Download Astronomy 114 Problem Set # 7 Due: 30 Apr 2007 SOLUTIONS 1

Astronomy 114
Due:
Problem Set # 7
30 Apr 2007
SOLUTIONS
1 What is the smallest angle (in arc seconds) that could theoretically be resolved
by the 10-m Keck telescope at a wavelength of 500 nm?
The angular resolution of a telescope with diameter D, at some wavelength
λ satisfies:
λ
radians
θ = 1.22
D
If λ = 500 nm, and for Keck, D = 10 m, we have
θ = 1.22
5 × 10−1 m
500nm
= 1.22×
= 6.1×10−8
10m
10m
radians = 0.0126 arcseconds
Unfortunately, due to atmospheric seeing, the angular resolution is limited to
a fraction of an arc second at best. Especially for ground-based telescopes,
the main goal is collecting photons!
2 How big would a radio telescope observing at 20 cm wavelength have to be in
order to resolve the same angle as the Keck telescope in the last problem?
Since 20 cm is in radio wavelength, we expect a very large diameter of the
telescope in order to resolve small angle, such as 0.0126 arc second. Use the
equation as previous question, we have
θ = 1.22
λ
D
radians,
which can be used to calculate the diameter of the telescope by solving for
D given θ. That is:
0.2m
λ
= 1.22 ×
= 4 × 106 meters
−8
θ
6.1 × 10 radians
= 4 × 103 km.
D = 1.22
This is a large fraction of the Earth’s surface and certainly much larger
than a common optical telescope, say, Keck. So radio astronomers use an
interferometer. This consists of widely separated individual antennas whose
signals are later combined to resolve such small angles.
3 The Sun orbits the center of the Galaxy at the speed of 220 km/s, 8500 pc from
the center. What is the mass enclosed by the Sun’s orbit?
Since the mass distribution of the Galaxy is roughly spherical symmetric, the
Sun’s motion is determined by the mass enclosed by its orbit. Therefore The
centrifugal force should be roughly balanced by the gravitational pull. Let
M(r0 ) denote the mass enclosed, we have
Fcentri =
2
M(r0 )M⊙
M(r0 )Vcirc
=G
= Fgrav .
r0
r02
Thus the enclosed mass
M(r0 ) =
=
=
=
=
2
r0 Vcirc
G
(8500pc) × (220km/s)2
G
(2.6 × 1022 cm) × (2.2 × 107 cm/s)2
6.67 × 10−8 cm3 g −1s−2
1.89 × 1044 g
9.5 × 1010 M⊙ ≈ 1011 M⊙
(1)
4 The rotation curve for the Sa galaxy NGC 4378 is shown in Figure 26-28 in your
book. Using the data from that graph, calculate the orbital period of stars 20
kpc from the galaxy center. What is the mass of the galaxy out to 20 kpc from
its center?
We can read from the plot that for NGC 4378, the circular velocity at 20 kpc
is roughly 280 km/s. Therefore the period of stars at 20 kpc from the center
takes
2 × π × 20kpc
2πR
=
Vcirc
280km/s
2 × π × 20 × 1000 × 3.09 × 1013 km
=
280km/s
16
= 1.4 × 10
seconds
8
= 4.4 × 10 yrs
P =
to complete one entire orbit.
]
When stars orbit about the center of a galaxy, their centrifugal force should
be roughly balanced by the gravitational force by the galaxy. We can approximate the galaxy’s mass distribution as spherically symmetric, so the
gravitational pull at some radius R is induced by the mass within R, we
denote it M(R). For the centrifugal force, we have
Fcent =
2
mstar Vcirc
,
R
for gravitational force, we have
Fgrav =
GM(R)mstar
.
R2
Equal these two forces, we have
M(R = 20kpc) =
2
Vcirc
×R
= 2.6 × 1041
G
kg = 3.65 × 1011 M⊙
[Alternatively, we could use the M(r) relationship from Newton’s Laws that
we derived earlier.] This mass is much larger than determined from the
luminous matter, so there should be a lot of invisible “dark” mass in the
NGC 4378.
5 What types of galaxies are most likely to have new star forming? List observational evidence to support your answer.
Irregular galaxies are possible sites for new star formation. There are often
knots of star forming region, indicated by OB association and HII regions,
seen throughout the irregular galaxies. Spiral galaxies are also sites of new
star formation. The observational evidence for this includes HII regions in
spiral arms, bluer color of stellar light on the leading edges of spiral arms, and
the overall bluer color of spiral galaxies. By contrast, elliptical galaxies show
little or no star formation and low relative gas fraction. Elliptical galaxies
would be the worst place to look for new star formation.
6 Freedman & Kaufmann Chap. 26, Problem 28, pg. 605
The light collecting power of a telescope is proportional to its area; and area
is proportional to the sqaure of its diameter. So by using larger telescope we
can dramatically increase the ability to see very remote (dimmer) galaxies.
Note that Hubble law is a function of distance, so the farther we observe, the
better accuracy we can gain in the Hubble law.
7 Freedman & Kaufmann Chap. 26, Problem 29, pg. 605
The image shows very evident bright regions that may be sites of OB associations and HII regions (note the Hydrogen α color), which indicate the
recent star formation. The inner part of the galaxy is rather blue, indicating
a young population of stars.
8 Freedman & Kaufmann Chap. 26, Problem 30, pg. 605
The distance of galaxies are often measured using standard candles. However,
these methods must be calibrated by accurately determining their properties
from very nearby standard candles. One example is the Cepheid Variable that
we often use to determine the distance of galaxies. By accurately measuring
the parallax of nearby Cepheid Variables we can then determine the correlation between their luminosity and luminosity variability period. Using this
as a standard candle, we can measure the distance of galaxies by observing
the variability of the Cepheid variables in them. A new, improved calibration
of the Cepheid period–luminosity relation shifted one of the “rungs” in the
distance ladder and caused the revision of all the larger distance estimates
that depended on Cepheids.