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E1 Astronomy and Space Science
4
Chapter 4 Stars and the Universe
Stars and the Universe
Practice 4.1 (p. 98)
7
1
A
2
D
3
(a) Star X is closer to the Earth.
(a) Distance to the star
1
1
= =
= 83.3 pc
p 0.012
(b) In this case, we have
30 .1 AU
= tan p
d
(b) Distance to star X
1
=
= 10 pc
0 .1
 p (in arcsecs) =
Distance to star Y
1
=
= 20 pc
0.05
Therefore,
parallax measured on Neptune
30 .1
=
(in arcsecs) = 0.361
83 .3
Star X is 10 pc closer than star Y.
4
Star
Parallax
Distance
Distance
Distance
(in )
(in pc)
(in AU)
(in ly)
0.379
2.64
5.44 
8.60
Sirius A
8
0.194
5.15
1.06 
(a) The parallaxes of stars farther away are
too small to be measurable by the
parallax method.
(b) The distance between the two
105
Altair
measurements is larger (about 3 AU).
16.8
The parallax measured would be larger
106
Vega
0.125
8
1.65 
and thus more accurate.
26.1
9
106
5
tan
 half the width of paper
=
length of arm
2
Distance to the star
1
=
= 6.67 pc
0.15
Time between sending and receiving signal
(Put in their own data into the formula.)
= 6.67  3.26  2 yr
Sample result:
21.5
cm

tan = 2
80 cm
2
= 43.5 yr
10
 = 15.3
This angle is much larger than the parallax of
6
30 .1
d (in parsecs)
Parallax of the farthest star measured
1
=
= 0.005
200
Proxima Centauri.
Practice 4.2 (p. 116)
The angular shift  of the star is about half
1
B
the width of the photo, i.e.  = 4.
2
C
3
A
4
C
5
A
p=

2
= 2
1 1
Distance to the star = = = 0.5 pc
p 2
New Senior Secondary Physics at Work
1
 Oxford University Press 2011
E1 Astronomy and Space Science
Chapter 4 Stars and the Universe
Apparent magnitude of star X
 100 
= 1 + 5 log 
 =4
 10 
(b) Energy received
= intensity received  area  time
 6  10 3
= 1.68  104    
2

Apparent magnitude of star Y
 10 
= 3 + 5 log   = 3
 10 
= 4.75  109 J
10
Since the apparent magnitude of star Y is
D
7
(a) Star X
light.
(b) Object A
11
(b) Star A
8
(a) Object A mainly emits ultra-violet
radiation. Object B mainly emits red
smaller, it appears brighter.
6
The x-axis represents surface temperature or
spectral class, while the y-axis represents
(a) Venus orbits the Sun. Its distance to us
luminosity, absolute magnitude, or luminosity
changes and therefore its apparent
relative to that of the Sun.
magnitude varies. Also, Venus has
12
different phases. With a phase showing a
From the H-R diagram, the surface
temperature of Betelgeuse is about 3000 K
larger part, it looks brighter.
and its luminosity is about 70 000 times the
(b) Sirius looks brighter.
I
Ratio of intensity S
IA
Sun’s.
Radius of Betelgeuse =
= 2.512 m A  mS 
= 2.512 11.4
=
= 9.12
The ratio of the intensity of Sirius to that
(c)
2
 10   1 
=   
 1  2
The ratio of the maximum intensity of
70 000
 3000 


 5800 
2
4
= 6.25
Venus to that of Sirius is 15.9 : 1.
(a) Intensity received
energy received
=
area  time
14
Its luminosity is 6.25 times that of the Sun.
IB
(a)
= 2.512 mA mB  = 2.512 31 = 39.8
IA
A star of apparent magnitude 1 is 39.8
8  10 9
 5.5  10 3
π

2

T2
13
= 15.9
=
L
= 989 solar radii
of Antares is 9.12 : 1.
I
Ratio of intensity V
IS
= 2.512 mS  mV 
= 2.512 1.44.4
9
2

 1


times brighter than another one of
2

 2


apparent magnitude 3.
= 1.68  104 W m2
New Senior Secondary Physics at Work
2
 Oxford University Press 2011
E1 Astronomy and Space Science
Chapter 4 Stars and the Universe
(b) A difference of 4 magnitudes means
The luminosity of Canopus is 13 700
difference in intensities by a factor of
times that of the Sun.
39.8, as calculated in (a).
L
(d) R =
T
As a result, a star of apparent magnitude
2
=
3 is 39.8 times brighter than another one
13 700
 7400 


 5800 
2
= 71.9 solar radii
of apparent magnitude 7.
15
Therefore, a star of apparent magnitude
Practice 4.3 (p. 128)
1 is (39.8  39.8 =) 1580 times brighter
1
D
than another one of apparent magnitude
2
B
7.
3
By
(a) Since I  T4,
intensity of the star
= 24 = 16
intensity of the Sun
radial velocity of the galaxy
 λ 
=
c
 λ 
Each cm2 of the star’s surface radiates
(b)
= 0.2  3  108
16 times more than the Sun does.
intensity  surface area star
intensity  surface area Sun
4
I
r 2
= star  star 2
I Sun rSun
5
= 16   
1
= 6  107 m s1 (away from the Earth)
λ v r

By
,
λ
c
 = λ 
2
vr
 25  10 3
= 500 
= 0.0417 nm
c
3  10 8
We see the spectral line at (500  0.0417 =)
= 400
5
The whole star radiates 400 times more
499.9583 nm.
λ v r

By
,
λ
c
800
vr
 = λ 
= 650  3.6 8 = 0.000 481 nm
c
3  10
than the Sun does.
16
λ v r

,
λ
c
1
= 96.2 pc
0.0104
d 
(b) Absolute magnitude = m  5 log 
 10 
(a) Distance of Canopus =
We would see it at (650 + 0.000 481 =)
= 0.62
 96 .2 
 5 log

 10 
6
650.000 481 nm.
λ v r

By
,
λ
c
 = λ 
= 5.54
vr
1000  10 3
= 656.3 
= 2.19 nm
c
3  10 8
We would see it at (656.3 + 2.19 =)
658.49 nm.
(c)
= 2.512 4.85.54
= 13 700
New Senior Secondary Physics at Work
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 Oxford University Press 2011
E1 Astronomy and Space Science
7
Chapter 4 Stars and the Universe
(a) Orbital speed of the Earth
2 πr
=
T
2π  (1.5 10 11 )
=
365  24  60  60
the wavelength increases when received
by us, causing the red shift.
The red shift suggests that the universe
is expanding.
9
= 29 900 m s1
(a)
Time / h
 / 1015 m
Radial velocity / m s1
76
114
52.1
= 100 000  29 900
85
77.9
35.6
= 70 100 m s1
94
15.4
7.04
100
30.6
14.0
Minimum speed of the Earth as seen by
the space traveller
Maximum speed of the Earth as seen by
(b)
the space traveller
= 100 000 + 29 900
= 129 900 m s1
λ v r

(b) By
,
λ
c
for minimum wavelength,
v
 = λ  r
c
129 900
= 700 
3 10 8
= 0.301 nm
Minimum wavelength = 700  0.301
= 699.699 nm
For maximum wavelength,
v
 = λ  r
c
70 100
= 700 
3  10 8
(c)
the time T for the planet to make one
complete revolution is 96 hours.
(d) By T2 =
= 0.164 nm
Maximum wavelength = 700 + 0.164
4π 2 r 3
,
GM
 GMT 2
r =
 4π 2

= 700.164 nm
8
The graph repeats in 96 hours. Therefore
(a) Red shift of a galaxy means that visible
1
3



light from it is slightly shifted to the
Radius of the planet’s orbit r
long-wavelength side when received by
 (6.67 10 11 )( 4 10 30 )(96  3600 ) 2  3
=

4π 2


1
us.
(b) When a galaxy recedes from our galaxy,
= 9.31  109 m
the visible light from it is stretched and
New Senior Secondary Physics at Work
4
 Oxford University Press 2011
E1 Astronomy and Space Science
Chapter 4 Stars and the Universe
Revision exercise 4
4
The whole spectrum would be shifted to
Multiple-choice (p. 132)
longer wavelengths compared with that if it is
1
D
at rest.
2
C
In this case, the star spears to emit radiation
By Stefan’s law, I  T , therefore the energy
4
4
emitted would become (3 =) 81 times its
5
present value.
3
A
4
D
5
C
6
D
(1A)
with longer wavelengths,
(1A)
so it would appear to be cooler.
(1A)
(a) Absolute magnitude
d 
= m  5 log  
 10 
 30 
= +4  5 log  
 10 
= 1.61
1
(a) Distance to the star =
0.08
= 12.5 pc
(b) 12.5 pc = (12.5  3.26) ly
= 40.75 ly
2
(1M)
(1A)
d = 1000 pc
(1M)
(1A)
The star is 1000 pc from the Earth.
(1A)
6
A difference of 5 magnitudes are different in
(a) Star B
LB
(b)
= 2.512 M A  M B 
LA
(1A)
4 .5
= 2.512 M A  M B 
0 .5
Apparent magnitude of star B
= +14  (5  3)
(1M)
= 1
(1A)
MA  MB = 2.39

 dA
m A  5 log
 10

 = 525.17  525 = 0.17 nm
λ v r

By
,
λ
c
radial velocity
 λ 
=
c
 λ 
(1M)
d 
5 = +5  5 log  
 10 
intensities by a factor of 100 times.
IB
= 1 000 000 = 1003
IA
3
(1A)
d 
(b) By M = m  5 log   ,
 10 
Conventional (p. 132)
1
(1M)
(1M)
 
 d B 
  m B  5 log
 = 2.39
 
 10 
d 
(mA  mB) + 5 log  B  = 2.39
 dA 
dB
= 3.01
dA
(1M)
(1A)
The distance to star B is 3.01 times that to
 0.17 
= 
  3  108
 525 
star A.
7
= 9.71  104 m s1 (away from the Earth)
(1A)
Apparent magnitude as seen on the Earth
d 
= M + 5 log  
 10 
1


= M + 5 log 

10

206
265


= M  31.6
New Senior Secondary Physics at Work
5
(1M)
 Oxford University Press 2011
E1 Astronomy and Space Science
Chapter 4 Stars and the Universe
Apparent magnitude as seen on Saturn
d 
= M + 5 log  
 10 
The angular diameter of the nebula is
5.72.
The size of the nebula is larger than that
9.54


= M + 5 log 

 10  206 265 
= M  26.7
of the Andromeda as seen from the
Earth.
(1M)
(1A)
(b)
Difference in apparent magnitude
= (M  31.6)  (M  26.7)
= 4.9
(1A)
The apparent magnitude as seen on the Earth
is smaller than that as seen on Saturn by 4.9.
8
(a) Distance of the star from the Earth
1
=
(1M)
0.05
= 20 pc
tan
(c)
(1M)
log 60
log 2.512
= 0.555
(1A)
(1M)
m = 29.3
(1A)
The apparent magnitude of the nebula is
29.3.
(1M)
10
d 
(a) By M = m  5 log   ,
 10 
(1M)
minimum apparent magnitude m
 100 000 
= +5 + 5 log 

 10 
(1M)
= +25
(1A)
(1A)
(1M)
(b)
Its apparent magnitude is 0.555.
9
from the Earth.
I
By
= 2.512 m0  m  ,
I0
10 = 2.512 26.8m 
 I 
log 
 I0 
m = m0 
log 2.512
=5
(1A)
The nebula would have to be at 3820 pc
 20 
= 5  5 log  
 10 
= 3.49
I
By
= 2.512 m0  m  ,
I0
(1M)
d = 3820 pc
(1A)
(b) Absolute magnitude
d 
= m  5 log  
 10 
3 100
=
d
2
(a)
= 5
log(3.8 10 7 )
log 2.512
= 13.9
tan
 100
=
2 2000
(1M)
 = 5.72
(1A)
New Senior Secondary Physics at Work
(1A)
The absolute magnitude of LBV
1806-20 is 13.9.
6
 Oxford University Press 2011
E1 Astronomy and Space Science
(c)
Chapter 4 Stars and the Universe
From (a), we know that the minimum
12
(a) (i)
apparent magnitude of stars that the HST
gravitational force
can see is +25.
R
π 
2
=    N m
πR 2
1
= Nm
4
2
d 
By M = m  5 log   ,
 10 
(1M)
d 
13.9 = +25  5 log  
 10 
d = 6.03  108 pc
11
Mass contributed to the
Orbital speed
(1A)
(a)
=
Star
L
T
R
1
1
1
0.0045
0.56
0.21
Sirius A
23
1.6
1.9
Vega
51
1.6
2.8
Arcturs
160
0.78
21
Sun
Barnard’s
GM
r
1

G Nm 
4

=
R
2
Star
=
GNm
2R
(1A)
(ii) Mass contributed to the
(3 correct answers.)
(1A)
gravitational force = Nm
(4 correct answers.)
(2A)
Orbital speed =
(b)
Star
GM
GNm
=
r
R
(1A)
Spectral class
(iii) Mass contributed to the
Sun
G
Barnard’s Star
M
gravitational force = Nm
Sirius A
A
Orbital speed =
Vega
A
Arcturs
K
GM
GNm
=
r
2R
(1A)
(3 correct answers.)
(1A)
(5 correct answers.)
(2A)
(b)
(c)
Star
Type
Sun
main sequence star
Barnard’s Star
main sequence star
Sirius A
main sequence star
Vega
main sequence star
Arcturs
giant
(3 correct answers.)
(1A)
(5 correct answers.)
(2A)
New Senior Secondary Physics at Work
7
 Oxford University Press 2011
E1 Astronomy and Space Science
Chapter 4 Stars and the Universe
(Correct labelled axes.)
(1A)
(Correct curve.)
(1A)
mass of the planet
=
(Passing through the three points in (a).)
(1A)
(c)
=
v2r
G
(3.15  10 4 ) 2 (85 000  10 3 )
6.67  10 11
The curve for RX  R will shift upwards,
depending on the density and
= 1.26  1027 kg
distribution of the invisible stars. For
Mass of the star = 1.99  10  0.32
14
(1A)
30
= 6.368  1029 kg
same density as that within the disc of
radius R, the rotational curve will be as
For the inner planet,
follow:
Tinner = 30  24  3600 = 2.592  106 s
Tinner2 =
4π 2 a inner 3
GM
(1M)
ainner
 GMT inner 2
=
 4π 2

1
3



 6.67  10 11  6.368  10 29  (2.59  10 6 ) 2 
=

4π 2


= 1.93  1010 m
By
Tinner2
ainner3
=
Touter 2
a outer 3
,
(1M)
1
13
2
 T
3

aouter =  outer   a inner 3 
 Tinner 



(Correct curve.)
(1A)
4.000 42  3.999 58
 =
= 0.000 42 nm
2
4.000 42  3.999 58
=
= 4.000 00 nm
2
λ v r

By
,
(1M)
λ
c
1
 2  2
3
=    (1.93 10 10 ) 3 
 1 

= 3.06  1010 m
(1A)
The semimajor axis of the orbit of the outer
planet is 3.06  1010 m.
radial velocity of the spacecraft
 λ 
=
c
 λ 
15
(a)
 0.000 42 
8
=
  3  10
4.000
00


= 3.15  104 m s1
By v =
GM
,
r
New Senior Secondary Physics at Work
(1M)
8
 Oxford University Press 2011
3
E1 Astronomy and Space Science
(All points plotted correctly.)
Chapter 4 Stars and the Universe
(2A)
(c)
(i)
d sin  = n
nλ
sin  =
d
(Points plotted correctly except one.)
(1A)
(Smooth curve through points.)
For diffraction grating,
(1A)
= 2  (590  109) 
(b) This is because the relative motion
(c)
between source and observer
(1A)
causes the Doppler effect.
λ v r
By
,

λ
c
(1A)
(1M)
(4.5  105)
 = 32.1
(1M)
(1A)
(ii)
(1M)
star’s orbital speed
 λ 
=
c
 λ 
 393 .69  393 .40 
8
=
  3  10
393.40


(1M)
= 2.21  105 m s1
(1A)
(d) (i)
Period measured from the graph
= 70 hours
(Appropriate diagram.)
(1M)
Light from slits is coherent. (1A)
= (70  3600) s
= 2.5  105 s
2πr
(ii) By v =
,
T
At positions with path difference
(1A)
equal to a multiple of one
(1M)
wavelength,
(a) (i)
(1A)
and constructive interference
2.21 10 5  2.5 10 5
=
2π
(1M)
= 8.79  109 m
(1A)
Doppler effect
(1A)
(ii) The universe is expanding.
(b)  = 650  590 = 60 nm
λ v r

By
,
λ
c
occurs to produce bright spectral
lines.
(2C)
(1A)
17
(a) The recessional speed increases with
distance.
(1A)
(b) Red shift / examined line emission
(1M)
spectra / Doppler effect
(c)
Earth
 λ 
=
c
 λ 
(1A)
Galaxies moving away from each other
(1A)
suggests that they started out from a
single point / explosion.
 60 
8
=
  3  10
 590 
New Senior Secondary Physics at Work
(1A)
(For effective communication.)
velocity of the galaxy relative to the
= 3.05  107 m s1
(1A)
the light waves arrive in phase,
radius of the orbit
vT
=
2π
16
(1A)
(d) 1000
(1A)
million light years
(1A)
9
(1A)
(1A)
 Oxford University Press 2011
E1 Astronomy and Space Science
18
(a) (i)
Galaxies A and B.
Chapter 4 Stars and the Universe
(1A)
(b) (i)
The wavelength is longer or the
light is red-shifted.
(1A)
(ii) Red shift /Doppler effect
(1A)
(iii) Milky Way Galaxy
(1A)
By
λ v r
,

λ
c
(1M)
change in wavelength
v 
=  r λ
 c 
(b) (i)
 6.1 
=
  656.3  10 9
 3  10 8 
(1M)
= 13.3  1015 m
(1A)
(4 points plotted correctly.)
(1A)
(ii)
(3 points plotted correctly.)
(1A)
(All points plotted correctly.) (2A)
(Best fit line.)
(1A)
(ii) The graph is a straight line through
speed
the origin, which means
is
distance
a constant.
19
(All points plotted correctly.) (2A)
(iii) (Reasonable curve.)
(1A)
(iv)
(1A)
(a) The planets orbit the Sun in ellipses,
with the Sun at one focus.
(1A)
The line joining the Sun and a planet
sweeps out equal areas in equal intervals
of time.
(1A)
(Correct position.)
The square of a planet’s orbital period is
(v) T = 72 h
proportional to the cube of its semimajor
(1A)
(1A)
axis of its orbit (or planet’s average
distance from the Sun).
New Senior Secondary Physics at Work
(1A)
10
 Oxford University Press 2011
E1 Astronomy and Space Science
3
(vi) By r =
GMT 2
4π 2
Chapter 4 Stars and the Universe
,
radius of the planet’s orbit
3
=
(6.67  10 11)( 4  10 30 )( 72  3600 ) 2
4π 2
(1M)
= 7.69  10 m
9
(1A)
Physics in articles (p. 137)
The intensity increases back to the
(a) From Figure b, the percentage drop in the
I
intensity during the eclipse
is about
I
normal value from t = 1 h to t = 1.5 h.
This represents the period for the partial
eclipse as shown in the figure below.
0.25%, where I = Ip+ Is is the total intensity of
the planet (Ip) and star (Is) together.
Is
 0.9975

I p  Is

I p  Is
Is
 1
Ip
Is

(1M)
(1M)
1
 1.0025
0.9975
 Ip : Is = 0.0025 : 1
(b) (i)
(1A)
(1A)
The intensity drops at the beginning
from t = 1.5 h to t = 1 h. This
represents the period for the partial
(Correct positions drawn.)
eclipse as shown in the figure below.
(ii) From (b)(i), we have:
Dp
= 0.5 hr
vp
(1A)
Ds  D p
vp
= 2 hr
(1A)
(1M)
(1M)
Therefore,
Ds  D p
2
=
0 .5
Dp
Ds  Dp = 4Dp
Dp 1
=
Ds 5
From t = 1 h to t = 1 h, the planet
travels a distance that is about the
(c)
diameter of the star as shown in the
intensity stays low.
New Senior Secondary Physics at Work
The star is hot and emits both visible and
infra-red radiation.
figure below. During this period, the
(1A)
(1A)
However, the planet is not hot enough to emit
(1A)
visible light.
11
(1A)
 Oxford University Press 2011
E1 Astronomy and Space Science
Chapter 4 Stars and the Universe
Hence, the contrast in visible light between
the star and the planet is enormous.
On the other hand, the planet is warm and
emits significant amount of infra-red radiation.
(1A)
Therefore, the contrast in infrared is much
lower.
(d) The hotter the object, the more infra-red
radiation it emits.
(1A)
Therefore, by measuring the infra-red
intensity of an extrasolar planet, one can get a
rough estimate of its surface temperature.(1A)
To get a more accurate result, one needs to
measure the distribution of the emission
intensity in wavelength  the spectrum, and
one can compare the emission spectrum of an
extrasolar planet to those of the objects heated
up to known temperatures in the laboratory.
(1A)
New Senior Secondary Physics at Work
12
 Oxford University Press 2011