Download PTYS/ASTR 206 – Section 2 - Lunar and Planetary Laboratory

PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #2 (Page 1/5)
NAME:_____KEY_____________________________________________________
Due Date: start of class 2/6/2007
• 5 pts extra credit if turned in before 9:00AM (early!)
(To get the extra credit, the assignment must be placed the box outside of room 330!)
• 5 pts taken off if turned in after the start of the class period and before 5PM of the due
date (late!)
• The assignment will not be graded if turned in later than 5PM 1/25/2007 (too late!)
Turn in your own work. It is not acceptable to turn in work that is identical to that of another
student. You may work together to understand the material, but the work you turn in must be in
your own words.
There are 3 parts, plus an extra-credit part. Please put your answers to all questions to PARTS 12 in the space provided (you may use the back of the page if necessary).
PART 1 (25 pts total): Conceptual. Please provide a concise short answer (not more than a few
sentences) for each of the following.
# 1. (5 pts) (Chapters 5) What is a blackbody? What is a blackbody spectrum? Why are these concepts
useful to planetary scientists?
A blackbody is an idealized object that does not reflect radiation at all (it can only absorb it) and
emits radiation in a manner given strictly by its temperature. A blackbody spectrum is the curve,
often graphed as intensity versus wavelength, which gives the intensity of the emitted radiation as
a function of the wavelength. The maximum of this curve occurs at a wavelength given by Wien’s
law, and the total intensity (area under the curve) is given by the Stefan-Boltzmann law. These
concepts are useful to planetary scientists because the radiation emitted by stars and other bodies
are often similar to a perfect blackbody and can tell us the temperature of the body.
# 2. (5 pts) (Chapter 7) The absorption lines in the spectrum of a planet or moon do not necessarily
indicate the composition of the planet or moon’s atmosphere. Why not?
Usually, the radiation measured at Earth from a planetary body or moon originated in the Sun
(reflected sunlight). This light has passed through the Sun’s atmosphere, the atmosphere of the
body we are observing, and finally the Earth’s atmosphere. Thus, the resulting absorption line
spectrum will have signatures of all three of these atmospheres – and not just that of the body we
are observing.
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #2 (Page 2/5)
# 3. (5 pts) (Chapter 6) Suppose that you are viewing the Moon through a small amateur telescope of
basic design (i.e. a primary lens and an eyepiece lens) and want to see more detail by magnifying the
image. How can you do this without buying a new telescope? Explain.
Remove the eyepiece that is in the telescope and replace it with one that has a smaller focal length.
The magnification is the ratio of the focal length of the primary lens (which cannot be replaced
unless you buy another telescope) to the eyepiece lens (most telescopes come with a variety of
eyepiece lenses). Note that the focal length of the eyepiece is normally printed somewhere
conspicuous on it.
# 4. (5 pts) (Chapter 5) Consider the four spectra below. The spectrum labeled (c) is an absorption line
spectrum from a star that is at rest. Blue light is shown on the left-hand side and red light is shown on
the right-hand side.
Blue
Red
a
.
b
c
←
Spectrum for the star at rest
d
a)
Which of the three spectra (a, b, or d) would be from the star that is moving the fastest?
Would this star be moving toward or away form the observer? Explain.
Star b is moving the fastest and directed away from the observer because the absorption
lines are shifted by the largest amount (from those for the stationary observer) and shifted
to longer wavelengths – i.e. redshifted.
b)
Which of the three spectra (a, b, or d) would be from the star that is moving the slowest?
Describe the motion of this star.
Star d is moving the slowest and directed towards the observer because the absorption lines
are shifted by the least amount and shifted to shorter wavelengths – i.e. blueshifted.
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #2 (Page 3/5)
# 5. (5 pts) (Chapter 3) Figure 1 shows Earth, the Sun, and five different possible positions for the
Moon during one full orbit (dotted line). It is important to recall that one half of the Moon’s surface is
illuminated by sunlight at all times. For each of the five positions of the Moon shown below, the Moon
has been shaded on one side to indicate the half of the Moon’s surface that is NOT being illuminated by
sunlight. Note that this drawing is not to scale.
Orbit of the Moon
moon phase
E
A
Earth
B
Sun
D
C
Figure 1
a) Which Moon position (A-E) best corresponds with the moon phase shown in the upper right
corner of Figure 1? Make sure that the moon position you choose correctly predicts a moon
phase in which only a small crescent of light on the left-hand side of the Moon is visible from
Earth.
Enter the letter of your choice:__D__ (this is the waning crescent phase)
b) In the blank boxes below, sketch how the Moon would appear from Earth for the four Moon
positions that you did not choose in part (a). Be sure to label each sketch with the corresponding
letter indicating the Moon’s position from Figure 1.
(A) Waxing Gibbous
(C) 3rd Quarter
(B) Waning Gibbous
(E) Waxing Crescent
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #2 (Page 4/5)
PART 2 (15 pts total): Quantitative
# 1. (5 pts) (Chapter 5) The solar constant is the amount of the solar energy flux (in the form of light)
that arrives at Earth’s orbit (1AU). Its value is 1370 W/m2. The solar energy flux obeys the inverse
square law – that is, it decreases with the square of the distance from the source. What is the amount of
solar energy flux that arrives at Jupiter’s orbit?
Jupiter is at a distance of 5.2AU from the Sun, while the Earth is 1AU from the Sun. Therefore,
the flux of solar energy arriving at the orbit of Jupiter is:
FJupiter = FEarth (1AU/5.2AU)2
= 1370 x (1/5.2)2 W/m2
= 50.7 W/m2
# 2. (5 pts) (Chapter 5) The Sun’s photosphere has a temperature of 5800 K. What will be the peak
wavelength of the spectrum of a star whose photosphere emits twice the Sun’s flux of energy?
We will ultimately use Wien’s law, but must first determine the temperature since we are only given the
flux of the star. This problem must be solved in steps.
Step 1: Get the flux of the star. What we know, the temperature of the Sun and the StefanBoltzmann law.
Fsun = σTsun4
Fstar = 2 Fsun = 2 σTsun4
Step 2: Get the temperature of the star from the Stefan-Boltzmann Law
Because Fstar = σTstar4 Î Tstar = (Fstar/σ)¼
From the last line of step 1, we see
Tstar = (2 σTsun4/σ)¼ = (2)¼Tsun = (1.189) x 5800 K = 6896 K
Step 3: Get the wavelength of maximum emission from Wien’s law
λmax = 0.0029/Tstar m = (0.0029/6896) m = 421 nm (this star will be bluish)
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #2 (Page 5/5)
# 3. (5 pts) (Chapter 5: #38 of the textbook) The wavelength of Hβ in the spectrum of the star Megerez
is 486.112 nm. Laboratory measurements demonstrate that the normal wavelength of this spectral line is
486.133 nm. Is the star coming toward us or moving away from us? At what speed?
The wavelength is shorter, or “blue shifted”, so the star is approaching us.
Δλ v
= →
λo c
Δλ
(486.112 nm − 486.133 nm)
, v = (3 × 108 m/s)
, v = -12.3 x 103 m/s = -12.3 km/s
v=c
λo
486.133 nm
PART 3 (10 pts total): Starry Night Backyard Observing Problems
Please provide your answers on a separate sheet of paper and attach it to the rest of your homework.
# 1. (5 pts) Chapter 5, #50 of the textbook
Using the info tab, the temperature of each star is given, as is the color. For each temperature, we can
use Wien’s law to get the wavelength of maximum emission. This is what I get:
Altair:
Procyon:
Epsilon Indi:
Tau Ceti:
Epsilon Eridani:
Lalande 21185:
T = 7758 K
T = 6800 K
T = 4604 K
T = 5655 K
T = 5136 K
T = 3483 K
λmax =
λmax =
λmax =
λmax =
λmax =
λmax =
Sun
T = 5800 K
λmax = 500 nm
374 nm
426 nm
630 nm
513 nm
565 nm
833 nm
Thus, using this to answer the questions, we get:
(a) the stars that have a longer wavelength of maximum emission than the Sun are: Epsilon Indi,
Tau Ceti, Epsilon Eridani, and Lalande 21185. The stars that have a shorter maximum
wavelength of emission than the Sun are Altair and Procyon
(b) the stars that have a reddish color are Epsilon Indi and Lalande 21185 – the two coolest among
this list. Note that the color charts showed that most appeared at least somewhat reddish (to
me), but the two that were the most red to me were these two. Also, the wavelength of
maximum emission for both of the these are near the red part of the EM spectrum.
(solutions continued on next page)
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #2 (Page 6/5)
# 2. (5 pts) Chapter 6, #53 of the textbook
(i)
(ii)
(iii)
(iv)
At 30 arcminutes, You can almost fit the entire Moon into your field of view. You can see
large craters, mare, and mountain ranges, but cannot make out any fine detail such as
rimae, rille (lava channels), or small craters.
At 30 arcminutes, you can easily see Jupiter as a planet (and not just a dot) and the 4
Galilean satellites. You may also be able to see two large bands of gas clouds on the planet,
but do not have enough magnification to make out the giant red spot.
At 30 arcminutes, you can easily see Saturn as a ringed planet. However, you would have
trouble making out the Cassini division in its rings.
At 30 arcminutes, the field of view is way too small to make out any detail in M31. You
would be looking into the center and only see a blob of material with no detail at all. M31
is best viewed at much larger field-of-views, such as at 5 degrees or so (try zooming out to 5
degrees to see what I mean).
PART 4: Extra Credit
Points available: 2 for one, 5 for both
Please provide your answers on a separate sheet of paper and attach it to the rest of your homework.
E.C. #1 Ch 5, #24 of the textbook
Assume the book is at room temperature, about 72 degrees Fahrenheit. This is about 22o C, which
is about 295 K. So, from Wien’s law, we have
λmax = (0.0029/295) m
= 9.83x10-6 m
= 9830 nm, which is infrared radiation
E.C. #2 Ch 7, #21 of the textbook
First of all, you must calculate Phobos’ orbital radius which is the altitude (above the surface) plus
the radius of Mars. This is
1
a = 5980 km + ( 6794 km ) = 9377 km
2
6
= 9.375 × 10 m
⎛ 4π 2
⎝ GM
2
From Box 4-2 we see that the appropriate form of Kepler’s third law is P = ⎜
⎞ 3
⎟ a , where M is
⎠
the mass of Mars. The orbital period of Phobos is P = 0.31891 days. Converting this to seconds, we
get P = 0.31891 days x (86400 seconds/day) = 2.755 × 104 s. Solving Newton’s form of Kepler’s Law
for the mass gives
⎛ 4π 2 ⎞ a 3
M =⎜
⎟ 2
⎝ G ⎠P
⎛
⎞ (9.375 × 106 m)3 .
4π 2
M =⎜
−11
2
2 ⎟
4
2
⎝ (6.67 × 10 N·m /kg ) ⎠ (2.755 × 10 s)
M = 6.43 x 1023 kg.
Now for the density, which is given by the following equation.
ρ=
Where V =
M
,
V
4 3
πR . Plugging in numbers to the volume gives
3
3
4 ⎛1
1000 m ⎞
V = π ⎜ × 6794 km ×
⎟
3 ⎝2
km ⎠ .
V = 1.64 × 10 20 m 3
With the mass and volume, we can now use the equation for density,
6.43 × 10 23 kg
1.64 × 10 20 m 3 .
ρ = 3920 kg/m 3
ρ=