grating be to resolve the two lines
of the eye is 6, derive an expression for the in first order? (b) Under these conditions
limiting magnitude for direct viewing with a what is the angular separation between the
telescope of diameter D. (Ignore the effects of two lines? (c) How would the results in (a) and
sky brightness.) (b) Use this result to derive an (b) change for second order?
4.19. A diffraction grating has N lines, a separation
expression for the farthest distance at which
a telescope of diameter D can be used to see d apart. The spectrum is projected on a screen
an object of absolute magnitude M. a distance D (Wd) from the grating. Two lines
4.12. (a) What is the diameter of a single telescope are and apart. How far apart are
with the same collecting area as the Multiple they on the screen?
Mirror Telescope? (b) Astronomers have pro- 4.20. If we want to observe at a wavelength of 10 m,
posed a new telescope with a total collecting what are the largest fluctuations that the
area equal to that of a single 25 m diameter mirror surface can have?
telescope. How many 4 m diameter telescopes 4.21. What are the angular resolutions of the KAO,
would be needed to make up this new tele- SOFIA, IRAS, ISO and SIRTF at wavelengths of
scope? (c) The Very Large Telescope, being 100 m?
built by ESO in Chile, has four telescopes, 4.22. Two infrared sources in the Orion Nebula are
each with an 8 m diameter area. What would 500 pc from us and are separated by 0.1 pc.
be the diameter of a single telescope with the How large a telescope would you need to distin-
same collecting area? guish the sources at a wavelength of 100 m?
4.13. Suppose you have a Cassegrain telescope at *4.23.Suppose we are observing an infrared source
home, with a 0.25 m diameter primary that is 500 pc away. It radiates like a 50 K
mirror and a secondary mirror with a diame- blackbody and is 1 pc in extent. (a) What is
ter of 5 cm. What fraction of the primary is the total energy per second per square meter
blocked by the secondary? reaching the Earth from this source? How
4.14. If we want to double the image size in a par- does that compare with the total amount of
ticular observation, by what amount would solar radiation reaching the Earth per second
we have to change the exposure length to per square meter. (b) Suppose we observe this
have a properly exposed photo? source using a satellite with a 1 m diameter
4.15. Scale the results in Example 4.2 to write an mirror, and we observe at a wavelength of
expression for the angular resolution, in sec- 100 m. What is the energy/s/Hz striking the
onds of arc ( ), of a telescope of diameter D for telescope? (c) Suppose the telescope radiates
viewing the middle of the visible part of the like a blackbody at 300 K, but with an effi-
spectrum. ciency of 1%. (That is, the spectrum looks like
4.16. What is the angular resolution of the HST at that of a blackbody but with an intensity
200 nm wavelength? reduced by a factor of 100.) What is the
4.17. Generally CCDs have fewer picture elements energy/Hz/s given off by the telescope at this
(pixels) than do photographic plates, so if you wavelength? How does your answer compare
want to image a large field, a single photo- with that in (b). (d) Redo part (c), assuming
graph might suffice, but you would need a that we can cool the mirror to 30 K (still with
number of CCD images. Suppose you had to a 1% emission efficiency).
make a 3 3 square of CCD images to cover 4.24. Suppose we are using an interference filter at
your single photographic field, and that your a wavelength of 10 m. (a) How far do you
photograph has a quantum efficiency of 5% have to move the plates to go from one order
4 TELESCOPES 81



maximum to the next? (b) For any given (power/surface area/Hz) in each case? (c) How
order, how far do you have to move the plates does this compare with the power you receive
for the peak wavelength to shift from 10.00 m from a 50 kW radio station 10 km from you?
to 10.01 m? 4.30. What are the angular resolutions of (a) the
4.25. What is the angular resolution (in arc VLA in its largest configuration (baselines up
minutes) of (a) a 100 m diameter telescope to 13 km) at a wavelength of 21 cm; (b) the
operating at 1 cm wavelength, and (b) a 30 m VLA in its most compact configuration (base-
telescope operating at 1 mm wavelength? lines up to 1 km) at a wavelength of 1 cm;
4.26. Two radio sources in the Orion Nebula are (c) the VLBA (with baselines up to 3000 km) at
500 pc from us and are separated by 0.1 pc. a wavelength of 21 cm; (d) the Millimeter
How large a telescope would you need to dis- Array (with baselines up to 1 km) at 1 mm
tinguish the sources at a wavelength of wavelength?
(a) 21 cm? (b) 1 mm? 4.31. How many pairs of telescopes are there in
4.27. We sometimes use as a measure of the quality (a) the VLA with 27 telescopes, (b) the VLBI,
of a radio telescope the diameter d, divided by with ten telescopes, (c) the proposed
the limiting wavelength min. (a) Why is this Millimeter Array, which may have 40 or 75
quantity important? (b) If a telescope has telescopes depending on the final design?
surface errors of size x, give an expression 4.32. What is the total collecting area of (a) the VLA
for this quantity in terms of d and x. (b) the ALMA with 75 10 m diameter tele-
4.28. How large a collecting area would you need to scopes? (c) For ALMA, how many 12 m diameter
collect 1 W from a 1 Jy source over a band- telescopes would you need to have the same
width of 109 Hz (1 GHz)? collecting area as 75 10 m diameter telescopes.
(a) Show that for an interferometer with N
4.29. If a radio source emits a solar luminosity 4.33.
(4 1033 erg/s) in radio waves, (a) what would telescopes, the number of independent pairs
of telescopes, at any instant, is N(N 1)/2.
be the power per surface area reaching us if we
were (i) 1 AU away, (ii) 1 pc away? (b) If that (b) Evaluate this for the VLA, the VLBA and
power is uniformly spread out over a frequency ALMA.
range of 1011 Hz, what is the flux density 4.34. Derive equation (4.9).


Computer problems

4.1. Draw a graph of the angular resolution vs. wave- Calculate the observed source intensity as a func-
length, over the infrared part of the spectrum, for tion of (one-dimensional) position on the sky,
IRAS, KAO, SOFIA, ISO. assuming that the source has a uniform intensity
4.2. Suppose we have a radio telescope whose (one- over a finite size of (a) 1 arc sec (b) 1 arc min
dimensional) beam pattern is a gaussian with a (c) 1 degree, and the source has zero intensity
full width at half maximum (FWHM) of 1 arc min. outside that region.
Chapter 5




Binary stars and stellar masses

Eclipsing binary. As we observe the light from
If we are to understand the workings of stars, it is
such a system, it periodically becomes
important to know their masses. The best way to
brighter and fainter. We interpret the dim-
measure the mass of an object is to measure its
ming as occurring when the companion
gravitational influence on another object. (When
passes behind or in front of the main star. For
you stand on a bathroom scale, you are measur-
us to see these eclipses, we must be aligned
ing the Earth™s gravitational effect on your mass.)
in the plane of the orbit (Fig. 5.1). A famous
For stars, we are fortunate to be able to measure
example of such a star is Algol (also known as
the gravitational effects from pairs of stars, called
binary stars. Persei, indicating that it is the second
brightest star in the constellation Perseus).
Astrometric binary. Astrometry is the branch of
5.1 Binary stars astronomy in which the positions of objects
are measured very accurately. In an astromet-
Many stars we can observe appear to have com- ric binary, we can only see the brighter star.
panions, the two stars orbiting their common However, when we follow its path on the sky,
center of mass. It appears that approximately half we see that, instead of following a straight
of all stars in our galaxy are in binary systems. By line, it ˜wobbles™ back and forth across the
studying the orbits of binary stars, we can meas- straight line path. This means that the star is
ure the gravitational forces that the two stars moving in an orbit, so we can infer the pres-
exert on each other. This allows us to determine ence of a companion (Fig. 5.2).
Spectroscopic binary. When we study the spec-
the masses of the stars.
We classify binaries according to how the trum of a star we may see that the wave-
companion star manifests its presence: lengths of spectral lines oscillate periodically
about the average wavelength. We interpret
Optical double. This is not really a binary star.
these variations as being caused by a
Two stars just happen to appear along
Doppler shift (discussed in the next section).
almost the same line of sight. The two stars
When the star is coming towards us in its
can be at very different distances.
orbit, we see the lines at shorter wave-
Visual binary. These stars are in orbit about
lengths, and when the star is moving away
each other and we can see both stars
from us in its orbit, we see the lines at
directly.
longer wavelengths.
Composite spectrum binary. When we take a
spectrum of the star, we see the lines of two It is possible for a given binary system to fit
different spectral type stars. From this we into more than one of these categories, depend-
infer the presence of two stars. ing on what we can observe.
84 PART I PROPERTIES OF ORDINARY STARS



Path of Unseen Companion
Primary
B
Orbit of Secondary
C

Secondary
A


Path of
Brightness




Center
C of Mass
B
A
Path of Visible Star

Fig 5.2. Astrometric binary.Two stars orbit about a com-
Time
mon center of mass, which in turn moves across the sky.The
(a) fainter star is too faint to see, so we only see the brighter
star, moving back and forth across the path of the center of
mass.



where the waves are moving through a particular
Counts




elastic medium (Fig 5.3).

5.2.1 Moving sources and observers
We first look at the case of the moving observer.
If the observer is moving toward the source, then
the waves will be encountered more frequently
than if the observer were standing still. This
0 10 means that the observed frequency of the wave
increases. If the frequency increases, then the
Time (min)
wavelength decreases. If the observer is moving
(b)
away from the source then the situation is
Fig 5.1. Eclipsing binary. (a) The binary system is shown
reversed. Waves will be encountered less fre-
above; the light curve below.The fainter secondary passes
quently; the frequency decreases; the wave-
alternately in front of and behind the primary. Most of the
length therefore increases. It should be noted
time, as at position C, we see light from both stars.When
that if the observer moves perpendicular to the
the secondary eclipses the primary, part of the primary light
line joining the source and observer, no shift will
is blocked and there is a dip in the intensity, as at point A.
When the secondary passes behind the primary, as at B, its be observed.
light is lost. Since the secondary is not as bright as the pri- We now look at the case of the moving source.
mary, the loss of brightness is not as great as at A. (b) Part of Each wavefront is now emitted in a different
the light curve for the eclipsing binary, NN Ser, around the
place. If the source is moving toward the observer,
eclipse part of the orbit. [(b) ESO]
the waves will be emitted closer together than if
the source were standing still. This means the
wavelength decreases. The decreased wavelength
results in an increased frequency. If the source is
5.2 Doppler shift moving away from the observer, the waves will be
emitted farther apart than if the source were
A Doppler shift is a change in the wavelength (and standing still. The wavelength increases and the
frequency) of a wave, resulting from the motion frequency decreases. Again, if the source is mov-
of the source and/or the observer. It is most easily ing perpendicular to the line joining the source
visualized for a sound wave or a water wave, and observer, no shift results.
5 BINARY STARS AND STELLAR MASSES 85



Moving Observer Moving Source




O S O O S O
Longer »
shorter »
Lower ν
Higher ν
(Lower ν)
(Higher ν)
(Longer »)
(shorter »)


(b)
(a)
Fig 5.3. Doppler shift for waves in an elastic medium, such
as sound waves. (a) Moving observer. On the left the
Suppose the source is moving with a speed vs
observer is moving toward the source, encountering wave
in a direction that makes an angle with the line
crests more frequently than for a stationary observer.The
of sight, and the observer is moving with a speed
frequency appears to increase (and the wavelength to
vo in a direction making an angle with the line
decrease). On the right, the observer is moving away from
of sight. Taking the components of the two veloc-
the source, encountering waves at a lower frequency, corre-
ities along the line of sight as vs cos and vo cos ,
sponding to a lower frequency, corresponding to a longer
and subtracting the get the relative radial veloc-
wavelength. (b) Moving source.The motion of the source dis-
torts the wave pattern, so the circles are no longer concen- ity, gives
tric.The observer on the left has the source approaching,
vr vs cos vo cos (5.2)
producing a shorter wavelength (and a higher frequency).
The observer on the right has the source receding, produc-
In astronomy we are interested in the Doppler
ing a longer wavelength (and a lower frequency).
shift for electromagnetic waves. The underlying
physics is a little different, because there is no
mechanical medium for these waves to move
It is possible for both the source and the
through. They can travel even in a vacuum. (We
observer to be moving. If their combined motion
will discuss this point further in Chapter 7.) For
brings them closer together, the wavelength will
sound waves, the actual amount of Doppler shift
decrease and the frequency will increase. If their
depends on whether the source or observer (or
combined motion makes them move farther
apart, the wavelength will increase, and the fre-
quency will decrease. If there is no instantaneous
vo
change in their separation, there is no shift in
wavelength or frequency.
The shift only depends on the component of the
relative velocity along the line joining the source
vs
and observer, since this is the only component that
can change the distance, r, between them. We call
this component the radial velocity (Fig. 5.4). We refer

θ
to the line joining the source and observer as the
line of sight. From our definition of radial velocity,
vs cos θ vo cos •
vr, we can see that it is given by
Source Observer
vr dr dt (5.1)

Note that if the source and observer are mov- Fig 5.4. Radial velocity.The horizontal line is the line of
sight between the source and observer.The radial velocity is
ing apart, r is increasing, and vr 0. If the source
the difference between the line of sight components of the
and observer are moving together, r is decreasing,
observer and source velocities.
and vr 0.
86 PART I PROPERTIES OF ORDINARY STARS



We now add this to the original wavelength to find
both) is moving. For electromagnetic waves, only
the observed wavelength:
the relative motion counts.
As long as the relative speed of the source and
¢
0
the observer is much less than the speed of light,
656.28 nm 0.022 nm
the results for electromagnetic radiation are rela-
tively simple. If is the wavelength at which a sig- 656.30 nm
nal is received, and 0 is the wavelength at which
(b) If we take the negative of vr, we just get the neg-
it was emitted, called the rest wavelength, the
ative of ¢ , so
wavelength shift is defined by
0.022 nm
¢
(5.3)
¢ 0

This gives a wavelength of
The simple result is that the wavelength shift,
expressed as a fraction of the original wave- 656.26 nm
length, is equal to the radial velocity, expressed
If we observe two spectral lines, their wave-
as a fraction of the speed of light. That is
length shifts will be different, since each is
1vr V c 2
vr c (5.4)
¢ shifted by an amount proportional to its own rest
0

wavelength. Thus, the spacing between spectral
If vr 0, then 0. For a spectral line in the
lines will be shifted.
middle of the visible part of the spectrum, a shift
to longer wavelength is a shift to the red, so this
5.2.2 Circular orbits
is called a redshift. The name applies even if we are
We now look at Doppler shifts produced by a star
in other parts of the spectrum. A positive radial
in a circular orbit. The orbital speed is v, and the
velocity always produces a redshift. If vr 0, then
radius is r. The angular speed of the star in its
0 , and we have a blueshift.
v r.
orbit (in radians per second) is given by
We now look at what happens to the fre-
The situation is shown in Fig. 5.5. Suppose the
c/ , so
quency. We remember that
star is moving directly away from the observer at
2
c
dd time t 0. At that instant the radial velocity
vr v. As the stars moves, the component of its

1 2 ¢ . Substituting
This means that ¢
into equation (5.4) gives v(t = π/2ω) vr
vr c    1vr V c 2 (5.5)
¢ 0
θ v(t = 0)
The shift in frequency, expressed as a fraction of
the rest frequency, is the negative of the radial
velocity, expressed as a fraction of the speed of
light. (So ¢ 0 0. )
¢ θ
Example 5.1 Doppler shift
The H line has a rest wavelength, 0 656.28 nm.
What is the observed wavelength for a radial veloc-
ity (a) vr 10 km/s, and (b) vr 10 km/s?

v(t = π/ω) v(t = 3π/2ω)
SOLUTION
(a) We use equation (5.4) to find the wavelength shift: Fig 5.5. Doppler shift for a circular orbit.The speed v of

0 1vr c2
the source remains constant, but the direction changes, so
¢
the radial velocity vr changes.The angle keeps track of how
1656.28 nm2 110.0 km s2 far around the circle the source has gone.The source velocity
13.0
0.022 nm is shown for ¬ve different values of .
105 km s2
5 BINARY STARS AND STELLAR MASSES 87



Fig 5.6. Radial velocity curve
Radial Velocity (km/s)


for the spectroscopic binary
100 GRS1915 105, which has an
orbital period of 33.5 days.The
horizontal axis is the fraction
0
through that orbit. Data points
and error bars are indicated.The
smooth curve is the best ¬t to
’100
the data.[ESO]

0 1
Orbital Phase (33.5 days)


velocity along the line of sight is v cos . However, the orbit. This gives us a radial velocity for an
t, so orbiting star
v cos1 t 2 v sin i cos1 t 2
vr (5.6) vr (5.7)
The radial velocity changes sign every half-
cycle, and repeats periodically. This is shown in
5.3 Binary stars and circular orbits
Fig. 5.6. The period of the motion is the circum-
ference, 2 r, divided by the speed v, so P 2 .
If we substitute equation (5.6) into equation (5.4), In this section we will see how Newton™s laws of
we find that the spectral lines shift back and motion and gravitation can be applied to binary
forth, with a shift given by stars in circular orbits. Circular orbits are not the
1v c 2 cos 1 t 2 most general case of orbital motion, but the
¢ 0
analysis is most straightforward, and most of the
So far we have been considering the situation basic points are clearly illustrated. In the next
in which the observer is in the plane of the orbit. section we will go to the general case of elliptical
If the observer is not in the plane of the orbit, the orbits.
Doppler shift will be reduced (Fig. 5.7). If i is the We consider two stars, of masses m1 and m2,
angle between the plane of the orbit and the orbiting their common center of mass at dis-
plane of the sky, then the projection of any veloc- tances r1 and r2, respectively (Fig. 5.8). From the
ity in the plane of the orbit into the line of sight
is v sin i. The angle i is known as the inclination of
v2


i
CM
m1
m2
Line of Sight r1 r2
Plane
v1
of
Sky

Plane of Orbit

Fig 5.7. Inclination of an orbit.The orbit is an ellipse, Fig 5.8. Binary system with circular orbits. Both stars orbit
which lies in a plane.The plane makes some angle i with a the center of mass (CM).The more massive star is closer to
plane of the sky.The plane of the sky is de¬ned to be per- the center of mass.The center of mass must always be
pendicular to the line of sight. between the two stars, so the stars lie on opposite sides of it.
88 PART I PROPERTIES OF ORDINARY STARS



definition of center of mass, these quantities are Combining equations (5.13) and (5.14) gives
related by
m1v2 m1m2
1
1r1 r2 2 2
G (5.15)
m1r1 m2r2 (5.8) r1
The center of mass moves through space sub- Note that we can divide both sides by m1. If we
ject only to the external forces on the binary star also use equation (5.10) to relate v1 to P, we find
system. The forces between the two stars do not that
affect the motion of the center of mass. We will 2
r1 G m2
4
therefore ignore the actual motion of the center
1r1 r2 2 2
(5.16)
2
P
of mass, and view the situation as it would be
viewed by an observer sitting at the center of This can be simplified if we introduce the total
mass. distance R between the two stars:
Since the center of mass must always be along
R r1 r2
the line joining the two stars, the stars must
r1 11 r2 r1 2
always be on opposite sides of the center of mass. (5.17)
This means that the stars orbit with the same
Using equation (5.12), this becomes
orbital period P. In general, the period of the
r1 11 m1 m 2 2
orbit is related to the radius, r, and the speed, R (5.18)
v, by
1r1 m2 2 1m1 m2 2 (5.19)
P 2 rv (5.9)
Substituting this into equation (5.16) gives
Solving for v gives
1m1 m2 2P2
4 R3 G (5.20)
v 2 rP (5.10)
Let™s look at how equation (5.20) can be used
Since the periods of the two stars must be the to give us stellar masses. For any binary system,
same, equation (5.9) tells us that we can determine the period directly if we watch
the system for long enough. If the star is a spec-
r1 v1 r2 v2 (5.11)
troscopic binary, we can see how long it takes for
Combining these with equation (5.8) gives the Doppler shifts to go through a full cycle. If it
is an astrometric binary we can see how long it
v1 v2 r1 r2 m2 m1 (5.12)
takes for the ˜wobble™ to go through a full cycle. If
(Note: We could have also obtained m1v1 m2v2 it is an eclipsing binary, we can see how long it
directly from conservation of momentum. This takes the light curve to go through a full cycle. If
is not surprising since the properties of the we can see both stars, we can determine R. Once
center of mass come from conservation of we know R and P, we can use equation (5.20) to
momentum.) determine the sum of the masses, (m1 m2 ). We
We now look at the gravitational forces. The can also obtain the ratio of the masses m1/m2,
distance between the two stars is r1 r2 , so the either from r1/r2, if both stars can be seen, or v1/v2,
force on either star is given by Newton™s law of if both Doppler shifts are observed. Once we
gravitation as know the sum of the masses and the ratio of the
masses, the individual masses can be deter-
m1 m2
1r1 r2 2 2
F G (5.13) mined. The situation we have outlined here is
the ideal one, however. Usually, we don™t have all
This force must provide the acceleration associ- of these pieces of information (as we will see
ated with the change of the direction of motion below).
in circular motion, v2 r . For definiteness, we look
Example 5.2 Mass of the Sun
at the force on star 1, so
We can consider the Sun and Earth as a binary sys-
m1 v 2 tem, so we should be able to apply equation (5.20) to
F r1 (5.14)
1
5 BINARY STARS AND STELLAR MASSES 89



find the mass of the Sun. It turns out that this is the The values of are so small that radians are an
most accurate measure we have of the Sun™s mass. inconvenient quantity. We can convert to arc sec-
onds (equation (2.16)) to give
d1AU2 1 “ 2 12.06 105 2
SOLUTION
R1AU2
Since the mass of the Sun is so much greater than
The factor of 2.06 105 was to convert radians to
that of the Earth, we can approximate the sum of
the masses as being the mass of the Sun, M . arc seconds, but it is also the factor to convert
Equation (5.20) then becomes astronomical units to parsecs, so we have
d1pc 2 1“ 2
4 2R3 R1AU2
M
GP2
If we use equation (2.17) to relate the distance in
11.5
2 13 3
4 10 cm2 parsecs to the parallax in arc seconds, this
16.67 dyn cm2 g2 2 13.16 107 s 2 2
8
becomes
10
1“ 2 >p1“ 2
1033g R1AU2 (5.22)
2

This can then be substituted directly into equa-
We call this quantity a solar mass. It becomes a con-
tion (5.21).
venient quantity for expressing the masses of other
We will now look at the behavior of the Doppler
stars.
shifts. Applying equation (5.10) to both speeds, v1
From the Earth and Sun we know that for a
and v2, and remembering that the period of the
pair of objects orbiting with a period of 1 yr, at a
orbit is the same for both stars, we have
distance of 1 AU (defined in Section 2.6), the sum
1P 2 2 1v1 v2 2
of the masses must be one solar mass. This sug- r1 r2
gests a convenient system of units for equation
Using this to eliminate R in equation (5.20) gives
(5.20). If we express masses in solar masses, dis-
1P 2 G2 1v1 v2 2 3
tances in astronomical units, and periods in years, m1 m2 (5.23)
the constants must equal one to give the above
If the orbit is inclined at an angle i, then the
result. We can therefore rewrite equation (5.20) as
Doppler shifts only measure the components
R3 P2
c d c dc d
m1 m2 vr vsin(i). In terms of the radial velocities v1r
(5.21)
and v2r, equation (5.23) becomes
1M
1 AU 1 yr

1P 2 G2 1v1r v2r 2 3 sin3i
For the Solar System, we write the sum of the m1 m2 (5.24)
masses as one in these units, so the equation sim-
If a binary happens to be an eclipsing binary,
ply says that the cube of the radius (in AU) is
then we know that we are close to the plane of
equal to the square of the period (in yr). This is
the orbit, and i is close to 90 . Otherwise we don™t
also known as Kepler™s third law of planetary motion.
know i. If a circular orbit is projected at some
The law was originally found by Kepler observa-
angle on the plane of the sky it will appear ellip-
tionally, and Newton used it to show that gravity
tical. We will see in the next section that there
must be an inverse square law force. (See Problem
are ways to determine i if we can trace that pro-
5.10.) We will discuss planetary motions in more
jected orbit on the sky. If i is unknown all we can
detail in Chapter 22.
do is solve equation (5.24) with i 90 . This will
For a visual binary, we don™t directly measure
give us a value of m1 m2 that is a lower limit to
R, the linear separation. We actually measure the
the true value. The true value would be this lower
angular separation on the sky, . If d is the dis-
limit divided by sin3i, and since sin3i is less than
tance to the binary, then R is equal to (rad)d,
or equal to unity, the value assuming i 90 is
where (rad) is the value of measured in radi-
less than the true value. Finding lower limits is
ans. When we use this relation, R and d will come
not as useful as finding actual values. However,
out in the same units. Therefore
if we study enough binary systems, we will
d1AU2 1rad2 encounter a full range of inclination angles.
R1AU2
90 PART I PROPERTIES OF ORDINARY STARS



These statistical studies can be used to relate If we now substitute into equation (5.24), we find
mass to spectral type.
m3 sin3 i
a b 1vir 2
P 2
3
1m1 m2 2 2
(5.25)
Example 5.3 Binary star Doppler shifts 2G
A binary system is observed to have a period of
The quantity on the right-hand side of equa-
10 yr. The radial velocities of the two stars are deter-
tion (5.25) is called the mass function. If we can
mined to be v1r 10 km/s and v2r 20 km/s, respec-
measure only one Doppler shift, we cannot
tively. Find the masses of the two stars (a) if the
determine either of the masses. We can only
inclination of the orbit is 90 , and (b) if it is 45 .
measure the value of the mass function. We can,
however, obtain information on the masses of
SOLUTION
various spectral types through extensive statisti-
From equation (5.24), we have
cal studies.
m1 m2
We can also look at energy of a binary system
M
110 yr 2 13.16 107 s yr2 3 110 202 11 105 cm s 2 4 3
with circular orbits. It is the sum of the kinetic
energies of the two stars plus the gravitational
2 16.67 dyn cm2 g2 2 12 1033 g2 1sin3i 2
8
10 potential energy, which we take to be zero when
the two objects are infinitely far apart. The
10.2 M sin3i
energy is
This means that
11 2 2m1 v2 11 2 2m2 v2
E Gm1 m2 R (5.26)
1 2
10.2 M sin3i
m1 m2
From equation (5.15), we have
90 , sin3i
If i 1, so
m1 m2
m1 m2 10.2 M m1v2
1R2 2
Gr1
1

We find the ratio of the masses from the ratio of
and, using equation (5.12), we obtain
the radial velocities
m1 m 2 v2 v1 m 1m 2
m2v2
1R2 2
Gr2
2
2.0
Substituting these into equation (5.26) and
This means that m1 = 2m2, so
simplifying gives
m2 10.2 M
2m2
11 22 Gm1m2 R
E (5.27)
giving m1 6.8 M and m2 3.4 M . If i 45 ,
1/sin3i 2.8. The ratio of the masses does not change, Compare this with equation (3.4), which has
since the sin i drops out of the ratio of the radial the energy for circular orbits with electrical
velocities. This means that we can just multiply each (rather than gravitational) forces.
mass by 2.8 to give 19.2 and 9.5 M , respectively. The negative energy means that the system
is bound. We would have to add at least
It is often the case that only one Doppler shift
(1/2)Gm1m2/R to break up the binary system. (We
can be observed. Let™s assume that we measure v1
can think of this as being analogous to the
but not v2. We must therefore eliminate v2 from
binding energy of a hydrogen atom.) For any pair
our equations. We can write v2 as
of masses, as you make the orbits of the binary
v1 1m1 m2 2
v2 system smaller, the energy becomes more
negative.
The sum of the velocities then becomes
v1 11 v2 v1 2 Example 5.4 Binding energy of a binary system.
v1 v2
v1 11 m1 m2 2
What is the binding energy of a binary system with
two 1 M stars orbiting with each 100 AU from the
1v1 m2 2 1m1 m2 2 center of mass?
5 BINARY STARS AND STELLAR MASSES 91



r r is a constant. We can see that for a point on
SOLUTION
the semi-major axis (and the ellipse), this sum is
Using equation (5.27)
2a, so it must be 2a everywhere. That is
ab
1 Gm1m2
E
r r¿ 2a (5.28)
R
2

1 16.67 dyn cm2 g2 2 12 1033 g 2 2
8 The eccentricity of an ellipse is the distance
ab
10
12002 11.5 between the foci, divided by 2a. A circle is an
1013cm2
2
ellipse of eccentricity zero (both foci are at the
1043 erg
4.5 same point, the center of the circle). The eccen-
tricity of any ellipse must be less than unity. From
the point where the curve crosses the minor axis,
5.4 Elliptical orbits
r r¿ a, so
1ae 2 2
5.4.1 Geometry of ellipses b2 a2
In general, orbiting bodies follow elliptical paths.
a2 11 e2 2 (5.29)
A circle is just a special case of an ellipse. In this
section, we generalize the results from the previ- In a binary system, the center of mass of the
ous section from circular orbits to elliptical two stars will be at one focus on the ellipse. The
orbits. The basic underlying physical ideas are the farthest point from that focus is called the apas-
same. tron. From the figure, we see that the distance
We first review the geometry of an ellipse, as from the focus to this point is
shown in Fig. 5.9. We describe the ellipse by its
r1apastron 2 e2
a11 (5.30a)
semi-major axis a and semi-minor axis b. Each
point on the ellipse satisfies the condition that The closest point to the focus is called the peri-
the sum of the distances from any point to two astron. Its distance from the focus is
fixed points, called foci (singular focus), is con-
r1periastron 2 e2
a11 (5.30b)
stant. If r and r are these two distances then
The average of these two values is a, the semi-
major axis. This is the quantity that replaces the
Major Axis radius of a circular orbit in our study of binary
stars.
It is useful to have an expression for the
v ellipse, relating the variables r and . From the
law of cosines, we see that
12ae 2 2 2r12ae 2 cos
r¿ 2 r2 (5.31)
r
Using equation (5.28) gives
r'
Minor Axis




e2 2
a11
a(1 - e) F θ ae r (5.32)
e cos
1
a F'
5.4.2 Angular momentum
b a(1 + e) in elliptical orbits
a
The gravitational force between two objects always
acts along the line joining the two objects. The
center of mass also lies along this line. This means
that the force on either object points directly from
Fig 5.9. Geometry of an ellipse.The length of the semi-
major axis is a; the length of the semi-minor axis is b.The that object towards the center of mass. Therefore,
two foci are at F and F .The eccentricity is e, and the dis- these forces can exert no torques about the center
tances from the two foci to points on the ellipse are r and r .
of mass. If there are no torques about the center of
92 PART I PROPERTIES OF ORDINARY STARS



The major consequence of angular momen-
v(dt)
tum conservation is therefore that the objects
• move slower when they are farther apart and



in
faster when they are closer together. (As we will
)s
dt
see, the Earth™s orbit is only slightly eccentric but
v(

it moves faster when it is closer to the Sun, in
r
January, and slower when it is farther away. This
results in winter and spring in the northern
hemisphere being shorter than summer and win-
F (CM) ter. Check a calendar to see this.)
Fig 5.10. Angular momentum in an elliptical orbit. In this
5.4.3 Energy in elliptical orbits
case the object is a distance r from the focus F (which is also
We next look at the total energy of the binary sys-
the center of mass). Its velocity makes an angle with the
line from F to the object. (If this were a circle, would be tem. Adding the kinetic energies of the two stars
90 .) The time interval over which we mark the motion is dt. and their gravitational potential energy (defined
as zero when the stars are infinitely far apart)
gives
mass, then the angular momentum about the cen-
11 2 2m1 v2 11 2 2 m2 v2
ter of mass is conserved. E Gm1 m2 R (5.34)
1 2
To consider the angular momentum of an
In this expression v1 and v2 are the speeds relative
object in an elliptical orbit, we look at Fig. 5.10. For
to the center of mass. Remember, from the defi-
an object of mass m, the angular momentum about
nition of the center of mass, we found that
the center of mass (which corresponds to one of the
m1v1 m2v2. We introduce the relative speed of
foci) is r times the component of its linear momen-
the two stars
tum perpendicular to the direction of r. That is
v v1 v2 (5.35a)
L mvr sin (5.33)
The two speeds are added because the two
where is between the line from the center of
stars are always moving in opposite directions, so
mass to the object (of length r) and its direction of
their relative speed will be the sum of the magni-
motion (the direction of the velocity vector, v).
tudes of their individual speeds. In terms of v, the
We look at the area swept out by the line from
two speeds are
the center of mass to r, in the time interval dt. The
m2v
area is the thin triangle shown in the figure. The
v1 (5.35b)
m1 m2
long side of the triangle is r and the short side is
v dt. The small right triangle shows that the
m1v
v2 (5.35c)
height of the larger triangle is v (dt)sin . The
m1 m2
area is then half the base height, so
11 2 2 r1v dt 2 sin
Substituting into equation (5.32) gives
dA
v2
m1 m2 c d
G
m2 2
E (5.36)
The rate at which the area is swept out is
R
21m1
11 2 2r v sin
dA dt
Since energy is conserved, we can evaluate it
at any point we want. For simplicity, we choose
However, we can use equation (5.33) to eliminate
apastron (speed va) and periastron (speed vp). We
r v sin , giving
11 2 2 1L m 2
can relate the speeds va and vp by conservation of
dA dt
angular momentum. Angular momentum con-
Since L is constant, the rate at which area is swept servation is easy to apply at the apastron and peri-
out is constant. Equal areas are swept out in equal astron because the velocities are perpendicular to
times. (When applied to planetary motions, this is the line from the focus to the star. Since 90
known as Kepler™s second law, as we will discuss in at these points, the angular momentum (equa-
Chapter 22.) tion (5.33)) is just mvr. Using equations (5.30a) and
5 BINARY STARS AND STELLAR MASSES 93



(3) Even if you are in the plane of the orbit, the
(5.30b) we find that
e 2va e 2vp
radial velocity curve depends on where you
a11 a11 (5.37)
are relative to the major axis of the ellipse.
Solving for the ratio vp/va gives
These points are illustrated in Fig. 5.11.
vp e
1
vr
(5.38)
va e
1
v3
3
Now that we have the ratio vp/va, we need
another relation between them to be able to solve
for va and vp individually. We can use conserva- t
2 4
tion of energy to equate the energies at the apas-
tron and periastron. Using equation (5.36) gives
v2
v2 G G
b
a
m2 2 e2 m2 2 e2
a11 a11
21m1 21m1
(5.39)
1 1
Rearranging gives
(a)
m2 2
c da b 1v v2 2
G1m1 1 1 12 To Observer
2p a
a e e
1 1
2
1 2 vp
v c a b 1d
2 a v2 a

We can use equation (5.38) to eliminate the
ratio vp/va. Solving for va gives v2
m2 2
c da b
G1m1 e
1 2
v2 (5.40a) v1
a
a e
1
m2 2
c da b
G1m1 e
1
v2 (5.40b)
b
a e
1
3 1
If we put these into equation (5.36), the total CM
v3
energy simplifies to
E Gm1m2 2a (5.41)
4
We can now use this in the left-hand side of
v4
equation (5.36). We can then solve for v at any
(b)
point r:
m2 2 12 r 1 a2
v2 Fig 5.11. (a) Radial velocity vs. time, t, for an elliptical orbit.
G1m1 (5.42)
(b) In contrast to the circular orbit both the magnitude and
direction of v change throughout the orbit. (We assume for
5.4.4 Observing elliptical orbits this ¬gure that the observer is in the plane of the orbit, or
In studying the Doppler shifts of elliptical orbits that i 90 .) Four points are shown in the orbit and in the
radial velocity curve. At points 2 and 4 the motion is perpen-
as compared with circular orbits, there are three
dicular to the line of sight, so vr 0. For point 1 the motion
important differences:
is directly toward the observer, producing the maximum
(1) The speed along an elliptical orbit is not con- negative vr, and, for point 3, the motion is away from the
stant. observer, producing the maximum positive vr .The motion is
also faster at 1 than at 3. In addition, going from 4 to 1 to 2
(2) In an elliptical orbit the velocity is not per-
takes less time than going from 2 to 3 to 4.This accounts for
pendicular to the line from the center of mass
the distorted shape of the radial velocity curve.
to the orbiting object.
94 PART I PROPERTIES OF ORDINARY STARS



An even more stringent constraint is the rela-
As we said above, we must correct any Doppler
tionship between mass and temperature on the
shift for the inclination of the orbit, i. If we take
main sequence. The cooler stars are less massive
a tilted ellipse and project it onto the sky, we still
and the hotter stars are more massive. We have
have an ellipse. However, that ellipse will have a
already said that the existence of the main
different eccentricity than the true ellipse. When
sequence implies a certain relationship between
we look at an elliptical orbit, how can we tell if it
size and temperature. This means that if a star is
is tilted or not? For a tilted orbit the foci will not
on the main sequence, once its mass is specified,
appear in the right place for the projected ellipse.
its radius and temperature are determined.
Therefore, if we see two stars orbiting a point dif-
Another way of looking at this to say that a star™s
ferent from the center of mass, we will know that
mass determines where on the main sequence it
the orbit is inclined. We can determine the incli-
will fall.
nation from the degree to which the foci appear
Since the mass determines the radius and
to be displaced. In this technique, we don™t actu-
temperature of a main sequence star, it should
ally know where the center of mass is, so a
not be surprising that it also determines the
process must be used in which we try one posi-
luminosity. The exact dependence of the lumi-
tion for the center of mass and try to match the
nosity on mass is called the mass“luminosity rela-
projected orbit, repeating the process until a
tionship. This relationship is also explainable
good fit is achieved.
from theories of stellar structure. This relation-
ship is shown in Fig. 5.12. We can summarize it
5.5 Stellar masses by saying that the luminosity varies approxi-
mately as some power, , of the mass. If we
express luminosities in terms of solar luminosi-
As a result of studying many binary systems,
ties, and masses in terms of solar masses, this
astronomers have a good idea of the masses of
means that
main sequence stars. These results are summa-
1M M 2
rized in Table 5.1. Just as the Sun™s temperature
LL (5.43a)
places it in the middle of the main sequence, its
mass is in the middle of the range of stellar
Intermediate
masses. The lowest mass main sequence stars
Mass
have about 0.07 of a solar mass, and the most
massive stars commonly encountered have about
Low High
60 solar masses. When we think of how large or
Mass Mass
small stars might have turned out to be, the
106
observed range of stellar masses is not very large.
This range is an important constraint on theories
104
of stellar structure.

102
.
L/L




Mass and spectral type (MS).
Table 5.1.
1
Spectral type M/M
10-2
O5 40.0
B5 7.1
A5 2.2
0.1 1 10 100
F5 1.4
G5 0.9
M/M .
K5 0.7
M5 0.2 Fig 5.12. Mass“luminosity relationship.
5 BINARY STARS AND STELLAR MASSES 95



3 3
NGC 581 NGC 683
Slope = ’1.78 Slope = ’1.06
Log (Number)




2 2



1 1



0 0
’0.5 ’0.5
2 1.5 1 0.5 2 1.5 1 0.5
0 0
Log (Mass/solar)

3
NGC 103
Slope = ’1.47
Log (Number)




2



1



0
’0.5
0.5 0
2 1.5 1
Log (Mass/solar)
Fig 5.13. Initial mass function for three young clusters.
Note that masses are in solar masses, so log(mass) 0 cor-
responds to one solar mass. [John Scalo, University of Texas, because we can see brighter stars to greater dis-
Austin] tances than faint stars. This is called a selection
effect, because it makes it difficult to select an
unbiased sample.
It is of interest to study the number of stars in
A single value of does not work for the full
different mass ranges. This is called the mass
range of masses along the main sequence. The
function. This is a good check on various theories
approximate values are:
of star formation, which we will discuss further
1.8 for M 6 0.3 M (low mass)
in Chapter 15. When we discuss stellar evolution
(Chapters 10 and 11), we will see that low mass
4.0 for 0.3 M 6 M 6 3 M (intermediate mass)
stars live longer than high mass stars. So, if we
(5.43b)
study the mass function in a group of stars (usu-
2.8 for 3 M 6 M (high mass)
ally clusters, as we will discuss in Chapter 13), it
In understanding how stars are formed will change with time. As the higher mass stars
(Chapter 14) we would like to know the distribu- die, they leave behind a cluster with a depleted
tion of stellar masses. That is, we would like to number of high mass stars. So, if we want to
know the proportions of stars of various masses. really see what the mix of masses are when the
Since we now know how to relate mass and spec- cluster forms, we need to look at young clusters.
This allows us to study what we call the initial
tral type, we can carry out such studies by look-
mass function (IMF). Data for some sample clusters
ing at the relative numbers of different spectral
types or luminosities. These studies are difficult, are shown in Fig. 5.13. Note: the number of low
96 PART I PROPERTIES OF ORDINARY STARS



mass stars is so much greater than the number of SOLUTION
high mass stars, that we show the results on a We can express the various quantities in solar units.
log“log plot. In general, we find that we can fit Taking ratios, we can use equation (5.44) to give
the data with power laws of the form N1m 2 m.
1L L 2 1 2 1T T 2 2
RR
The best fit values of are shown as slopes in the
1802 1 2 11.69 2
figure. These figures don™t have much data for 2

stars much less massive than the Sun, but other
3.1
studies indicate that the function doesn™t rise as
fast as it does for the mass range shown in Fig. 5.13.
Eclipsing binaries (such as in Fig. 5.1) provide
We can also look at the number of stars in dif-
us with another means of determining stellar
ferent luminosity ranges, or the luminosity func-
radii. This method involves analysis of the shape
tion. We find that there are also many more low
of the light curve and a knowledge of the orbital
luminosity stars than high luminosity stars.
velocities from Doppler shift measurements. (In
However, the luminosity of each high mass star is
an eclipsing binary, we don™t have to worry
so much greater than that for each low mass star,
about the inclination of the orbit.) Particularly
that most of the luminosity of our galaxy comes
important is the rate at which the light level
from a relatively small number of high mass stars.
decreases and increases at the beginning and
end of eclipses.
We can also estimate the radii of rotating
5.6 Stellar sizes
stars. If there are surface irregularities, such as
hot spots or cool spots, the brightness of the star
We have alluded so far to stellar sizes, but we have will depend on whether these spots are facing us
not discussed how they are determined. In this or are turned away from us. The brightness varia-
section, we will look at various methods for meas- tions give us the rotation period P. From the
uring stellar radii. broadening of spectral lines, due to the Doppler
The star whose size is easiest to measure is the shift, we can determine the rotation speed v. This
Sun. This is actually quite useful. We have seen speed is equal to the circumference 2 R, divided
that the Sun is intermediate in its mass and tem- by the period. Solving for the radius gives
perature, so its radius is probably a fairly repre-
sentative stellar radius. The angular radius of the R Pv 2 (5.45)
Sun, ¢ , is 16 arc min. The Sun is at a distance
Sometimes the Moon passes in front of a star
d 1.50 108 km, so its radius, R , is given by
bright enough and close enough for detailed
R d tan ¢ study. An analysis of these lunar occultations tells
us about the radius of the star. The larger the star
105 km
6.96
is, the longer it takes the light to go from maxi-
The Sun is the only star whose disk subtends mum value to zero as the lunar edge passes in
an angle larger than the seeing limitations of front of the star. Actually, since light is a wave,
ground-based telescopes. We therefore need there are diffraction effects as the starlight
other techniques for determining radii. If we passes the lunar limb. The light level oscillates as
know the luminosity (from its absolute magni- the star disappears. The nature of these oscilla-
tude) and the surface temperature (from the tions tells us about the radius of the star.
spectral type) of a star, we can calculate its radius There is another observational technique,
using equation (2.7). Solving for the radius gives called speckle interferometry, that has been quite
1L 4 T4 2
successful recently. If it were not for the seeing
1>2
R (5.44)
fluctuations in the Earth™s atmosphere, we would
be able to obtain images of stellar disks down to
Example 5.5 Luminosity radius
the diffraction limits of large telescopes. However,
Estimate the radius of an A0 star. (Use Appendix E
the atmosphere is stable for short periods of
for the stellar properties.)
5 BINARY STARS AND STELLAR MASSES 97




F0G0
A0
B5
B0



O5

(a)

G0
K0
M0

M5




(b)
Fig 5.15. Stellar sizes for different spectral types. (a) The
largest, down to G0. (b) The smaller ones, from G0 down,
Fig 5.14. Hubble Space Telescope image of the red giant but blown up.The G0 circle is repeated to give the relative
star Betelgeuse.You can see that it is barely resolved, but by scale for the two parts of the ¬gure.
understanding how the telescope response smears the
image, we can achieve a very accurate estimate for the size light coming through slightly different paths in
of the star. [STScI/NASA] the atmosphere. The final images must be recon-
structed mathematically.
time, of the order of 0.01 s. The blurring of Even more recently, work on stellar sizes has
images comes from trying to observe for longer been done on the Hubble Space Telescope, as
than this. If an image were bright enough to see shown in Fig. 5.14. The results of these various
in this short time, we could take a picture with techniques are shown in Fig. 5.15. We can see
diffraction-limited resolution. Unfortunately, that, on the main sequence, stars become larger
0.01 s is not long enough to collect enough pho- with increasing surface temperature. This is why
tons from a star. However, we can collect a series the luminosity of stars increases with increasing
surface temperature at a rate greater than T4.
of 0.01 s images, observing interference between


Chapter summary
We saw in this chapter how the gravitational We saw how the Doppler shift can be used to
interactions in binary systems can be studied to determine the radial velocities of objects. In
determine the masses of stars. binary systems, we can observe the wavelengths
98 PART I PROPERTIES OF ORDINARY STARS



of spectral lines varying periodically as the stars elliptical orbits, and saw how the kinetic energy
orbit the center of mass. We saw what informa- varied as the speed varies.
tion could be obtained even if we don™t know the We saw that the range of masses for stars
orbital inclination. along the main sequence is much less than the
We saw how the masses of orbiting objects are range of luminosities. There is also a close rela-
related to the orbital radii and speeds, and the tionship between mass and luminosity for main
period of the system for circular orbits. We sequence stars.
extended these ideas to elliptical orbits. For ellip- We saw how eclipsing binaries can be used to
tical orbits, we saw how conservation of angular tell us something about stellar sizes. We also
momentum means a change in the speed with looked at other techniques for determining stel-
distance from the center of mass, and how this lar sizes, including knowing the luminosity and
affects the Doppler shift we see in different parts temperature, using lunar occultations and
of the cycle. We also found the total energy for speckle interferometry.


Questions
5.1. Under what conditions can we determine the (g) Source moving perpendicular to line of
masses of both stars in a binary system? sight. Observer moving towards source.
Think of as many combinations of situations (h) Source moving and observer moving per-
as you can. pendicular to line of sight but in opposite
5.2. For a binary that is only detected as an astro- directions.
metric binary, what are the conditions under 5.8. Why must the center of mass of two stars be
which we can determine the masses of both along the line joining the two stars?
stars? 5.9. Why is it not possible for the orbits of the
5.3. If you observe two stars close together on the two stars in a binary system to have different
sky, how would you decide if they were an periods?
optical double or a true binary? 5.10. Is the Earth“Sun center of mass closer to the
5.4. If we see an eclipsing binary, how do we center of the Earth or the center of the Sun?
know the inclination of the orbit? *5.11. Suppose that a binary system is moving
5.5. Describe situations in which a source and/or under the external influence of other stars
observer are moving and no Doppler shift for (in a cluster for example) and we observe the
electromagnetic waves is observed. center of mass to be accelerating as a result.
5.6. When you hear the Doppler shift in a train Can we still apply the analyses in this chapter
whistle, the effect is most obvious as the to such a system?
train goes by you. Why is that? 5.12. Why do we say that for the Earth“Sun system,
5.7. For the following situations, indicate whether the sum of the masses is approximately equal
the radial velocity is positive, negative or zero: to the mass of the Sun?
(a) Observer moving towards (stationary) source. 5.13. In a binary system, the gravitational force that
(b) Source moving towards (stationary) star 1 exerts on star 2 must produce an accel-
observer. eration. How is that acceleration manifested?
(c) Source moving away from (stationary) 5.14. Discuss how the Sun is a ˜typical™ main
observer. sequence star.
(d) Source and observer moving towards each 5.15. How would you measure the mass of the
other. Earth?
(e) Source and observer moving away from 5.16. Use a calendar to find out how much longer
each other. it takes to go from the first day of Spring to
(f) Source moving away from observer at the first day of Autumn than from the first

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