Film Review of Grass for SOC 1010

Grass did a good job of using statistics and numbers to persuade the viewers to adopt an “anti-prohibition” standpoint on marijuana.  The use of government costs on drug wars was very influential in portraying the “outlandish” and “desperate” attempts of the US Government to control the issue and spread anti-drug propaganda.

The film’s use of verbiage and editing in portraying the government exposed its own propagandist goals.  While those opposing the enforcement of anti-marijuana laws were described as “skeptical,” and “impartial.”  Officials enforcing the laws were described as emotionally driven (i.e. “furious,” “saw the entertainment industry as degenerate”, etc..) which leads the audience to believe that marijuana laws are based solely on misinformation and misdirected malcontent.  At the same time, when anyone was shown broadcasting something anti-marijuana, there were several cut scenes showing the errors of these “supposedly trustworthy” media sources.  The outtakes attempt to portray the uncertainty, ineptness and general buffoonery of these officials and media representatives.

Grass follows the stigmas of marijuana from social pre-deviance to crime, while at the same time trying to claim that its government enforced criminalization is, in itself, deviant.  The movie claims that, at one time, marijuana was socially acceptable.  According to our textbook (pg. 118), “a stigma is a negative label that devalues a person and changes her or his self-concept and social identity.”  The film portrays the negative stigma towards marijuana and its use as deriving from bigotry, prejudice, and/or racism.  In this way, the film argues that the negative stigma of marijuana is founded on the negative stigmas of prejudice.  It’s a veritable I’m not stupid, YOUR stupid” argument which most likely only appeals to those who already have a preference towards one side or the other.

Crime is defined as, “a violation of societal norms and rules for which punishment is specified by public law.”  The use of marijuana did become a crime and did become punishable by public law.  The movie argues that it should have never happened in the first place and that the law itself is criminal because of its inefficiency in being enforced.  It argues that the vast expenses incurred by the war on marijuana could be construed as a crime against society.


In summation, the film begs the viewers to ask who the “real” criminals and deviants are, the government or the marijuana users.  Unfortunately, the argument also seems to be propaganda vs. propaganda and there seems to be little impartial evidence to support either side.

Diameter of the Sun for AST 1040

Statement of Purpose:  Find the diameter of the Sun

Statement of Procedure:  Cover a mirror with a piece of cardboard with a 7mm circle cut out of the center.  Mount the mirror to a camera tripod.  Position the tripod and mirror 6 meters (6000mm) from a shaded wall.  Adjust the mirror so that the sun reflects through the hole in the cardboard and projects onto the shaded wall.  Measure and record the diameter of the reflection on the wall.  Repeat 3 times.  Using these measurements, calculate the diameter of the sun using the formula D/L = d/l or D = d/l*L.  D is the diameter of the Sun in kilometers, L is the distance to the Sun in kilometers (150000000km), d is the diameter of the projection in millimeters, l is the distance to the projection (from the mirror) in millimeters.  Example (using data for first measurement): D = d/l*L  -- D = 55/6000*150000000 --  D = 1375000.  Using the 3 calculated diameters, find the average diameter of the Sun by adding all three and dividing by three.

Data:

	"d" (in mm)	"l" (in mm)	D=d/l*150000000 km
1	55	6000	1375000
2	56	6000	1400000
3	58	6000	1450000
Average			1408333

Statement of Conclusion:  According to Google, the actual diameter of the Sun is 1391000km.  So my measurements weren’t too far off (1.25%).  My measurement from mirror to wall may have been slightly more or less than 6000mm.  Also, my projected image may not have been exactly straight on with the wall, which may have caused some skewing.  I’m pretty sure the mirror I used wasn’t magnifying, but it was a cheap mirror so there may have been some slight concave/convex parts to its surface.  My circular hole in my cardboard may not have been quite 7mm and also may not have been absolutely circular.  But even with all those imperfections, I’m pretty satisfied with my results.

1)      Throughout the year, the Earth varies in its distance from the Sun.  Would it make a significant difference (say, 10 percent or more) if you did this again six months from now?

a.       According to science.nasa.gov, when Earth is closest to the Sun (perihelion) the distance is 147500000km.  When Earth is farthest from the Sun (aphelion) the distance is 152600000km.  If I use these distances in my equation, the resulting diameters vary about 2% increase or decrease.  The projected image will also increase or decrease with distance to the Sun, but only by about 1mm, so the difference is almost too small to be measurable.  If I attempted this project again six months from now, I would expect my answers to be nearly the same with variations of less than 10% difference.

2)      Does it make a difference if you do this in the morning, noon or afternoon?  Why or why not?

a.       As long as the Sun is completely over the horizon, the time of day shouldn’t matter significantly.  Because of the curvature of the Earth, the Sun is slightly further away before or after noon (just as you are further from the sun at midnight rather than noon), but this distance is negligible in our calculations. The ideal time to take measurements would be at the Sun’s apex in the sky because this is the time that you would be closest to the Sun during Earth’s rotation. 

3)      Could you use this technique with the Moon?

a.       Providing you can get a reflected image, this method could only be used on the Moon during a Full-Moon phase.  In any other phase, you could not measure the diameter of the projected image.

4)      What is the one essential piece of data – that you do not directly measure – that you must have to determine the Sun’s diameter?

a.       The distance to the Sun.

5)      If you already knew the Sun’s diameter, could you determine the piece of data mentioned in the question above from the process in this activity?  Why or why not?

a.       Yes.  The equation would change from D/L = d/l  to  L =D/(d/l).  Because we have the measurements of three of the four variables, the equation is easily solved.

Example (using data for first measurement & actual diameter of the Sun): L = D/(d/l)  --  L = 1391000/(55/6000) --  L = 1391000/.0091666667 --  L = 151745454km.

Radar Ranging of Venus for AST 1040

Statement of Purpose:  Determine the distance and diameter of Venus based on radar data and arc seconds recorded on 3 separate dates.

Statement of Procedure:  The data given was in minutes and seconds.  The first step in this procedure was to convert those times into seconds by multiplying the minutes by 60 and adding the remaining seconds.  These were the times of a round-trip to Venus and back on radar beams (which travel at the speed of light).   The time it takes light to travel to and back from Venus can tell us the distance to Venus.  Light travels at 300,000km/s and our first trip took 574s.  So we can find the distance by multiplying our recorded time by the speed of light and dividing by two (for half the round-trip): (574x300000)/2 = 86100000km.  Then using the graphic we can count the apparent arc seconds of Venus at the same time the radar beams were sent.  The arc seconds are then converted into degrees by dividing by 3600 (since there are 3600 arc seconds in a degree).   Using this Angular Size and the Distance, we can find the true size of Venus by multiplying the two and dividing the sum by 57.3 degrees (Radian).  Using the data of 3 dates, we can average our measurements by adding all three and dividing by three.

Data:

Date	One-way Timing (sec)	True Distance (km)	# of ticks	Angular Size in degrees	True Diameter(km)
11/15/2005	574	8.61 x 107	29	0.0080555556	12104
11/30/2005	426	6.39 x 107	36	0.01	11152
12/15/2005	364	5.46 x 107	47	0.0130555556	12440
Average					11899

Statement of Conclusion:  The actual diameter of Venus is 12103.6km.  So my results weren’t too far off (204.6km difference).  Considering I was using data recorded by someone other than myself, I’m not exactly sure what I could have done differently to get a more accurate result.  I did consider rounding the angular size but noticed quickly that my results became further off after rounding.

1)      Could this procedure be used with other astronomical bodies such as the Sun, Mars or Jupiter?  Why or Why not? 

a.       Depending on the body’s composition will determine whether or not this method may or may not work.   If an object has an atmosphere, radar (radio) waves may or may not be able to penetrate the atmosphere depending on its composition and density.   This procedure can only be used if the wavelength of the sent beam will penetrate and return through the atmosphere of the object (if it has one).  This method will almost always work with objects in our solar system which do not have an atmosphere and have a measurable angular size.

2)      Why would this method not be used with distant stars?  (There may be more than one answer here.)

a.       The closest star (Proxima Centauri) is 4.2 light years away.  That means a round trip at the speed of light from Earth to Proxima Centauri would take 8.4 years.  Considering the Earth’s rotation and orbit, this procedure would take precise timing and calculations to send a signal and receive the bounced signal 8.4 years later.  And then only if Proxima Centauri (or another object) has a penetrable atmosphere.

b.      The other problem with this method is getting an accurate reading on the angular size of a distant star.  Even Proxima Centauri is so far away that getting an accurate reading on its angular size would be incredibly difficult if not improbable.

c.       Another problem would be that the signal would gradually lose intensity as it travels further.  The returning echo (if there is a return echo) would be so faint, it would be nearly impossible to detect.

d.      Another problem might be finding an unobstructed direct path to a distant star.  The space between may be filled with particles, gasses, ice, or other objects that may absorb, deflect, scatter or even bend our sent light signal.

3)      What would be different if we used data for different dates?

a.       If the dates were different in this exercise the distance of, and therefore the angular size of Venus would be different.   As both Earth and Venus travel around the Sun in their respective orbits, Venus may appear to have a larger or smaller angular size depending on its respective distance to Earth.  While the angular size might change, the radar beams should also change, respectively.  If Venus is closer to Earth, Venus will have a larger angular size and a shorter radar time.  If Venus is farther from Earth, it will have a smaller angular size and a longer radar time.  The radar time and angular size of Venus will change in proportion to each other.

4)    Would this procedure work with other forms of electromagnetic radiation such as visible light or infrared light?  Why or why not?

a.       This procedure will work with other forms of electromagnetic radiation given the object being measured.  The moon is currently under scrutinous observation with laser beams (visible light).  However, depending on the atmosphere of the object being measured will determine whether this method will or will not work (see question 1).   The wavelength of the light must be able to reflect off of the object’s surface and return to Earth.  Unless, of course, the object’s atmosphere is what is being studied.  If the object has no atmosphere, this method would probably work with most forms of light.