Abstract: This lab will use the program CLEA andthe simulation The Revolutions of the Moons of Jupiter to track the moons ofJupiter for 12-hour intervals. These points will be plotted, and a sine wavewill be fit to each of the moons location curves. The period and amplitude ofthis wave will then be used to calculate the mass of Jupiter. This will becompared to the actual mass of Jupiter and discussed.

Introduction: The mass of Jupiter can becalculated by observation of the moons that orbit the planet. The physicalobservation of Jupiter is time consuming and almost impossible in Irelandbecause of the weather. A telescope is also needed for actual observation. Toovercome these issues a simulation program CLEA was used to simulate the moonsof Jupiter as they orbit. Kepler’s Third Law (M = a³/p²) is used to calculatethe mass of Jupiter.Method: TheCLEA program simulation The Revolution of the Moons of Jupiterwas opened. The moons positions were logged 18 times at 12-hour intervalsmaking note of the cloudy days.

The program then plots thepoints for one moon on a graph. The period, amplitude and t-zero (the pointwhere the curve crosses the Y axis going from negative to positive) were foundand input into the fit sine curve. The curve was then manipulated to best fitthe points. The period and amplitude of this curve was then noted and convertedto years and A.U.

This was repeated for all the moons of Jupiter.Kepler’s Third Law Mass =amplitude³/period² was then used to calculate the mass of Jupiter according to eachmoon in Solar Masses. Results: Table 1: Location Data forthe 4 moons of Jupiter Figure 1: Sine curve fitto Callisto location pointsPeriod of Callisto =16.688 days/365 = 0.0457 yearsRadius of Callisto = 13.2Jupiter diameters/1050 = 0.

01257 A.U.Mass of Jupiter FromCallisto = 0.01254³(A.U.)/0.0457²(years) = 0.

0009509 M? Figure 2: Sine curve fitto Ganymede location pointsPeriod of Ganymede = 701days/365= 0.0195 yearsRadius of Ganymede = 7.49Jupiter diameters/1050 =0.00713 A.U.Mass of Jupiter From Ganymede= 0.00713³(A.U.

)/ 0.0195²(years) =0.000953 M? Figure 3: Sine curve fitto Europa location pointsPeriod of Europa = 3.584 days/365= 0.00982 yearsRadius of Europa = 4.7952Jupiter diameters/1050 =0.

004567 A.U.Mass of Jupiter From Europa= 0.004567³(A.U.

)/ 0.00982²(years) = 0.0009878 M? Figure 4: Sine curve fitto Io location pointsPeriod of Io = 1.768 days/365= 0.004839 yearsRadius of Io = Jupiterdiameters/1050 = 0.002868 A.U.

Mass of Jupiter From Io = ³(A.U.)/²(years) = 0.001007 M? Discussion/Conclusion: Average mass of Jupiter =(0.0009509+0.000953+0.

0009878+0.001007)/4 = 0.0009751 M?Mass of Jupiter in Earthmasses = 0.0009751/3*10?? = 325.058 Earth massesActual mass of Jupiter = 0.0009546M? or 317.

83 Earth masses 1 The calculated mass ofJupiter is close to the actual mass of Jupiter. Callisto and Ganymede’s massesof Jupiter came out as the closest with Europa close behind. The least accurateof the moons is Io.

This may have been due to the small orbit of this moon.Many times, it was impossible to mark the location as the moon is behindJupiter. The period of this moon is also very small resulting in difficultywhen trying to fit the sine curve to the limited points found. Jupiter has moons that arefurther away than Callisto. These moons will have larger periods than Callistobecause the period is the time it takes to orbit the planet. The further awayfrom a planet a moon is the longer the orbit will take. A 10% error in amplitudewould cause a larger error than a 10% error in period because in the formula amplitudeis cubed where as period is only squared.

Amplitude ends up as the largernumber. The moons orbit aroundEarth: 27.3 days/365 = 0.07479 yearsMass of Earth (question4): 0.00256³ A.U./0.07479² years = 0.000002999 or 2.999×10?? M?Actual mass of Earth = 3.003467×10??M? 2 1 https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Jupiter_mass.html2 https://www.wolframalpha.com/input/?i=mass+of+earth+in+solar+mass+units