Download Exercise 4

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.) Assumption: the stars
are blackbodies.
(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 
(ii) Applying I 
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

4π  4.24  9.46  10
15

2
 3.23  10 11 W m 2
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 ratio 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) By Stefan's law:
L⊙ = 4πR⊙ 2 σT⊙ 4
Luminosity of object A:
LA = 200L⊙ = 4πRA2 σ(0.8T⊙ )4
200 (4πσR⊙ 2 T⊙ 4) = 4πσR2 (0.8T⊙ )4
RA = 22R⊙
Assumption: Object A is well approximated as a blackbody.
(b) C > D > E > A > B (Largest m to smallest m, i.e. dimmest to brightest, i.e. least luminous to
most luminous since the objects are at the same distance from the Earth)
(c) B > A > E > D > C (Using m–M = 5 log d/10, the largest m-M corresponds the farthest star)
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
(e)
LA
= (RA / R☉ )2 (TA / T☉ )4
L⊙
5.407 = (RA / R☉ )2 (6900/5800)4
The radius of Altair is 1.64 R☉.
7.
(a)
(i)
LS  TS4 (4 Rs2 )
L   T 4 (4 R 2 )

LS TS4 RS2
 4 2
L
T R
T 
R S
T 
 
2
1
 L 2
  RS
 LS 
(ii) Sun:
TS = 5780 K
Betelgeuse: T = 3650 K ;
L = 126000 LS
1
2
 5780   126000 LS  2
R
 
 RS = 890 RS
LS
 3650  

(b)
(i)
(ii)
(c)
Brightness =
L
L
or  2 or brightness increases with L and decreases with d.
2
4π d
d
L of Betelgeuse is larger if the distance d is (197+45) pc. Thus R is also larger.
The parallax measurement (accurate to within ~ 100 pc) is too small,
(~ (1/200)” = 5 milliarcsec).
http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit1/distances.html
L = 109 Ls brightness =
(0.01  10 9 Ls )
4π d 2
d = 200 × 206265 AU = 41253000 AU
Brightness =
Ls
(0.01  10 9 )
(0.01  10 9 )
=
brightness of the Sun
41253000 2
41253000 2 4π (1AU) 2
= 5.88 × 109 brightness of the Sun
MC
1-5 A A C B A
6-10 C D B C D
11-15 B C B C A
16-20 D C A C B
Explanations to selected mc
2. d = 1/p = 1/0.001 = 1000 pc
3. Absolute magnitude is related to luminosity. Apparent magnitude depends on brightness which in
turn depends on luminosity and distance.
4. Recall: L = 4R2 T4.Thus luminosity depends on size R and surface temperature T.
d 
6. By M = m  5 log   ,
 10 
8. The presence of hydrogen absorption lines only indicates that there are hydrogen lines in the
outer atmosphere of star, but the amount of hydrogen is unknown.
9. For Statement 1, from the H-R diagram, luminosity of W = luminosity of X, and temperature of
W is higher than that of X, so according to L = 4R2 T4, radius of W must be smaller than that of X.
Thus (1) is correct.
For statement 2, temperature of W = Temperature of Y, and luminosity of W is higher than that
of Y, so by L = 4R2 T4, radius of W > radius of Y. Thus (2) is correct.
For statement 3, luminosity of Y = luminosity of Z, since temperature of Y > temperature of Z,
thus by L = 4R2 T4, radius of Y < radius of Z.
10. Star X has shifted 2 grids, which corresponds to 0.2”.
Thus parallax p = 0.2 / 2 = 0.1” (refer to diagram on the right)
Distance required d = 1/p = 1/0.1 = 10 pc
L
4d 2
P and Q have the same luminosity, and P is 2.5 times brighter than star Q,
so P must be nearer to the earth than Q, and the Q’s distance must be 5 times
that of P.
11. Apparent brightness b 
12. d = 1/p = 1/0.0889 = 11.25 pc = 11.25 x 206265 = 2.32 x 106 AU
13. L = 4R2 T4 =A T4, thus L is directly proportional to surface area A.
The graph is a straight line through origin.
14. Surface temperature is known from H-R diagram. Radius can be found from L = 4R2 T4.
L
Apparent brightness is unknown because distance is unknown ( Apparent brightness b 
)
4d 2
d 
15. X: mX - MX = 5 log   => mX – (-1) = 5 log 100/10
=> mX = 7
 10 
d 
Y: mY - MY = 5 log   => mY – 3 = 5 log 10/10
=> mY = 3
 10 
Since the apparent magnitude of Y is smaller than that of X, Y appears brighter.
17. Since L = 4R2 T4. The luminosity is proportional to (temperature)4. 34 = 81.
18. Statement 1. A blackbody emits radiation in a wide range of frequencies (or wavelengths)
Statement 2. Recall max 
1
T
Statement 3. A difference of 10 in apparent magnitude corresponds to a ratio of brightness (or
intensity) of 2.51210 = 10004
19. A bluer star is hotter than a red star, so its spectrum peaks at a shorter wavelength.