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