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Homework #7 Solutions
Questions due at the beginning of lecture on May 9, 2000.

Chapter 15 Question 2.  Structure of Saturn's rings
Saturn's rings are made up of millions of individual pieces of ice (or ice covered rock) that range in size from 1 cm to 5 m.  The average size of 10 cm would make a good snowball.  These pieces of ice are in turn organized into ringlets by gravitational effects of Saturn and its moons (see figure).  Because each ringlet has a color that is distinct from its neighbor, it seems unlikely that the material in the ringlets migrates substantially in the radial direction.

Chapter 15 Question 6.  Compare atmospheres of Saturn and Jupiter.
Being gas giants, the atmospheres of Saturn and Jupiter are at first glance very similar, however, there are some subtle differences between them.  First, Saturn's outer atmosphere of molecular hydrogen appears to be much deeper than Jupiter's.  This is evidenced by Saturn's more oblate shape.  Saturn's the deeper atmosphere contributes to its low average density.  The manifestation of Saturn's lower density in the topmost regions of its atmosphere is a larger spacing between cloud layers, relative to Jupiter.  This is why it is harder to see deep into Saturn's upper atmosphere and it appears more washed out than Jupiter.  Another important difference between Jupiter and Saturn's atmospheres is the difference in their helium content.  Jupiter's helium content (relative to hydrogen) is very similar to the Sun's.  Saturn's helium content is significantly lower.  The reason for this is unknown, but one theory is that Saturn's helium condensed in the form of rain and sank to lower regions of the atmosphere.  This raining process may have also contributed to Saturn's unexpectedly large internal heat output.

Chapter 15 Question 12.  Titan's atmosphere.
Titan the thickest atmosphere of any moon in the solar system.  The atmosphere probably started out as ammonia (NH3), but in a process similar to the breaking up of molecules in other atmosphere's, UV sunlight split the ammonia into nitrogen and hydrogen.  Being light, the hydrogen can escape Titan's gravitational pull, which has left Titan with a predominantly nitrogen atmosphere.  Titan also has a lot of methane in its atmosphere.  Methane and sunlight interact to form complex hydrocarbon, like ethane, acetylene, and ethylene.  The ethane produced in these reactions may have rained down on the surface of Titan in such quantities as to produce rivers and lakes!  Hopefully the Huygens probe, due to arrive in 2004, will give us a better view of Titan's surface.

Chapter 16 Question 5.  Colors of the outer planets
[Optional background] For reasons which aren't exactly known, Uranus and Neptune have proportionally more heavy elements than Jupiter and Saturn.  One theory is that Uranus and Neptune were formed closer to the Sun (where heavy elements condensed in greater abundance) and then were moved by to their larger orbits by gravitational interactions with Jupiter.  Alternately, Uranus and Neptune formed a little later than Jupiter and Saturn, when some of the lighter elements had already been swept away by the solar wind.  [Now to answer the question]  Methane is one of the compounds present in the atmospheres of Uranus and Neptune in much greater quantities than found on Jupiter and Saturn.  Methane preferentially absorbs red light, leaving only blue light to reflect from Uranus and Neptune.  We are not exactly sure what causes the colors in Jupiter and Saturn's clouds, since the major constituents should all be white in color.  However, since the clouds are white, whatever other colors there may be can show through more easily.

Chapter 16 Question 12.  Triton captured by Neptune?
The fact that Triton has a retrograde orbit is the most convincing evidence that it was captured by Neptune.  Also, though it is now in a circular orbit and does not currently experience much tidal force, it has a relatively new surface, suggesting that internal heat was generated not long ago.  A likely source of heat would be tidal forces experienced in its initial capture orbit, which was probably highly elliptical.

Chapter 16 Question 15.  Pluto as a Kuiper belt object.
The retrograde rotation of pluto, its high inclination and eccentricity, as well as its distinct composition (as evidenced by its intermediate average density) all suggest that Pluto is in a class of objects distinct from the major planets.  Such a class of objects was proposed by Gerard Kuiper in 1951 to explain the source of short period comets.  These objects are probably left over icy planetesimals from the protoplanetary disk that never collided with a major planet.

Chapter 17 Question 11.  Ion and dust tails
The ion and dust tails of comets point in different directions because different physical effects act on them.  The ion tail is is pushed back by interaction with the solar wind.  The dust tail is pushed back by radiation pressure.  The solar wind is much more effective at moving ions than solar photons at moving dust.  Therefore, the ion tail sticks back straighter from the comet head.

Chapter 17 Question 13.  The Kuiper belt.
The Kuiper belt is a disk-shaped distribution of icy planetesimals that orbit the Sun with semi-major axes greater than that of Pluto.  Kuiper belt objects are like asteroids in that they never coalesced to form a planet, but they are primarily made of ice instead of rock.

Chapter 17 Question 14.  The Oort cloud
The Oort cloud is a spherical cloud of icy planetesimals with very large orbits.  The Oort cloud was proposed in 1950 to explain long period comets, in particular why long period comets don't have a preference for appearing in the ecliptic, like most other solar system objects.  The Oort cloud is probably made up of planetesimals that were ejected in random directions by gravitational interaction with the planets (particularly the Jovian planets).

Chapter 30 Question 2.  Playing with the Drake Equation

N = Rs fp ne fl fi fc L

Keeping all values except N and L at the values given in the book, what is the value of L if
(a) N = 1000
(b) N = 1 x 106

A nice way to solve this problem is to solve the Drake equation for L:

L = N / [Rs fp ne fl fi fc]

and then plug in.
 
Variable Book Value Alternate value (a) Alternate value (b)
Rs 1 1/year 1 1/year 1 1/year
fp 1 1 1
ne 0.1 0.1 0.1
fl 1 1 1
fi 1 1 1
fc 1 1 1
L 100 years 10,000 years 1 x 107years
N 10 1000 1 x 106


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