Analyzers: Moons of Jupiter
Objectives of this Lab:
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.
Background:
Kepler's third law for a moon orbiting a much larger body is
C = r^3/T^2, where C is a constant, r is the length of the semi-major axis of the elliptical orbit in units of the mean Earth-Sun distance, 1 A.U. (astro-nomical unit), and T is the planetary orbit in Earth years.
If the orbit is circular (as will be assumed in this lab) the semi-major axis is the same as the radius of the orbit.
Newton expanded on Kepler’s Third Law, by using his Universal Law of Gravitation to solve for the constant C and derived:
r^3/T^2= GM/(4 pi^2)
where G is the gravitational constant 6.67x10^-11 Nm^2/kg^2 and M is the mass of the larger body.
C = r^3/T^2, where C is a constant, r is the length of the semi-major axis of the elliptical orbit in units of the mean Earth-Sun distance, 1 A.U. (astro-nomical unit), and T is the planetary orbit in Earth years.
If the orbit is circular (as will be assumed in this lab) the semi-major axis is the same as the radius of the orbit.
Newton expanded on Kepler’s Third Law, by using his Universal Law of Gravitation to solve for the constant C and derived:
r^3/T^2= GM/(4 pi^2)
where G is the gravitational constant 6.67x10^-11 Nm^2/kg^2 and M is the mass of the larger body.
Procedure:
Data:
| jupsatdata4.csv |
Data Analysis:
Slope of line = 3E15 3.21344E+15
Slope = GM/4p2
4p2/G= 5.9188E+11
Therefore, M = Slope*4p2/G
M (calc) = 1.90197E+27
M (accepted) = 1.8986E+27
% error = 0.17769102
Slope = GM/4p2
4p2/G= 5.9188E+11
Therefore, M = Slope*4p2/G
M (calc) = 1.90197E+27
M (accepted) = 1.8986E+27
% error = 0.17769102
| glassicdataanalysis.xlsx |
| io_results30daysng.bmp |
| ganymede_results20days.bmp |
| ganymede_results30days_good.bmp |
| callisto_resultsbest.bmp |
| callisto_results20days.bmp |
| europa_results30days.bmp |
Conclusion:
1. Calculate the percentage error with the accepted mass of Jupiter (1.8986 × 10^27 kg).
M (calc) = 1.90197E+27
M (accepted) = 1.8986E+27
(1.90197E+27-1.8986E+27)/(1.8986E+27) x 100
% error = 0.17769102
2. There are moons beyond the orbit of Callisto. Will they have larger or smaller periods than Callisto? Why?
The period of an orbit is the total time that is takes for the object to make one complete revolution. They will have larger periods than that of Callisto. Due to Kepler's relationship between period, T, and radius, R, it can be seen that T=sqrt(4pi^2R^3/GM). If the radius from the mass of orbit (M), is larger than that of Castillo, the period will increase as well. As the radius increases, the time it takes for one revolution around Jupiter will increase; the period will increase.
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?
I believe that a ten percent error in "r" would have a greater error in the mass of Jupiter. Since the graph is r^3 vs. T^2, the error will be greater since the radius value is cubed compared to squared, and therefore changing the slope which is a direct proportion of the line. Since the mass is proportional to the slope, it will cause the mass to have a greater error.
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's observations of the orbits of Jupiter's moons showed that they directly obey Kepler's Laws themselves. This would support the heliocentric model of the universe because the larger the mass, the grater the gravitational pull, and therefore that the sun is the center of our solar system. This model accounts for retrograde motion and the observed size and brightness variations of the planets in a way that makes more sense to even the less-educated population than the geocentric model.
M (calc) = 1.90197E+27
M (accepted) = 1.8986E+27
(1.90197E+27-1.8986E+27)/(1.8986E+27) x 100
% error = 0.17769102
2. There are moons beyond the orbit of Callisto. Will they have larger or smaller periods than Callisto? Why?
The period of an orbit is the total time that is takes for the object to make one complete revolution. They will have larger periods than that of Callisto. Due to Kepler's relationship between period, T, and radius, R, it can be seen that T=sqrt(4pi^2R^3/GM). If the radius from the mass of orbit (M), is larger than that of Castillo, the period will increase as well. As the radius increases, the time it takes for one revolution around Jupiter will increase; the period will increase.
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?
I believe that a ten percent error in "r" would have a greater error in the mass of Jupiter. Since the graph is r^3 vs. T^2, the error will be greater since the radius value is cubed compared to squared, and therefore changing the slope which is a direct proportion of the line. Since the mass is proportional to the slope, it will cause the mass to have a greater error.
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's observations of the orbits of Jupiter's moons showed that they directly obey Kepler's Laws themselves. This would support the heliocentric model of the universe because the larger the mass, the grater the gravitational pull, and therefore that the sun is the center of our solar system. This model accounts for retrograde motion and the observed size and brightness variations of the planets in a way that makes more sense to even the less-educated population than the geocentric model.
