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Solutions to Physics I Gravity and Keplerβs Laws Practice Problems 1.) Titan, the largest moon of Saturn, has a mean orbital radius of 1.22x109 m. The orbital period of Titan is 15.95 days. Hyperion, another moon of Saturn, orbits at a mean radius of 1.48x109 m. Use Keplerβs third law of planetary motion to predict the orbital period of Hyperion in days. ππ = 1.22π₯109 π ππ = 15.95 πππ¦π ππ» = 1.48π₯109 π ππ» =? ππ 2 ππ 3 ( ) =( ) ππ» ππ» 3 15.95 πππ¦π 2 1.22π₯109 π ( ) =( ) ππ» 1.48π₯109 π 15.95 πππ¦π 2 ( ) = 0.8243 ππ» 254.4 πππ¦π 2 ππ» 2 ππ» = β = 0.560 254.4 πππ¦π 2 0.560 ππ» = 21.3 πππ¦π 2.) The mass of Earth is 5.97x1024 kg, the mass of the Moon is 7.35x1022 kg, and the mean distance of the Moon from the center of Earth is 3.84x105 km. Use these data to calculate the magnitude of the gravitational force exerted by Earth on the Moon. ππΈ = 5.97π₯1024 ππ ππ = 7.35π₯1022 ππ π = 3.84π₯105 ππ = 3.84π₯108 π πΊ = 6.673π₯10β11 π β π2 βππ2 πΉππΈ =? πΉππΈ = πΊ ππΈ ππ π2 (5.97π₯1024 ππ)(7.35π₯1022 ππ) πΉππΈ = (6.673π₯10β11 π β π2 βππ2 ) [ ] (3.84π₯108 π)2 4.39π₯1047 ππ2 πΉππΈ = (6.673π₯10β11 π β π2 βππ2 ) ( ) 1.47π₯1017 π2 πΉππΈ = (6.673π₯10β11 π β π2 βππ2 )(2.98π₯1030 ππ2 βπ2 ) πΉππΈ = 1.99π₯1020 π 3.) The planet Mercury travels around the Sun with a mean orbital radius of 5.8x1010 m. The mass of the Sun is 1.99x1030 kg. Use Newtonβs version of Keplerβs third law to determine how long it takes Mercury to orbit the Sun. Give your answer in Earth days. ππ = 5.810π₯1010 π ππ = 1.99π₯1030 ππ ππ =? ππ 2 4π 2 3 =( )π πΊππ ππ 2 = [ ππ 2 = [ (6.673π₯10β11 39.5 ] (5.810π₯1010 π)3 π β π2 βππ2 )(1.99π₯1030 ππ) 39.5 ] (1.96π₯1032 π3 ) π β π2 βππ 1.33π₯1020 ππ 2 = (2.96π₯10β19 π 2 βπ3 )(1.96π₯1032 π3 ) ππ 2 = 5.82π₯1013 π 2 ππ = β5.82π₯1013 π 2 1 βππ’π 1 πππ¦ ππ = 7.63π₯106 π ( )( ) = 88.3 πππ¦π 3600 π 24 βππ’ππ 4.) Earth has an orbital period of 365 days and its mean distance from the Sun is 1.495x108 km. The planet Plutoβs mean distance from the Sun is 5.896x109 km. Using Keplerβs third law, calculate Plutoβs orbital period in Earth days. ππΈ = 365 πππ¦π ππΈ = 1.495π₯108 ππ ππ = 5.896π₯109 ππ ππ =? ππΈ 2 ππΈ 3 ( ) =( ) ππ ππ 3 365 πππ¦π 2 1.495π₯108 ππ ( ) =( ) ππ 5.896π₯109 ππ 365 πππ¦π 2 ( ) = (2.54π₯10β2 )3 ππ 1.32π₯105 πππ¦π 2 ( ) = 1.63π₯10β5 2 ππ 1.32π₯105 πππ¦π 2 ππ = β 1.63π₯10β5 ππ = 9.00π₯104 πππ¦π 5.) The planet Venus orbits the Sun with a mean orbital radius of 1.076x1011 m. The mass of the Sun is 1.99x1030 kg. Using Newtonβs version of Keplerβs third law, calculate the orbital period of Venus. ππ = 1.076π₯1011 π ππ = 1.99π₯1030 ππ ππ =? ππ 2 4π 2 =( )π 3 πΊππ π 2 ππ = [ (6.673π₯10β11 ππ 2 = [ 4π 2 ] (1.076π₯1011 π)3 π β π2 βππ2 )(1.99π₯1030 ππ) 39.5 ] (1.25π₯1033 π3 ) π β π2 βππ 1.33π₯1020 ππ 2 = (2.97π₯10β19 π 2 βπ3 )(1.25π₯1033 π3 ) ππ 2 = 3.17π₯1014 π 2 ππ = β3.17π₯1014 π 2 ππ = 1.93π₯107 π