Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch
ID: 2011965 • Letter: Y
Question
Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a 2.50-kg stone thrown upward from the ground at 14.0 returns to the ground in 9.00 ; the circumference of Mongo at the equator is 1.00×105 ; and there is no appreciable atmosphere on Mongo.Part A:
Question:
The starship commander, Captain Confusion, asks for the following information: what is the mass of Mongo?
Answer:
1.18*10^25
Part B:
Question:
If the Aimless Wanderer goes into a circular orbit 1.00×104 above the surface of Mongo, how many hours will it take the ship to complete one orbit?
Not sure where Im going wrong with part b, but that is where i need help.
Explanation / Answer
the free fall acceleration on mongo, g m = 14 m/s/4.5 = 3.11 m/s2 the radius of mongo, r = circum/ 2 = 1.00 x 108 meters / 2 = 1.59 x 107 meters (a) mass of planet is found with g = GM /R^2 ==> M = R^2g/ G = (1.59 x107)^2 * 3.11 / 6.673 x 10-11 = = 1.18x 1025 kg (b) The radius of the orbit would be height +radius of planet = 10000 + 15909 = 25909.09 km and using keplers third law T2 = (42 / GM) r3 = (42 / 6.673 x 10-11 * 1.18x1025 ) * (2.5909x 107 )3 = = 872.67 x 106 T = 29.53 x 103 seconds the free fall acceleration on mongo, g m = 14 m/s/4.5 = 3.11 m/s2 the radius of mongo, r = circum/ 2 = 1.00 x 108 meters / 2 = 1.59 x 107 meters (a) mass of planet is found with g = GM /R^2 ==> M = R^2g/ G = (1.59 x107)^2 * 3.11 / 6.673 x 10-11 = = 1.18x 1025 kg (b) The radius of the orbit would be height +radius of planet = 10000 + 15909 = 25909.09 km and using keplers third law T2 = (42 / GM) r3 = (42 / 6.673 x 10-11 * 1.18x1025 ) * (2.5909x 107 )3 = = 872.67 x 106 T = 29.53 x 103 secondsRelated Questions
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