### Astro 501 Midterm This test is open notes. Answer all questions and

```Astro 501
Midterm
This test is open notes. Answer all questions and PLEASE BE NEAT. If possible, write out
the formulae before plugging in the numbers. If the questions require assumptions, state and
justify them clearly! Here are some constants that may (or may not be useful):
1. (10 pts) There are several regions of the sky that have extremely deep data from the
Hubble Space Telescope and follow-up spectroscopy from 8-m class telescopes on the ground.
The table below summaries some of them:
Field Name
Size
α(2000)
δ(2000)
UDS
GOODS-S
COSMOS
GOODS-N
AEGIS
2.4′ × 2.4′
10′ × 16′
1◦ × 1◦
10′ × 16′
10′ × 67′
02:18:00
03:32:30
10:00:28
12:36:55
14:17:43
−05:00:00
−27:48:20
+02:12:21
+62:14:15
+52:28:41
The Hobby Eberly Telescope, which is located in west Texas (latitude ϕ = 30◦ 40m 48.5s ), is
a fixed altitude telescope which always observes 35◦ from the zenith. The spectrographs for
the Hobby Eberly Telescope Dark Energy Experiment are currently scheduled to come on
line toward the end of June. Which of the above fields should we plan to observe first? (You
don’t need to do all the calculations in detail; you can just argue this question as if someone
Since the telescope is at a latitude of ϕ ∼ 30◦ and can only observe 35◦ from the zenith,
the furthest south it can point is ϕ − 35 = −5◦ . So the telescope cannot reach GOODS-S
and can only barely reach the Ultra Deep Field (UDS). Next, consider the hour angle. The
conversion between the horizon system and the equatorial system gives
sin a = sin δ sin ϕ + cos δ cos ϕ cos H
so
cos H =
sin a − sin δ sin ϕ
cos δ cos ϕ
which implies that the HET can only observe COSMOS at an hour angle of H = ±1.4 hours
for COSMOS, GOODS-N at H = ±1.6 hours for GOODS-N, and AEGIS at H = ±2.5 hours.
At the end of June (the summer solstice), 18 hours is on the meridian at midnight; if you
estimate about 8 hours of dark time per night, then 14 hours is on the meridian at dusk, and
22 hours at dawn. At a right ascension of 10 hours, COSMOS is gone, since the latest it
can be observed is 1.4 hours past the meridian, i.e., when α(2000) = 11.4 hours transiting.
GOODS-N can be observed 1.6 hours past the meridian, so it is just barely visible at the
beginning of the night. AEGIS is the only real option, since it can be observed when 16.8
hours is transiting, which is 2.8 hours after twilight.
2. (10 pts) A telescope collects 100 photons a second from an m = 18 star observed through
a 2 arcsec diameter circular aperture. The sky brightness is 20 magnitudes per square arcsec.
How long does one need to integrate to achieve a precision of 0.01 mag. You may assume
that the detector is noiseless and that all the light from the star is falling into the aperture.
A precision of ∆m = 0.01 implies and error in the signal of
F + ∆F
∆F
F
= 10∆m/2.5 = 1.009 =⇒
= 0.009 =⇒ SNR =
= 109
F
F
∆F
(Alternatively, you can use the approximation that for small values of ∆m, σ(m) = 0.XX is
roughly an XX% error.)
Now let N be the count rate from the star and n be the count rate from the sky. After t
seconds, the total number of counts from the star will be N t, while that from the sky will be
n tA, where A = πr2 is the area of the aperture. For a 2 arcsec diameter aperture, r = 1, so
A = π, and, for a ∆m = 2 mag diﬀerence between the star and sky, n = N 10−∆m/2.5 =
15.8. The signal-to-noise of an observation is then
Nt
Nt
SNR = √
=√
N t + nt
N + N A 10−∆m/2.5
so
(
t=
SNR2 1 + A 10−∆m/2.5
N
)
= 178 s
3. (10 pts) Estimate the number of neutrinos going through you each second. (Assume that
your cross-sectional area is 1 square meter.)
Hydrogen fusion produces Q = 6.3 × 1018 ergs gm−1 , and during the fusion, two neutrinos
are produced per four hydrogen atoms. Therefore, the Sun produces
Lν = 0.5
L⊙
neutrinos s−1
QmH
and the neutrino flux over a 1 m2 region at a distance of d = 1 A.U. is
L⊙
Fν = 0.5
QmH
(
104
4πd2
)
= 6.5 × 1014 neutrinos s−1
4. Below is a list of some of the stable isotopes of selenium, bromine, and krypton.
Element
# protons
Selenium
Bromine
Krypton
34
35
36
Stable Isotopes
76, 77, 78, 80, 82
79, 81
78, 80, 81, 82, 83, 84
a) (5 pts) Of the isotopes listed above, which would you expect to be cosmically least
abundant? Why?
78
80
Kr
Kr
76
77
Se
82
Kr
79
Br
78
Se
Br
Se
81
Kr
Se
83
Kr
84
Kr
81
80
82
Se
78 Kr does not lie on the s-process chain, and it is shielded from the r-process. It can therefore
only be formed via the p-process. Since this requires overcoming a substantial electrostatic
repulsion, we can expect that it is quite rare. (And indeed it is.)
b) (5 pts) How is 81 Br produced? How is 79 Br produced?
According to the table, 80 Br is unstable (it has a half-life of 17.6 min), so 81 Br cannot be
created via the s-process. Conversely, there is nothing stopping it from being the product of
a series of β-decays from a very neutron-rich isotope. It is therefore an r-process element.
79 Br can be made via the s-process and the r-process.
5. (10 pts) The brighter component of a binary star has log L = 1.5, log Teﬀ = 3.9, and a
mass of M = 0.7M⊙ . The two stars are separated by 0.5 A.U. Has mass transfer played a
role in the evolution of the system? How do you know?
The primary of the system is not on the main sequence, and, in fact, its position on the HR
diagram is consistent with the star being on the horizontal branch. This suggests the star
began life as a ∼ 1M⊙ star. If this is correct, then, while at the top of the RGB, its size
was ∼ 100R⊙ , or ∼ 0.5 A.U. Mass transfer must have occurred. On the other hand, if the
star began life with M > 2M⊙ , then, in order to get down to M = 0.7M⊙ , it must have
lost an abnormal amount of mass, i.e., via mass transfer. So this system must have evolved
via binary evolution.
6. A 1M⊙ protostar sequence star collapses from a radius of ∼ 100R⊙ to 5R⊙ .
a) (5 pts) If the star has an initial luminosity of ∼ 50L⊙ , very roughly, how long will this
collapse take?
The star will collapse on a thermal timescale, so
GM2
∆t ∼
∼ 2 × 1013 s ∼ 6 × 105 yr
RL
b) (5 pts) What is the star’s mean luminosity over this time period? (For simplicity, you
may assume that the star acts as a uniform density, ideal gas.)
During the collapse, the gravitational potential of the star changes by
(
3
1
1
∆Egrav = GM2
−
5
R1 R2
)
= 4 × 1047 ergs
For an ideal gas star to stay in virial equilibrium, half of this energy must go into heating the
star; the other half is radiated away. So during this period, the average luminosity of the star
must be
1 ∆Egrav
LT =
∼ 6L⊙
2 ∆t
Table of Constants
c = 3 × 1010 cm sec−1
h = 6.62 × 10−27 erg sec
σ = 5.7 × 10−5 ergs cm−2 K−1 s−1
M⊙ = 2 × 1033 g
L⊙ = 3.8 × 1033 ergs s−1
R⊙ = 7 × 1010 cm
1 MeV = 1.6 × 10−6 ergs
1 A.U. = 1.5 × 1013 cm
k = 1.38 × 10−16 ergs K−1
NA = 6.02 × 1023 atoms
a = 7.6 × 10−15 ergs cm−3 K−4
G = 6.67 × 10−8 dyne cm2 g−2
QH = 6 × 1018 ergs g−1
mH = 1.67 × 10−24 gm
1 A.M.U = 1.66 × 10−24 gm
1 pc = 3.1 × 1018 cm
```