# Exercise 4 (Stars and the universe) Suggested answers

```Exercise 4 (Stars and the universe) Suggested answers
1. (a) The star is nearly a blackbody, the spectrum of a star can be approximated as a blackbody
radiation curve. On the curve, there is a peak which shifts to shorter wavelength when the
temperature of the blackbody increases. From the position of the peak, astronomers can
deduce the surface temperature of the star. In general, the spectra of hot stars peak at shorter
wavelengths and therefore hot stars appear bluer. (More precisely, according to Wien’s
displacement law, λmax and T of a black body are related by  max T  2.90  10 3 m K . Thus,
we can calculate the surface temperature of a star from the spectrum.)
(b) When the spectrum of a star is examined closely, specific absorption lines can be found. The
positions (i.e. wavelength) and strengths of the absorption lines in the spectrum are “atomic
fingerprints” of the elements and are unique for each element, thus the chemical composition
of the atmosphere of the star can be obtained.
2. (a) Luminosity L of a celestial body is the actual amount of electromagnetic radiation emitted by
the body in each second.
(b)
(i)
Since intensity I is the radiant power received per unit area at a position with a radial
distance d from the source, we have I 
L
L

.
total surface area 4πd 2
L
, the intensity of Proxima Centauri measured on the Earth is
4 πd 2
6.53  10 23
I
 3.23  10 11 W m 2
15 2
4π  4.24  9.46  10
(ii) Applying I 



3. (a) Since a large portion of the spectrum of star X lies around the wavelength 700 nm, the star
appears red.
(b) From the graph, the spectrum of the Sun peaks at 5 × 10−7 m while the spectrum of star X
peaks at 7 × 10−7 m. Applying Wien’s displacement law to the spectrum of the Sun and the
star X,
max T  (5  10 7 )  (5800)  7  10 7  T
T  4142.8  4140 K
(c) The radiant flux or intensity I of star X is given by
I  T 4  (5.67  10 8 )  (4142.8) 4  1.67  107 W m 2
(d) The student may not be correct. Luminosity L of a star is given by L = 4R2 T4, that is, the
luminosity depends on both the surface temperature of the star and the size of the star.
Although the Sun has a higher surface temperature, the radius of star X is unknown, in other
words, star X may be larger than the Sun. Therefore, there is not enough information to
determine whether the Sun is more luminous than star X or not.
4. (a) Applying L = 4R2 T4
= 4  x (7 x 108 )2 x 5.67 x 10-8 x 60004
= 4.52 x 1026 W
The luminosity of star A is 4.52 x 1026 W.
(b) The rate of their luminosities is
2
2
4
 7  108   6000 
L A R A TA

 2 4  
 4
9  
LB RB TB
 1.4  10   3000 
4
Star A’s luminosity is 4 times that of star B.
5.
(a)
luminosity / L
10 000
red giants
mainsequence
stars
1
Sun
0.000 1
white dwarfs
30 000
6000
3000
surface temperature / K
(b) (i) Spectral class shows the surface temperature of a star. The spectral classes are arranged
as O, B, A, F, G, K, M if stars are arranged from the highest surface temperature to the
lowest surface temperature (1A). W is of the class B while X and Y are of the same class
K. Thus, W has a higher surface temperature while X and Y are of the same lower surface
temperature (1A).
(ii) (Note the arrow on the axis of the absolute magnitude: it increases downward). A smaller
absolute magnitude of a star implies the star has a higher luminosity (1A). Thus, Y has the
highest luminosity while W and X have the same lower luminosity. From (b)(i), Y and X
have the same lower surface temperature. Thus, Y should be the largest according to the
equation L = 4R2 T4 (L the largest, but T the smallest)(1A). Since W and X have the
same luminosity and W has a higher surface temperature than X, thus W should be the
smallest (1A).
6.
(a) By
IA
= 2.512 m⊙  m A  ,
I⊙
IA = 2.512 m⊙  m A   I☉ = 2.512 26.740.77   I☉ = 9.90  1012 I☉
The brightness (or intensity) of Altair is 9.90  1012 I☉.
(b) Altair must be farther away from us.
It is because Altair has lower absolute magnitude than the Sun, so Altair
must be more luminous, however it has a higher apparent magnitude, in other
words, it is apparently dimmer than the Sun, so it must be farther away than the
Sun.
d 
(c) By M = m  5 log   ,
 10 
m M
= 10 5
0.77 2.21
= 10 5
d
 10
 10 = 5.15 pc
The distance between Altair and the Earth is 5.15 pc.
(d) By I = T4,
1.285 x 108 = 5.67 x 10-8 T4
T = 6900 K
The temperature of Altair is 6900K.
(e) RA =
LA
2
=
10.6
 1.64 R☉
(6900 / 5800) 2
TA
The radius of Altair is 1.64 R☉.
7. (a) Orbital speed of the Earth
=
2π  (1.5  10 11 )
2 πr
=
= 29 900 m s1
T
365  24  60  60
Minimum speed of the Earth as seen by the space traveller
= 100 000  29 900 = 70 100 m s1
Maximum speed of the Earth as seen by the space traveller
= 100 000 + 29 900 = 129 900 m s1
(b) By
λ v r

,
λ
c
for minimum wavelength,
 = λ 
vr
129 900
= 700 
= 0.301 nm
c
3 10 8
Minimum wavelength = 700  0.301= 699.699 nm
For maximum wavelength,
 = λ 
vr
70 100
= 700 
= 0.164 nm
c
3  10 8
Maximum wavelength = 700 + 0.164
= 700.164 nm
8. (a) d = 1/p = 1/0.545 = 1.83 pc
   0 v
(b) By


0
0
c

v
 106.8  103
  
 656.28 = - 0.234 nm
0 c
3  108
Negative sign means that the observed wavelength is smaller than the wavelength observed if
the object is at rest. This corresponds to blue shift. Thus the star must be approaching.

(c) Speed of the star = 902  106.82  140 km s -1
(d) To find the change in angular position, transverse direction should be considered.
Observed change in position s = vt = 90 km s-1 x 365 x 24 x 3600 = 2.838 x 1012 m
Required change in angular position
 = s/r
= 2.838 x 1012 / (1.83 x 3.09 x 1016) = 5.02 x 10-5 rad
9. (a)
Time / h
76
85
94
100
 / 1015 m
114
77.9
15.4
30.6
52.1
35.6
7.04
14.0
(b)
(c) The graph repeats in 96 hours. Therefore the time T for the planet to make one complete
revolution is 96 hours.
(d) By T2 =
4π 2 r 3
,
GM
 GMT 2
r =
 4π 2

1
1
11
30
2 3

3
 =  (6.67 10 )( 4 10 )(96  3600 )  = 9.31  109 m

4π 2



Radius of the planet’s orbit r is 9.31 x 109 m.
10. (a)
Star
Sun
Barnard’s Star
Sirius A
Vega
Arcturs
L
1
0.0045
23
51
160
T
1
0.56
1.6
1.6
0.78
R
1
0.21
1.9
2.8
21
Spectral Class
G
M
A
A
K
11. (a) The recessional speed increases with distance. (1A)
(b) Red shift / examined line emission spectra / Doppler effect
(c) Galaxies moving away from each other (1A)
suggests that they started out from a single point / explosion.
(d) 1000 million light years
MC
1-5 A A C B A
6-10 A D D B D
Type
main sequence star
main sequence star
main sequence star
main sequence star
giant
(1A)
(1A)
(1A)
11-13 C A D
Explanations to selected mc
2. d = 1/p = 1/0.001 = 1000 pc
4. Recall: L = 4R2 T4
9. Statement 2, the rotational velocities in the outer part of the galaxy are greater than expected.
Refer to the rotation curve.
11. Recall: L = 4R2 T4. The luminosity is proportional to (temperature)4. 34 = 81.
```
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