Predictions
Our predictions for this experiment include:
- Kepler’s Law will apply to the Galilean Satellites because by Newton’s law of Universal Gravitiation, the moons are constantly falling towards Jupiter. These moons also have an independant orbital motion that allows them to overcome this gravitational pull and move in an elliptical orbit around Jupiter.
- Using the chart provided by the Voyager II computer program, we can predict the proportional relationships of the moon’s distances from Jupiter in their furthest points in orbit from Jupiter. Using the furthest distance that Io moves from Jupiter in its orbital motion as our basic astronomical unit, we can predict several things. First, we can predict that Europa’s furthest point from Jupiter will be a distance of 2 AU. Second, Ganymede’s furthest point from Jupiter will be at a distance of 3 AU. Lastly, Callisto’s furthest point from Jupiter will be at a distance of 5.4 AU.
- From this same information, we can determine that Europa’s orbital path is twice that of Io, Ganymede’s is three times that of Io, and Callisto’s is 5.4 times that of Io.
Results
After calculating the math, we accepted our hypothesis that Kepler’s third law can apply to the Galilean satellites. From the Voyager program, we were able to determine that the data we collected was valid and used this information to help prove our thesis. With this data, we were also able to calculate the mass of Jupiter with a few equations.
Data Interpretation Strategies and Calculations
When we first sat down to decipher the data that we had collected, we realized that we had no idea which moon was which on our observational data recordings. This was because we were not looking at the rotation of the Galilean satellites around Jupiter from the right angle. We were observing the movement of these moons in a plane, rather than in a circular motion. Therefore, the four moons could easily be confused and misidentified, as shown below.
*
Figure 1
The Right half of the diagram is what we observed through the telescope,the left half is a diagram of how the satellites were really moving (not drawn to proportion or scale). In this observation, following the logic that the moons are arranged with Io being the closest moon to Jupiter (then Europa, then Ganymede, then Callisto) to determine which moon was which in our observations (on the right) would cause the moons to be mislabeled. In this example, Callisto would be misinterpreted for Io, Ganymede for Europa, Io for Ganymede, et. Cetera.
In order to decipher just what the data we recorded on our data sheets meant, we used the Voyager program graph printout in the diagram shown below. As this diagram shows, by comparing the Voyager chart data for 8:00 pm on November 23rd (for example) with our experimental data from that same day, we were able to decipher which moon was which.
**
Diagram 2
This diagram demonstrates how we used the graph that we obtained from the Voyager program to determine which of the dots on our data sheets were which moons. By looking at the graph at what we determined to be approximately 8:00 pm (the average time of our observations) for this example date of November 23, 1997 we were able to compare the graph data to our data and determine which dot was which moon.
In our thesis, we postulated that Kepler’s third law could be used to predict the orbits of not only celestial bodies around the sun (as it was intended to do), but also the rotation of the Galilean satellites around the planet Jupiter. We had intended to use our observations to trace the orbital paths of the satellites and thereby determine through observation the average distance of each moon from Jupiter and the period of revolution of each of these moons. However, the unexpected bad weather encountered in the months that we were performing observations prevented us from being able to do this. Because of this lack of data, we turned to the internet and Nasa for the period of revolution and average distances from Jupiter for each of the moons.
*
Once we obtained this information, we were able to set up a proportional system (a.k.a develop an AU measurement system) for these satellites’ rotations.
1 AU of revolution = 1.79 days, or Io’s revolutionary period around Jupiter
1 AU of distance from Jupiter = 421, 600 km, or Io’s average distance from Jupiter
*
These new numbers could then be entered into the equation that is Kepler’s third law, P2 = a3.
Io: Europa:
P2 = a3 P2 = a3
(1)2 = (1)3 (1.983240223)2 = (1.591318786)3
1 = 1 3.933241782 = 4.029689367
Ganymede: Callisto:
P2 = a3 P2 = a3
(3.994413408)2 = (2.537950664)3 (9.38547486)2 = (4.459203036)3
15.95533847 = 16.34743151 88.08713835 = 88.66898583
As shown above, in our calculations we discovered that our original hypothesis was correct (within an acceptable degree). The numbers were not exactly correct, but the degree to which they differed was so small as to be almost unnoticeable, and probably attributable to human error or rounding of the numbers we pulled off of the NASA web pages. This shows that Kepler’s third law can be applied to the prediction of the orbits of the Galilean satellites around Jupiter as well as the prediction of the orbits of celestial bodies (such as planets) around the sun.
Another equation, T2 = (4*2/GM)r3, is commonly used to determine the mass of celestial bodies. Using our AU proportional system, we also attempted to determine the mass of Jupiter.
Io:
T2 = (4*2/GM)r3
(1 AU)2 = [4(3.14)2/(6.67 x 10-11 newton m2/kg2)M](1 AU)3
1 = [(39.4784176)/(6.67 x 10-11 )M](1)
1 = [(39.4784176)/(6.67 x 10-11 )M]
(6.67 x 10-11 )M = 39.4784176
M = 5.918803239 x 1011 kg
Europa:
T2 = (4*2/GM)r3
(1.983240223 AU)2 = [4(3.14)2/(6.67 x 10-11 newton m2/kg2)M](1.591318786 AU)3
3.933241782 = [(39.4784176)/(6.67 x 10-11 )M](4.029689367)
.9760657519 = [(39.4784176)/(6.67 x 10-11 )M]
(.9760657519)(6.67 x 10-11 )M = 39.4784176
(6.51035857 x 10-11)M = 39.4784176
M = 6.06393906 x 1011 kg
Ganymede:
T2 = (4*2/GM)r3
(3.994413408 AU)2 = [4(3.14)2/(6.67 x 10-11 newton m2/kg2)M](2.537950664 AU)3
15.95533847= [(39.4784176)/(6.67 x 10-11 )M](16.34743151)
.9760150064 = [(39.4784176)/(6.67 x 10-11 )M]
(.9760150064)(6.67 x 10-11 )M = 39.4784176
(6.51002009 x 10-11)M = 39.4784176
M = 6.064254342 x 1011 kg
Callisto:
T2 = (4*2/GM)r3
(9.38547486 AU)2 = [4(3.14)2/(6.67 x 10-11 newton m2/kg2)M](4.459203036 AU)3
88.08713835= [(39.4784176)/(6.67 x 10-11 )M](88.66898583)
.993437982 = [(39.4784176)/(6.67 x 10-11 )M]
(.993437982)(6.67 x 10-11 )M = 39.4784176
(6.55669068 x 10-11)M = 39.4784176
M = 6.064254342 x 1011 kg
Average:
5.918803239 x 1011 + 6.06393906 x 1011 + 6.064254342 x 1011 + 6.064254342 x 1011 = 2.411125098x 1012
2.411125098 x 1012/4 = 6.027812746 x 1011 kg
The actual mass of Jupiter is 1899.728 x 1024 kg.
We then tried to calculate Jupiter’s mass a slightly different way a derivative of Kepler’s third law and abandoned our AU proportions and returning to the established Au system:
I0:
(1.77 days)/(365days/years) = .0048 years
(421,600 km)/(150,000,000 km/AU) = .0028 AU
(.0028)^3/(.0048)^2 = .00095278 solar masses
.00095278 x (1.984 X10^30) = 1.895 x 10^27
(1.895 x 10^27) - (8.94 x 10^22) = 1.894 x 10^27 kg = mass of jupiter
Europa:
(3.55 days)/(365 days/years) = .0097 years
(670900km)/(150,000,000 km/AU) = .00447 AU
(.00447)^3/(.097)^2 = .000949 solar masses
.000949 x (1.989 x 10^30) = 1.88 x 10^ 27
(1.88 x 10^ 27) - (4.8 x 10^ 22) = 1.88 x 10^27 kg = mass of jupiter
Ganymede:
(7.15 days)/(365 days/year) = .0195890411 years
(1,070,000 km)/(150,000,000 km/AU) = .0071333333 AU
(.00713)^3/(.0196)^2 = .009444 solar masses
.000944 x (1.989 x 10^30) = 1.8766 x 10^27
(1.87 x 10^27)-(1.48 x 10^23) = 1.8765 x 10^27 kg = mass of jupiter
Callisto:
(16.8 days)/(365 days/year) = .046 years
(1.88 x 10^ 6 km)/ (150,000,000 km/AU) = .01255 AU
(.00447)^3/(.097)^2 = .000949 solar masses
.000949 x (1.989 x 10^30) = 1.88 x 10^27
(1.88 x 10^27) - (4.8 x 10^22) = 1.88 x 10^27 km = mass of jupiter
We then averaged the four masses of jupiter that we got from our calculations and our resultant calculated mass of Jupiter is: 1.877 x 10^27 kg. This proves to be only a 1.2% difference between our calculated mass and Jupiter’s known mass.
We are not clear as to why we were unable to obtain Jupiter’s mass using Kepler’s law of periods. Maybe a broader background of physics could have helped.
We would have liked to use our observational data to set up an AU system and thereby calculate proportional numbers to plug into these equations. However, because we only were able to take measurements on three nights we were not able to determine observationally what the revolutionary periods and average distances from Jupiter of each moon. Therefore, the only way we have of knowing that our observations were somewhat accurate, and that if the weather had permitted we would have been able to set up an AU system using our data, is the fact that when we compare our data sheets to the Voyager program graphs they match up as well as the example shown in figure 2.
Things that could have been done better
1. We could have had a better understanding of the physics involved with our lab from the beginning. This way, we could have comprehended the results we were getting throughout the lab--we could have known whether we were proving or disproving our thesis before the first week of December.
2. It would have been nice if we could all have understood the math of our lab, as several of us have a hard time comprehending the final equation to be able to interpret our data.
3. We should have considered weather conditions. We were not aware of nor forewarned about the greyness of the sky in winter, and as a result we have little data due to unpermitting weather conditions for the past two months.
4. It would have been nice to be able to follow through with the original plan of taking pictures of what we were seeing through the lens of the telescope, but first we needed a good understanding of how the telescope functions. Of course, due to the lack of dates that we could look at Jupiter, we never developed this familiarity.
5. We all wish we had a better idea of what we were getting into in the first place. It seemed that the closer to the end we got in this lab, the more complicated it got, and the less we understood what we were doing.
6. We should have talked to several people experienced in this field before trying to do this lab in order to have a better understanding of what we were getting into and how we could have initially improved it.
Questions we now have
1. Do all of Jupiter’s moons move in an elliptical pattern?
2. If Kepler’s third law applies to the orbits of the Galilean satellites, do his first and second laws also apply to these satellites?
3. Do Kepler’s Laws apply to all satellite systems within our visible realm?
4. If Kepler’s laws apply to satellite systems as well as solar systems, how do they apply to other orbital systems (such as the orbits of comets)?
5. What about Newton’s Laws of gravitation? How could they help us understand revolutions of planets and moons, as opposed to planets around stars (a.k.a. solar systems)?
6. How does Hays find Jupiter so quickly in the telescope?
7. How does the rotation of the Earth on its axis effect viewing Jupiter through the telescope?
8. How do you adjust the telescope so that it is easier to find and keep track of what you’re looking at?
9. How so you take pictures with the telescope?
References
The Environment of Space. http://www.jpl.nasa.gov/basics/bsf3-3.html
Humphrey, Colin, 1992. The Amateur Astronomers Pathfinder. Pp. 12-15, 18-23, 40-41.
Jupiter. http://seds.lpl.arizona.edu/nineplanets/jupiter.html
Kepler, Johannes, 1997. The Haromony of the World. Pp. VII- XXXVIII.
Kepler, Johannes, 1981. The Secret of the Universe. Pp. 7-9, 17-31.
Kazhathadan, Job, 1994. The Discovery of Kepler’s Laws, the Interaction of Science, Philosophy and Religion. Pp. 1-109.
Morrison, David, 1982. Satellites of Jupiter. Pp. 556-597.
Newton’s Laws of Motion and the Law of Universal Gravitation. http://angus.kingsu.ab.ca/~brian/astro/neut.htm
Physics 252, Feb. 23, 1996. http://zebo.uoregon.edu/~kevan/ph252/022396/022396.html
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