PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #5 (Page 1/4)
NAME:____KEY__________________________________________________________
Due Date: start of class 4/12/2007
• 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 4/12/2007 (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) (Chapter 14) Describe Jupiter’s internal structure and compare it with the internal structure
of Earth.
From the center outward, Jupiter has a rocky core (also containing perhaps some metals). Then
there is a layer of liquid “ices,” and a layer of helium and liquid metallic hydrogen, then a layer of
helium and ordinary liquid hydrogen, and finally the upper atmosphere we see. The interior of the
Earth is also layered, but except for the atmosphere and oceans, the Earth is made of iron and
rock.
# 2. (5 pts) (Chapter 14) Why are Saturn's rings conspicuous at times (as seen from Earth) and very
difficult to see at other times?
Because Saturn is tilted relative to the ecliptic plane, depending on where it is in its orbit, we see
either the underside of the rings, the top side of the rings, or, occasionally, we don’t see the rings
at all. This is because Saturn is at a point in its orbit where the rings are seen to be “edge on.”
This is called a ring-plane crossing. During this time, and dince the rings are very thin, we do not
see them from Earth. When we see the topside or underside of the rings, they are easily seen
because they have a high albedo and are quite large.
# 3. (5 pts) (Chapter 14) How do we know that Saturn’s rings are not solid disks (or sheets) of
material? What is the direct observational evidence of this?
We know this because of the laws of physics. The inner parts of the rings experience a larger
gravitational force than the outer parts. If they were sheets of material, then the differences in
gravitational forces would destroy the rings (this was realized first by James Clerk Maxwell). The
direct evidence of this came from the study by James Keeler in 1895 by measuring the Doppler
shift of reflected sunlight from the inner and outer portions of the rings.
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #5 (Page 2/4)
# 4. (5 pts) (Chapter 15) Explain why the Jovian system is often referred to as being a mini Solar
System. Give TWO ways in which the Galilean satellites resembles the solar system and TWO ways in
which they differ.
Any 2 of the following will do:
(1) One way the system of Galilean satellites resembles our solar system in that smaller objects
orbit the large object as the gravitational center (Jupiter in the case of the Galilean satellites and
the Sun for the case of the planets in the solar system)
(2) Another way is that Jupiter and the Galilean satellites condensed from the same part of the
solar nebula, mimicking the way the solar system formed. Thus, the Galilean satellites formed as
a mini solar system, or, more precisely as a “Jupiter” system.
(3) Yet another is that the moons all orbit Jupiter in the same direction, and on the same plane.
Just like the planets in the solar system.
(4) Yet another is that the density of Galilean satellites decrease with distance from Jupiter, like
the planets in our solar system
Any 2 of the following will do:
(1) One difference is that Jupiter is not a star that can warm the surface of its “planets.”
(2) Another is that the Galilean satellites are all terrestrial like worlds with a significant amount
of rock and ice. Thus, there are no “gas giants” orbiting Jupiter.
(3) Yet another is that the least geologically active worlds are the largest moons (Ganymede and
Callisto), and the most geologically active worlds (Io and Europa) are the smallest moons. This
is opposite to the case of the terrestrial planets in our solar system.
# 5. (5 pts) (Chapter 15) Why are Ganymede, Callisto, and Titan all composed of about half water ice
(H2O), while Io and The Moon have very little to no water ice?
Ice is an important constituent of Ganymede, Callisto, and Titan because water condensed in that
part of the solar system because of the low temperature. For the case of Ganymede and Callisto,
they were sufficiently far from Jupiter that water condensed there too. The Moon, conversely, is
smaller and warmer, which caused it to lose any water vapor that might otherwise have condensed.
The same is true for Io because it is very close to Jupiter where it was much warmer during the
formation.
PART 2 (15 pts total): Quantitative
# 1. (5 pts) (Chapter 15) The mean density of Europa is 3 g/cm3. Assuming that it is made up of
uncompressed rock (density = 3.5 g/cm3) and pure water (density = 1 g/cm3), calculate the percent of
volume that must be rock and the percent of volume that must be water ice. (HINT: this is probably
easiest done using Excel or a similar type of program. You are required to guess an initial percentage
for each, compare the mass that those percentages produce to the actual mass of Europa and then scale
your values as required.)
Note that the density is mass divided by volume. Therefore, because the total mass of rock and the
total mass of ice must be the total mass of Europa, we must have that the density of rock, times the
fraction of the volume that is rock + the density of ice times the fraction of the volume that is ice
must be equal to the mean density of Europa.
But, we also know that the fractions must add to one because we have assumed that Europa is only
made of rock and ice. Thue, in addition to the equation above, we also have
By either solving these two equations simultaneously, or using trial and error (as in excel), you will
find that the correct fraction is 80% rock and 20% water (i.e. frock = 0.8, fice = 0.2)
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #5 (Page 3/4)
# 2. (5 pts) (Chapter 14) It has been claimed that Saturn would float if placed in a big enough bathtub.
Using the mass and size of Saturn given in the textbook, determine its mean density and comment on the
truth of this fanciful claim (note that in order to get full credit, you must show your work!).
The density is the mass divided by the volume. Saturn’s mass is 5.69x1026 kg. To get its volume,
we must assume that it is spherical and use the formula for the volume of a sphere. We get
Note that we used Saturn’s equatorial diameter (and divided it by 2 to get the radius!) in the above
formula; therefore, that our answer is going to be different from the actual density given in the
book. This is because Saturn is oblate. The density is then
Which is less than water – so, Saturn would float in a large enough bathtub of water.
# 3. (5 pts) (Chapter 15)
1
⎡ (1 − A) ⋅ S ⎤ 4
=
⎢ 4 ⋅ σ ⎥ describes the surface temperature of an object due to heating
⎦
⎣
caused completely by incoming solar energy, where A is the albedo, S the solar constant
(W/m2), and σ is the Stefan-Boltzmann constant. Using this equation, calculate the surface
temperature of both Earth (albedo of 0.4, S of 1370 W/m2) and Titan (albedo of 0.2, S of 15
W/m2). Side note, the S at Titan can be found using the inverse square law.
a) The equation T
Inserting the quantities into the above formula, we obtain
b) The actual surface temperature of Earth is 288K, while that of Titan is 90K. Why is Earth’s
and Titan’s temperature different from what you found in part a)?
Earth has an atmosphere that has a significant greenhouse effect (contains lots of water vapor and
some carbon dioxide) – thus, as we discussed a few weeks ago, this leads to an equilibrium
temperature for Earth that is much higher than the value computed above which does not take
into account the greenhouse effect. Note that the Titan’s actual temperature is a bit higher than
what we computed above, indicating that a greenhouse effect is probably taking place, but it is not
particularly strong because we are not far from the actual value.
PTYS/ASTR 206 – Section 2 – Spring 2007
Homework #5 (Page 4/4)
PART 3 (10 pts total): Starry Night Backyard Observing Problem
Please provide your answer in the space below
# 1. (10 pts) Ch 15, #69 of the textbook
(a) yes, in fact when you start them, they are all one the same side of Jupiter
(b) yes, you can tell that they all orbit in the same direction because they all pass in front of
Jupiter going to the right, and all pass behind Jupiter when going to the left. This is, in
fact, a very useful thing to remember if you ever go out and observe the moons of
Jupiter through a telescope for a period of several hours. If you are fortunatele enough
to observe a transit or occultation, just prior to the event you will be able to predict
whether the moon will pass in front of, or behind Jupiter, simply by noting which it way
it is moving!
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.
EC #1 Ch 14, #52 of the textbook
(a) The easiest way to do this problem is to use the fact that all Saturn’s satellites obey Kepler’s
third law as they orbit the planet. For instance, Titan’s period (382.69 h) and semimajor axis (1.22
× 109 m) are in accord with Kepler’s third law. Thus, for any particle orbiting Saturn with a period
P and a semimajor axis a, we can write the proportion
P2
a3
=
2
3
( 382.69 h ) (1.222 × 109 m )
At the outer edge of the A ring, a = 136,600 km = 1.366 × 105 km = 1.366 × 108 m, so
P2
( 382.69 h )
2
⎛ 1.366 × 108 m ⎞
=⎜
⎟
9
⎝ 1.222 × 10 m ⎠
= (1.118 × 10–1 )
3
3
1/ 2
3
P = ( 382.69 h ) ⎡⎢(1.118 × 10 –1 ) ⎤⎥
⎣
⎦
= 14.3 h ( 3600 s/h )
= 5.15 × 104 s
At the inner edge of the B ring, a = 92,000 km = 9.20 × 107 m, so
P2
( 382.69 h )
2
⎛ 9.20 × 107 m ⎞
=⎜
⎟
9
⎝ 1.222 × 10 m ⎠
3
= 4.267 × 10–4
P = 382.69 h ( 4.267 × 10 –4 )
1/ 2
= 7.91 h
= 2.85 × 104 s
(b) The rotation period of Saturn is 10h 13m 59s = 10.2 hours. Therefore the outer edge of the A ring
would appear to drift eastward while the inner edge of the B ring would move westward.
EC #2 Ch 15, #49 of the textbook
(a) Because the temperature is higher, elements that might be liquid or frozen at Titan’s distance
would be gaseous or liquid at 1 AU from the Sun. The atmospheric gasses would have a higher
velocity. To answer the question completely, we must compare the escape velocities:
v Titan = v Moon
mTitan RMoon
RTitan mMoon
Using the data in Table 7-2,
v Titan = v Moon
(1.34 × 1023 kg) (3476 km)
(5150 km) (7.35 × 1022 km)
v Titan = (1.11) v Moon
Titan’s slightly higher escape velocity means it would retain more of heavier gaseous molecules.
(b) Because Titan has a larger diameter, it has a larger angular size in the sky, and could cause
more total eclipses as opposed to partial eclipses.