Question

The table on the figure gives the mean distance, period, and mass for the four largest moons of Jupiter (those discovered by Galileo in 1609). (Figure 1) Par Deter Expre Figure 1 of 1 > MJupi Submit Principal Moons of Jupiter Part B Moon Period Mean distance Mass (kg) (Earth days) from Jupiter (km) 8.9 X 10 1.77 422 x 10 4.9 X 10 671 x 10 15 x 10" 11 X 102 1070 x 10 16.7 1883 x 10 lo Europa Ganymede Callisto Determine ti 3.55 7.16 O no (dit O yes (di Submit Esc

Answer 1

3 To calculate the mass of Jupiter you Simply use Newslow's version of Kepler's Third law. 412 T2 = - Mim where, M is the mass of Jupiter and m is the mass of the moon and G is the universal Pravitational Constant (6.67x10-" m3/kg-seo"). Because the mass of the moon (m) 75 so small compared to the moss of Jupiter (m) we are able to remove m from the equation Rearranging we get: To period a = distance from jupiter to the moon a M = 47² G

Part A: For To: M = 4X (3.14)2 (422 x 103 x 103) 3 6.67810-11 (1.77* 86400)2 as. I ewith day = 86400 see - 1.902x10°7 Boston 1.902 x 1047 kg So, mass of the Jupiter 1.902 x to??KA

For Europa: ME Home = 4 x3, 14)² (671x 10° x 203) 3 wirectory - 6.67x10-11 X (3.55 x 86400) 2. = 1.901 x 1027 kg So, the mass of Jupiter is 1.901 x 10 kg Difference of mass with the 1st one is 1.902x10 27 – 1.90 1x1027 = 9 10²4 kg So, the percentage = 1,902x1027 X1007.-0.05, 26 de 7 <1 % 94 to 24 For Ganymede ma 402 4x (3.14)2 (1070 x 10 x 10°) 6.67 X 10^11 (7:16 x 86400)2 2 1.895 x 1027 kg So mass of Iupiter is 1.895 × 10 27 kg Difference of mass with the Ist one is 1.902x10 27 – 1,895x1027 = 7x1024 So, the percentage - 7X 1024 1.902x1027 * [00 =0. 3,68 164 <1%

For Callisto : m = 4112 - a 2 = 4x (3.14)² (1883x102103) 3 - 6.67X 10-11 (16,7x86400) 2 = 1.898x10 27 kg mass of Jupiter is 1.898 x1027 kg Differenee in mass with Ist one is (1.902-1-898) x10279 2 4x1024 kg Percentage = ter x 100% = 0.21% 4x1024 1.902x1027 So, answer will be yes (difference between them <1%)