Purpose
1. Determine the mass of Jupiter.
2. Gain a deeper understanding of Kepler's third law.
3. Learn how to gather and analyze astronomical data.
2. Gain a deeper understanding of Kepler's third law.
3. Learn how to gather and analyze astronomical data.
BackgroundFrom Kepler, we are able to derive celestial information like mass without actually placing a planet on a scale. We are able to do this because of his Third Law:
R^3 / T^2 = C Now this alone would not be enough to solve, however Newton used Kepler's Third Law to derive the following equation: R^3 / T^2 = GM / 4(3.14)^2 where G is the Gravitational constant 6.67e-10 and M is the mass of the body that the celestial object is orbiting. This is how we will solve for the mass of Jupiter using the orbits of its moons. |
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Procedure
The details of the lab procedure are all in the public domain. Click here for the web address.
Data
The following is the data acquired using the CLEA software. Below the image are the estimated periods and orbital radii of each moon.
The average period of Io is approximately 1.78 days and its orbital radius is 3 Jupiter Diameters (JD)
The average period of Europa is approximately 3.5 days and its orbital radius is 4.67 JD
The average period of Ganymede is approximately 7.2 days and its orbital radius is 7.67 JD
The average period of Callisto is approximately 16.9 days and its orbital radius is 13.3 JD
The average period of Europa is approximately 3.5 days and its orbital radius is 4.67 JD
The average period of Ganymede is approximately 7.2 days and its orbital radius is 7.67 JD
The average period of Callisto is approximately 16.9 days and its orbital radius is 13.3 JD
Data Analysis
The above calculations were determined by fitting a sine graph to each of the orbits. The amplitude of the graph is its orbital radius and the period is surprisingly enough the period. Below are the sine graphs used.
Using the information from these graphs, we were able to plot a straight line curve using the linearized R^3 = [GM / (4(3.14)^2)] * T^2.
In this format, [GM / (4(3.14)^2)] becomes the slope of the graph. In our graph, the slope is about 3e+15. Using algebra, our equation becomes
M = [slope(4(3.14)^2)] / G = [3e+15(4(3.14)^2)] / 6.67e-11 = 1.7756 e+27
Percent error is calculated as (observed - expected) / (expected). In our case the calculation would be (1.7756e+27 - 1.89e+27) / 1.89e+27 = .0605 or 6.05% which is within the 10% allowed error.
In this format, [GM / (4(3.14)^2)] becomes the slope of the graph. In our graph, the slope is about 3e+15. Using algebra, our equation becomes
M = [slope(4(3.14)^2)] / G = [3e+15(4(3.14)^2)] / 6.67e-11 = 1.7756 e+27
Percent error is calculated as (observed - expected) / (expected). In our case the calculation would be (1.7756e+27 - 1.89e+27) / 1.89e+27 = .0605 or 6.05% which is within the 10% allowed error.
Conclusion
Using Newton's derivation of Kepler's Third Law, we discovered that the estimated mass of Jupiter is 1.7756e+27 by setting the slope of the linearized graph:
R^3 = [GM / (4(3.14)^2)] * T^2 equal to [GM / (4(3.14)^2)]. We may then solve M = [slope(4(3.14)^2)] / G.
We also learned the process of measuring the orbit of a celestial body using its relative position to the mass it is orbiting. We used the CLEA software to accomplish this goal.The "maximum distance" from the central mass is actually the radius at all times. Each full orbit results when the moon reaches the same "maximum distance."
1. Calculate the percentage error with the accepted mass of Jupiter (1.8986 × 10^27 kg).
Percent error is calculated as (observed - expected) / (expected). In our case the calculation would be (1.7756e+27 - 1.89e+27) / 1.89e+27 = .0605 or 6.05% which is within the 10% allowed error.
2. There are moons beyond the orbit of Callisto. Will they have larger or smaller periods than Callisto? Why?
They will have larger periods do to the equation R^3 / T^2 = C. Since C is a constant, if R increases, T must also increase to compensate for the change.
3. Which do you think would cause the larger error in the mass of Jupiter calculation: a ten percent error in "T" or a ten percent error in "r"? Why?
10% error in R would be more significant because the R value is cubed while the T value is only squared.
4. Why were Galileo's observations of the orbits of Jupiter's moons an important piece of evidence supporting the heliocentric model of the universe (or, how were they evidence against the contemporary and officially adopted Aristotelian/Roman Catholic, geocentric view)?
Galileo proved that the Jupiter system had orbits of its own, and based his model of the heliocentric model of the universe off of the theory that smaller masses will be attracted to and orbit larger masses. This theory was evidence against the geocentric view of the universe that believed the entire universe revolved around the earth.
R^3 = [GM / (4(3.14)^2)] * T^2 equal to [GM / (4(3.14)^2)]. We may then solve M = [slope(4(3.14)^2)] / G.
We also learned the process of measuring the orbit of a celestial body using its relative position to the mass it is orbiting. We used the CLEA software to accomplish this goal.The "maximum distance" from the central mass is actually the radius at all times. Each full orbit results when the moon reaches the same "maximum distance."
1. Calculate the percentage error with the accepted mass of Jupiter (1.8986 × 10^27 kg).
Percent error is calculated as (observed - expected) / (expected). In our case the calculation would be (1.7756e+27 - 1.89e+27) / 1.89e+27 = .0605 or 6.05% which is within the 10% allowed error.
2. There are moons beyond the orbit of Callisto. Will they have larger or smaller periods than Callisto? Why?
They will have larger periods do to the equation R^3 / T^2 = C. Since C is a constant, if R increases, T must also increase to compensate for the change.
3. Which do you think would cause the larger error in the mass of Jupiter calculation: a ten percent error in "T" or a ten percent error in "r"? Why?
10% error in R would be more significant because the R value is cubed while the T value is only squared.
4. Why were Galileo's observations of the orbits of Jupiter's moons an important piece of evidence supporting the heliocentric model of the universe (or, how were they evidence against the contemporary and officially adopted Aristotelian/Roman Catholic, geocentric view)?
Galileo proved that the Jupiter system had orbits of its own, and based his model of the heliocentric model of the universe off of the theory that smaller masses will be attracted to and orbit larger masses. This theory was evidence against the geocentric view of the universe that believed the entire universe revolved around the earth.