Lab 4 Glileo and Romer

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Dec 6, 2023

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Name: Jack Steibel TA: Natalio Partner’s Name: Astronomy 1101 Galileo’s Observations of Jupiter and Romer’s Speed of Light Activity Part 1: The Moons of Jupiter – Use a Calculator The diagrams on the next page are inspired by Galileo’s drawings of the moons of Jupiter. He observed four points of light very close to Jupiter, all in a line, and that their positions changed from night to night, and during a night, but always staying close to Jupiter. In each diagram, the black circle marks the position of Jupiter, and the asterisks mark the positions of the moons. The angular scale is marked in degrees, as shown below: In this picture, the third moon is separated from Jupiter by an angle of about 0.1 degrees. Each diagram shows an observation at a different time, separated by intervals of 0.5 days, as marked to the left or right. The picture above (the first observation) shows a time when each moon is at its maximum angular separation from Jupiter. From inner to outer, the moons are marked with a different symbol: Io ( ), Europa ( ), Ganymede ( ), and Callisto ( ). 1. What are the angular separations ( ) of the four moons from Jupiter, in degrees? Io: .04 Europa: .06 Ganymede: .1 Callisto: .17 2. These observations correspond to a time when Jupiter is closest to the Earth, a distance (D) of 6.3 10 8 km. Use the equation R = D ( / 57.3 degrees) to compute the orbital radii (R) of the moons’ orbits around Jupiter in kilometers. Io: 4.4 x 10^5 km Europa: 6.6 x 10^5 km Ganymede: 1.1 x 10^6 km Callisto: 1.9 x 10^6 km 3. From the diagram, determine the orbital periods of the four moons (i.e., how long they take to go around Jupiter), in days. Try to measure the orbital period to at least the nearest 0.5 days, or to a finer degree (e.g., 0.1 days) if you can. [ Hint: Start with Callisto and work inwards to Io.] Io: 1.75 days Europa: 3.5 days Ganymede: 7 days Callisto: 16.5 days 4. You have been conveniently given simulated observations separated by regular intervals of 0.5 days, which makes it easier to measure the orbital periods. Why would it have been impossible for Galileo to obtain an actual series of observations of Jupiter every 0.5 days? Galileo lacked technology and Jupiter is not visible every .5 days, only at night. 5. Convert the period (P) for each moon from days to seconds. Use the fact that there are 60 60 24=86,400 seconds in a day. Io: 1.5 x 10^5 seconds Europa: 3.0 x 10^5 seconds Ganymede: 6.0 x 10^5 Callisto: 1.4 x 10^6 seconds 6. Assume that the moons are on circular orbits (not a bad assumption). Use the formula: Speed = 2 R/P to compute the orbital speeds of the four moons in units of km/sec. 1
Io: 18.43 km/sec Europa: 13.82 km/sec Ganymede: 11.52 km/sec Callisto: 8.53 km/sec As you saw in lecture, if one body is on a circular orbit around a much more massive body, the orbital speed, V orbit , at a distance R away from a central body of mass M , is given by 2
V orbit = GM / R If you know the orbital speed V orbit and the orbital radius, R , you can solve for the mass of the central body: V ( ¿¿ orbit 2 ×R )/ G M Jup = ¿ – Use this Equation In both equations, G, is the Newtonian Gravitational Constant, one of the fundamental physical constants. The value of G is G = 6.67 × 10 11 m 3 kg 1 s 2 – Gravitational Constant with Units Important: because the value of G contains units of meters (m) instead of kilometers (km), you will need to convert your orbital radii, R, from km to m, and your orbital speeds from km/sec to m/sec using the facts that 1 km = 1000 m, and 1 km/sec = 1000 m/sec. If you do this, then the mass formula above will give you an estimate of the mass of Jupiter! 7. Use the mass formula to estimate the mass of Jupiter from the orbital speed and orbital radius of Callisto. Show/explain your calculation in the space below (i.e., write down the numbers that you are multiplying or dividing, and show the result). Orbital Radius in Meters: (1.9 x 10^6 km) x 1000 m/km= 1.9 x 10^9 meters Orbital Speed in m/sec: (8.53 km/sec) x 1000 (m/sec)/(km/sec) = 8530 m/sec (((8530 m/sec)^2) x (1.9 x 10^9 m)) / (6.67 x 10^-11 m^3kg^-1s^-2) = 2.07 x 10^27 kg 3
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Part 2: Romer’s Speed of Light (Credit: https://lco.global/spacebook/light/speed-light/ ) In 1676 a Danish astronomer named Ole Rømer was studying the orbits of the moons of Jupiter and making tables to predict when eclipses of the moons would occur. He noticed that when Jupiter and Earth are far apart (near conjunction), the eclipses of the moons occurred several minutes later than when Jupiter and the Earth are closer (near opposition.) He reasoned that this could be because of the time light takes to travel from Jupiter to Earth. 8. Rømer found the maximum variation in timing of these eclipses to be 16.6 minutes. He interpreted this to be the amount of time it takes light to travel across the diameter of Earth's orbit. If 1 AU = 1.4959 × 10 8 km, Earth’s orbital radius, calculate the speed of light in km/s based on his measurement. (16.6 minutes) x (60 sec/minute) = 996 seconds (1.4959 x 10^8 km) / (996 seconds) = 1.5 x 10^5 km/sec 9. Compare this to a “rounded” but modern value of 299,792 km/s. Is your answer close? My answer is awfully close to half the modern value. That is far off, however in 1676 this would be a solid measurement. 4