Download Solutions Assignment #3

Astronomy 2020
Fall 2010
Assignment # 3
Solutions
Chapter 14 problem 50
a.
The surface area of a sphere 1 AU (= 1.5 1011 m) in radius is:
4 r 2 4 (1.5 1011 m) 2 2.83 1023 m 2
b.
The flux of solar radiation at the surface of this imaginary sphere is the luminosity of the
Sun divided by the surface area of the sphere:
3.8 1026 watts
1344 watts per square meter
2.83 1023 m2
c.
The average power per square meter collected by a solar collector on the ground will
always be less because of absorption by the atmosphere, the angle of incidence not being 90°,
the weather (cloud cover), nighttime, and varying amounts of daylight.
d.
To optimize the amount of power collected, a solar collector should be pointed south in
the Northern Hemisphere, and north in the Southern Hemisphere (and up toward the celestial
equator). To achieve even more optimization, one might rotate the face of the collectors east to
west to follow the Sun’s daily path across the sky, and north to south to match the Sun’s
changing path with the seasons.
Chapter 14 problem 52
Wien’s law states:
2,900,000
nm
max
T (Kelvin)
Plugging in the average temperature of the Sun’s photosphere, 5800 K, gives:
2,900,000
nm 500 nm
max
5800
Thus, the Sun’s thermal spectrum peaks at a wavelength of 500 nanometers, which is in the
green part of the visible spectrum. However, because the Sun also radiates other colors of the
visible spectrum and the atmosphere scatters the bluer light, it appears white or yellow to our
eyes.
Chapter 15 problem 40
a.
Sirius appears brightest in our sky because it has the smallest (most negative) apparent
magnitude.
b.
Regulus appears faintest of the stars on the list because it has the largest apparent
magnitude.
c.
Antares has the greatest luminosity of the stars on the list because it has the smallest
(most negative) absolute magnitude.
d.
Alpha Centauri A has the smallest luminosity of the stars on the list because it has the
largest absolute magnitude.
e.
Sirius has the highest surface temperature of the stars on the list because its spectral
type, A1, is hotter than any other spectral type on the list.
f.
Antares has the lowest surface temperature of the stars on the list because its spectral
type, M1, is cooler than any other spectral type on the list.
g.
Alpha Centauri A is most similar to the Sun because it has the same spectral type and
luminosity class, G2 V.
h.
Antares is a red supergiant; its spectral type M means it is red, and its luminosity class I
indicates a supergiant.
i.
Antares has the largest radius because it is the only supergiant on the list.
j.
Aldebaran, Antares, and Canopus have luminosity classes other than V, which means
that they have left the main sequence and are no longer burning hydrogen in their cores.
k.
Spica is the most massive of the main-sequence stars listed because it has the hottest
spectral type of the main-sequence stars; thus, it appears higher on the main sequence of an
H-R diagram, where masses are larger.
l.
Alpha Centauri A, with spectral type G2, is the coolest and therefore the longest-lived
main-sequence star in the table.
Chapter 15 problem 44
If a star doubled in size with no change in luminosity, its surface temperature would go down,
because the surface area of the star would increase but its energy output would stay the same,
so a lower temperature would be required to maintain its energy output. Mathematically, the
surface area would go up by a factor of four, so the temperature would go down by a factor of
( x)4 14 or x 1/ 2 0.71
x, where
to maintain a constant luminosity.
Chapter 16 problem 48
a.
From Appendix E, we learn that the mass of the Sun is 2 1030 kilometers, or
2 1033 grams. A 1000 MSun cloud then has a mass of 2 1036 grams. If 1% of this material is
in the form of dust, then there are 2 1034 grams of dust in the cloud. Each particle has a
mass of 10–14 grams, so we get the number of grams by dividing:
2 1034 g/cloud
2 1048 grains/cloud
14
10 g/grain
So there would be around 2 1048 grains in such a cloud.
b.
To get the area in square light-years, we will first convert the radius of the individual
particles to light-years. Appendix E tells us that a light-year is 9.46 1012 kilometers, or 9.46
1015 meters. The grains have radii of 10–7 meter and this converts to 1.06 10–23 light-years.
The area of a single grain is πr2, so putting in the radius we get an area of 3.51 10–46 square
light-year. From part (a) there are 2 1048 grains, so multiplying the number of grains by the
area per grain, we get the total surface area of all of the grains in the cloud: 702 square lightyears.
c.
Since density is defined as mass over volume:
mass
density
–21
We know that the cloud’s density is about 10 grams per cubic centimeter and we know that
the mass is about 2 1036 grams. Therefore, the volume of the cloud must be 2 1057 cubic
centimetrs. If the cloud is spherical, the radius can be expressed in terms of the volume by
turning around the formula for the volume of a sphere:
3 volume
r 3
4
Putting in the numbers, we find that the radius of the spherical cloud is about 7.82 1018
centimeters or 7.82 1016 meters. Converting to light-years gives us 8.27 light-years. The
2
cross-sectional area is r , like a circle, so the area of the cloud is 214 square light-years (200
light-years, rounded off).
volume
d.
Since the area of the dust grains is more than three times the cross-sectional area of
the cloud, any photon entering the cloud will probably collide with at least one grain as it
passes through.
Chapter 16 problem 50
From Mathematical Insight 16.1, the critical mass needed for a cloud to collapse is:
T3
M balance 18M Sun
n
We are told that the temperature, T, is 10 K and the density, n, is 100,000 molecules per cubic
centimeter, so we can find the critical mass: 1.8MSun. The mass of the Sun is 2 1033 grams,
so the cloud’s mass in grams is 3.6 1033 grams. To get the radius, we will first need to
convert this mass to a volume and then from volume to radius. The density of the cloud in
grams per cubic centimeter can be found by taking the density in molecules per cubic
centimeter and multiplying by the mass per molecule. We are told to assume that all of the
molecules are hydrogen with a mass of 3.3 10–24 gram so the density is 3.3 10–19 gram per
cubic centimeter. Since density is mass over volume, we can solve to get
mass
volume
density
:
The volume of the cloud is therefore 1.09 1052 cubic centimeters.
Assuming that the cloud is spherical, we can solve the formula for the volume of a sphere to
get the radius:
3 volume
r 3
4
Using our volume, this leads to a radius of 1.38 1017 centimeters. Converting to meters gives
us 1.38 1015 meters. Converting to light-years gives 14.5 light-years. So the cloud would
need to be at least 14.5 light-years in radius to start star formation.