Download ASTR0 100 HW #4 – SOLUTIONS – 2 points each

ASTR0 100 HW #4 – SOLUTIONS – 2 points each
Chapter 5
24. Thanks to adaptive optics, the telescope on Mount Wilson can now make ultraviolet
images of the cosmos.
False. While Mount Wilson does have adaptive optics, it is a ground based telescope,
and so it cannot do observing in ultraviolet because of atmospheric absorption – only radio
waves, visible light, and small parts of the infrared spectrum can be observed from the
ground.
30. Blue light has higher frequency than red light. Blue light therefore has (a) higher energy
and shorter wavelength than red light.
Energy is directly proportional to frequency, so higher frequency means higher energy.
Frequency, though, is inversely proportional to wavelength ( since c =£ so higher
frequency means shorter wavelength.
34. The set of spectral lines that we see in a star's spectrum depends on the star's (b)
chemical composition.
All types of atoms (and also ions and molecules) have their own distinct set of energy
levels and a corresponding distinct set of energy transitions when photons are emitted or
absorbed by those atoms. These transitions mean that each atom has a unique set of
spectral lines, and so when we examine a star's spectrum, we can determine what atoms,
ions, and molecules the star is composed of.
35. A star whose spectrum peaks in the infrared is (a) cooler than our Sun
Wien's Law tells us that temperature is inversely proportional to peak wavelength. We
know the Sun's peak wavelength is in the visible, since we are able to see it! Infrared
wavelengths are longer than visible wavelengths, so we are dividing 2,900,000 by a larger
number, meaning temperature must go down.
50. Suppose the surface temperature of the Sun were about 12,000 K rather than 6000 K.
a. How much more thermal radiation would the Sun emit?
Initial power = ü Ti)4
Final power = ü Tf)4
In this case we want the ratio of final power to initial power:
ratio = ü Tf)4 / ü Ti)4
Sigmas cancel, leaving:
ratio = Tf)4/ Ti)4 = (12000 K)4 / (6000 K)4
= 16 times more radiation
(note that units cancel!)
b. What would happen to the Sun's wavelength of peak emission?
Wien's Law: ümax [nm] = 2,900,000 / T [K]
Initial peak: ümax_i [nm] = 2,900,000 / Ti [K]
Final peak: ümax_f [nm] = 2,900,000 / Tf [K]
Again we want the ratio of final to initial:
ratio = (2,900,000 / Tf ) / (2,900,000 / Ti )
= (1 / Tf ) / (1 / Ti )
= (Ti / Tf) = (6000 K/12000 K)
= ½ of the initial peak wavelength
(note that units again cancel!)
c. Do you think it would still be possible to have life on Earth?
No, at least not life as we know it. With rare exceptions, life on Earth is not
capable of surviving at extreme climates, and a sun with double the temperature
and 16 times the power would certainly produce extreme climates. At those
temperatures, it's likely that the polar ice caps would melt and cause extreme
flooding. Additionally, with the sun's peak wavelength dropping by ½, most of
the light would shift out of the visible and into the ultraviolet. All of this combines
to make conditions extremely non-conducive to life on Earth.
Chapter 6
27. The fact that we have not yet discovered an Earth-mass planet tells us that such planets
must be very rare.
This should be considered surprising rather than reasonable. The lack of lower-mass
planets discovered thus far is because lower-mass planets are harder to detect, and our
current technology is too limited to effectively discover things so small, especially since other
stellar systems are so far away. Thus we cannot conclusively say whether Earth-mass
planets at this time.
33. Planetary orbits are (b) fairly circular and in the same plane.
According to the solar nebula hypothesis, collisions in the solar nebula caused the gas
cloud to flatten and caused orbits to become more circular. This is why planets today are
believed to all move in nearly circular orbits about nearly the same plane (section 6.3)
34. The composition of the solar nebula was 98% (c) hydrogen and helium.
The Big Bang was responsible for the production of hydrogen and helium, and thus
those two gases dominate the composition of the universe. Other elements are produced in
massive stars, but our solar system is still dominated by H and He.
Chapter 9
20. A mission to Pluto finds that it has lakes of liquid water on its surface.
Surprising. Pluto is far too distant from the sun to ever maintain liquid water, as its
temperature is much too low.
25. The asteroid belt lies between the orbits of (b) Mars and Jupiter.
While there are other asteroids in the solar system, the majority are found in the main
asteroid belt, located between Mars and Jupiter.
Extra Credit:
42. If a million asteroids 1 kilometer across were combined into one object, how big would it
be?
First, find the volume of one 1 kilometer across asteroid. Here 1 km is the diameter, so
the radius is 0.5 km, and we know for a sphere, V = (4/3) 
< .

Volume of 1 asteroid: V = (4/3) 
À


 
V = 0.524 km3
To find how big an object made of a million 1 km asteroids would be, we can multiply
the volume of one 1 km asteroid by 1 million:
Volume of new asteroid: V = 1,000,000 * (0.524 km3)
= 5.24 x 105 km3
For comparison, that asteroid would have a radius of 50 km, meaning it would be 100
kilometers across.
How many 1 km asteroids would it take to make an object as large as the Earth? First,
what is the volume of the Earth?
Earth's radius is 6378 km (Appendix E – A-14)
Volume of Earth: V = (4/3) <(6378  

= 1.09 x 1012 km3
To figure out how many 1 km asteroids it would take to make an object this big, we can
divide the volume of the Earth by the volume of one 1 km asteroid.
# of asteroids = Earth's volume / 1 km asteroid volume
= (1.09 x 1012 km3) / (0.524 km3)
= 2.07 x 1012 asteroids
*Note: to solve the extra credit problem in simpler fashion, note that in dividing two volumes, the (4/3) <will
cancel. Thus, to do volume ratios, we can simply solve by taking (rnew / rold)3 Try testing to see that this
works!