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Unformatted text preview: homework 07 GAUTHIER, JOSEPH Due: Oct 19 2007, 1:00 am 1 Question 1, chap 9, sect 1. part 1 of 1 10 points According to some nineteenthcentury geo logical theories (now largely discredited), the Earth has been shrinking as it gradually cools. If so, how would g have changed over geo logical time? 1. It would decrease; the Earths radius is decreasing. 2. It would not change; the mass of the Earth remained the same. 3. It would increase; g is inversely propor tional to the square of the radius of the Earth. correct Explanation: The acceleration g of gravity would increase if the Earth shrank, but its mass stayed the same, since g 1 r 2 . Question 2, chap 9, sect 1. part 1 of 2 10 points Given: G = 6 . 67259 10 11 N m 2 / kg 2 In Larry Nivens science fiction novel Ring world , a ring of material rotates about a star. n Star F g The rotational speed of the ring is 1 . 27 10 6 m / s, and its radius is 1 . 06 10 11 m. The inhabitants of this ring world experience a normal contact force vector n . Acting alone, this normal force would produce an inward accel eration of 9 . 01 m / s 2 . Additionally, the star at the center of the ring exerts a gravitational force on the ring and its inhabitants. What is the total centripetal acceleration of the inhabitants? Correct answer: 15 . 216 m / s 2 (tolerance 1 %). Explanation: The centripetal acceleration of the inhabi tants is a c = v 2 r = (1 . 27 10 6 m / s) 2 1 . 06 10 11 m = 15 . 216 m / s 2 . Question 3, chap 9, sect 1. part 2 of 2 10 points The difference between the total acceler ation and the acceleration provided by the normal force is due to the gravitational at traction of the central star. Calculate the approximate mass of the star. Correct answer: 1 . 04504 10 33 kg (tolerance 1 %). Explanation: The difference a between the total accel eration and the acceleration provided by the normal force is due to the gravitational at traction of the central star and is equal to a = a c a i . On the other hand, this difference can be expressed from Newtons law of gravity a = GM r 2 , where M is the mass of the central star. Using the values for a c and a i , the mass of the star is M = r 2 a G = r 2 ( a c a i ) G = (1 . 06 10 11 m) 2 (15 . 216 m / s 2 ) (9 . 01 m / s 2 ) 6 . 67259 10 11 N m 2 / kg 2 = 1 . 04504 10 33 kg . Question 4, chap 9, sect 99. part 1 of 1 10 points homework 07 GAUTHIER, JOSEPH Due: Oct 19 2007, 1:00 am 2 Given: The radius of the earth is R Earth = 6 . 37 10 6 m . How high does a rocket have to go above the Earths surface until its weight is 0 . 61 times its weight on the Earths surface?...
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 Fall '07
 Gilbert
 Logic

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