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Neptune's largest moon, Triton, has an orbital period of 5.88 days and is 3.55 x 105 km from the planet. At what distance (in m) would the moon need to be to escape the planet at its current cin velocity? Part 1 of 4 Newton's interpretation of Kepler's Third Law describes how the distance between any two bodies and the orbital period are related: p² -- (472),² GM where G = 6.67 x 10-11 m³/s2/kg. Part 2 of 4 We can use this expression to determine the mass of the planet-moon system. Convert kilometers to meters using 1 km = 1,000 m. (Enter your answer to at least three significant figures.) m = m Convert the period into seconds. (Enter your answer to at least three significant figures.) Psec = Pdays (24hr) (3,600 s) 1 day. 1 hr Psec = seconds Now solve the expression for M and plug in the period in seconds and the distance in meters. (47²)(rm) ³ G (Psec)² M = M = kg

Neptune's largest moon, Triton, has an orbital period of 5.88 days and is 3.55 x 105 km from the planet. At what distance (in m) would the moon need to be to escape the planet at its current cin velocity? Part 1 of 4 Newton's interpretation of Kepler's Third Law describes how the distance between any two bodies and the orbital period are related: p² -- (472),² GM where G = 6.67 x 10-11 m³/s2/kg. Part 2 of 4 We can use this expression to determine the mass of the planet-moon system. Convert kilometers to meters using 1 km = 1,000 m. (Enter your answer to at least three significant figures.) m = m Convert the period into seconds. (Enter your answer to at least three significant figures.) Psec = Pdays (24hr) (3,600 s) 1 day. 1 hr Psec = seconds Now solve the expression for M and plug in the period in seconds and the distance in meters. (47²)(rm) ³ G (Psec)² M = M = kg

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Neptune's largest moon, Triton, has an orbital period of 5.88 days and is 3.55 x 105 km from the planet. At what distance (in m) would the moon need to be to escape the planet at its current circular velocity? Part 1 of 4 Newton's interpretation of Kepler's Third Law describes how the distance between any two bodies and the orbital period are related: 477² GM where G = 6.67 x 10-11 m³/s2/kg. Part 2 of 4 We can use this expression to determine the mass of the planet-moon system. Convert kilometers to meters using 1 km = 1,000 m. (Enter your answer to at least three significant figures.) rm = m Convert the period into seconds. (Enter your answer to at least three significant figures.) 24 hr P sec Psec = p2 p² = = M = M = thr) (³,61 Pdays 1 day Now solve the expression for M and plug in the period in seconds and the distance in meters. (4π²) (rm) ³ 2 G (Psec)² 3,600 s 1 hr seconds kg
Transcribed Image Text:Neptune's largest moon, Triton, has an orbital period of 5.88 days and is 3.55 x 105 km from the planet. At what distance (in m) would the moon need to be to escape the planet at its current circular velocity? Part 1 of 4 Newton's interpretation of Kepler's Third Law describes how the distance between any two bodies and the orbital period are related: 477² GM where G = 6.67 x 10-11 m³/s2/kg. Part 2 of 4 We can use this expression to determine the mass of the planet-moon system. Convert kilometers to meters using 1 km = 1,000 m. (Enter your answer to at least three significant figures.) rm = m Convert the period into seconds. (Enter your answer to at least three significant figures.) 24 hr P sec Psec = p2 p² = = M = M = thr) (³,61 Pdays 1 day Now solve the expression for M and plug in the period in seconds and the distance in meters. (4π²) (rm) ³ 2 G (Psec)² 3,600 s 1 hr seconds kg
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Once we have the mass, we can determine the magnitude of the circular velocity of the moon using the expression: GM Vc = Vc m/s
Transcribed Image Text:Once we have the mass, we can determine the magnitude of the circular velocity of the moon using the expression: GM Vc = Vc m/s
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