Download The first nine names are R through Z, and 10,... RU. Therefore, you would call this star

Problem 5 (20 points): Give the appropriate name for the following variable stars. Use a reasonable abbreviation for the name of the constellation.
a. (10 pt): The 13th variable star discovered in the constellation Duntante.
The first nine names are R through Z, and 10, 11, 12, and 13 are RR, RS, RT, and
RU. Therefore, you would call this star
“RU Dun”
b. (10 pt): The 302nd variable star discovered in the constellation Iampetus?
In this case, it is easiest to count backwards:
Final 10 stars
Previous 11 stars
Previous 12 stars
(Numbers 325 - 334)
(Numbers 314 - 324)
(Numbers 302 - 313)
are QQ through QZ
are PP through PZ
are OO through OZ
so the 302nd star is OO. Therefore, you would call this star
“OO Iam”
Problem 4 (15 points)
a. (8 pt): Use Zeilik Figure 13-10 and the position of the RR Lyrae gap to estimate the distance to
the globular cluster M3.
Figure 13-10 shows the RR Lyrae gap to be at apparent magnitude m ≈ 15.5. These
stars all have absolute magnitude M ≈ 0.5, regardless of their period. Therefore
m – M = 15.0 = 5 log d – 5
and so
log d = 4
or
d = 10 4 pc = 10 kpc
which is the same answer you get with main sequence fitting. (See Exam 1.)
b. (7 pt): What is the ratio of the periods for population I and population II Cepheids with 1000
times the luminosity of the sun?
From Zeilik Figure 18-3, midway between the 102 and 104 tick marks, you read
log PI = 0.7
log PII = 1.3
so
log PI − log PII = log (PI/PII) = −0.4
or
PI/PII = 10−0.4 = 0.4
Problem 3 (25 points): Consider the X-ray binary system Cygnus X-1.
a. (5 pt): What is the Schwarzschild radius of the black hole companion in Cygnus X-1.
2GM BH
 M BH 
R BH = --------------------- = 3km ×  ---------------- = 48km
c2
 M SUN 
where we used Eq.17-7b in Zeilik, and the 16 solar mass value for Cyg X-1.
b. (15 pt): Matter falls from a large distance onto the black hole emitting 5% of its gravitational
potential energy as X-rays. What rate of mass accretion (i.e. kg/sec) is necessary to emit with the
X-ray luminosity observed for Cygnus X-1?
Let a small amount of matter ∆m fall into the black hole in time ∆t. The amount of
energy radiated into X-rays in this time is just L∆t = [ ( GM BH ∆m ) ⁄ R BH ] × 0.05
so we can write
R BH
L
∆m
-------- = ---------- × ----------------- = ( 40L ) ⁄ c 2
0.05 GM BH
∆t
(Notice how much drops out in this equation!) With L=2×1030 W (Zeilik Tab. 18-4
or lecture notes), this gives
14
–8
∆m
-------- = 9 ×10 kg/sec = 1.4 ×10 Solar Mass/year
∆t
c. (5 pt): Is this consistent with the expected mass loss rate for the visible companion star
observed in Cygnus X-1? (You might want to use the converstion that one solar mass per year
equals 6.3×1022 kg/sec.) Explain.
The companion to Cygnus X-1 is an O9 supergiant. The mass loss in such massive
stars is on the order of 10−7 solar masses, or more, per year. (See Zeilik page 309,
or class notes.) This can easily meet the requirements as the mass source falling
onto the black hole to produce X-rays.
Problem 2 (15 points): Consider two white dwarf stars. Star #1 is twice as massive as star #2.
The stars have the same surface temperatures. Ignoring relativistic effects, what is the ratio of
their luminosities?
The two important relations you need are
L ∝ R2T 4
and
1
R ∝ -------------M1 ⁄ 3
So you can write
2
4
2⁄3
L1
 R 1  T 1
 M 2
1 2⁄3
------ =  ------  ------ =  --------
=  ---
= 0.63
 2
L2
 R 2  T 2
 M 1
since T1=T2.
Problem 1 (25 points): Consider a star identical to our sun.
a. (20 pt): Assume the star converts 10% of its hydrogen to helium while on the main sequence,
and then arrives on the “helium main sequence” as a red subgiant, burning helium in its core. If
the star then burns all of this helium to carbon, estimate its lifetime in this stage. The atomic
weights of 4He and 12C are 4.0026 and 12.000 times the mass of hydrogen, respectively. Refer to
figures in Zeilik for other information you may need.
The mass available for helium burning is 10% of the solar mass or 2×1029 kg.
This translates to a number N=2×1029/(4×1.67×10−27)=3.0×1055 helium nuclei.
In helium burning, three 4He are converted to one 12C releasing
(3×4.0026−12.000)×1.67×10−27×c2 = 1.17×10−12 J
of energy. The total available energy for helium burning is therefore
(3.0×1055 / 3) × 1.17×10−12 = 1.17×1043 J
Zeilik Figure 16-5 shows that the star’s luminosity at this point in its life is 100
times the solar luminosity, or 3.9×1028 J/sec. Therefore, the star’s lifetime in this
stage is just
1.17×1043 / 3.9×1028 = 3.0×1014 sec = 10 million years
b. (5 pt): List the final three stages of this star’s life, which follow the core helium burning stage.
The final three stages of this star’s life, after exhausting helium in the core, are
• Asymptotic red giant / variable star
• Planetary nebula
• White dwarf
Exam #2
79205 Astronomy
Fall 1996
NAME:
Solution Key
You have two hours to complete this exam. There are a total of five problems and you are
to solve all of them. Note that not all the problems are worth the same number of points.
You may use your textbook (Zeilik), workbook (Hoff), and class notes and handouts, or
other books. You may not share these resources with another student during the test.
Indicate any figures or tables you use in your calculations. Show all Work!
GOOD LUCK!
Problem
Score
Worth
1.
25
2.
15
3.
25
4.
15
5.
20
Total Score:
100