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hw7 solutions

# hw7 solutions - homework 07 VENNES ROSS Due 1:00 am...

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homework 07 – VENNES, ROSS – Due: Oct 19 2007, 1:00 am 1 Question 1, chap 9, sect 1. part 1 of 1 10 points According to some nineteenth-century 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 Earth’s 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 Niven’s 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 . 38 × 10 6 m / s, and its radius is 1 . 3 × 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 . 27 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: 14 . 6492 m / s 2 (tolerance ± 1 %). Explanation: The centripetal acceleration of the inhabi- tants is a c = v 2 r = (1 . 38 × 10 6 m / s) 2 1 . 3 × 10 11 m = 14 . 6492 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 . 36242 × 10 33 kg (tolerance ± 1 %). Explanation: The difference Δ 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 Δ a = a c a i . On the other hand, this difference can be expressed from Newton’s law of gravity Δ a = G M 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 Δ a G = r 2 ( a c a i ) G = (1 . 3 × 10 11 m) 2 × (14 . 6492 m / s 2 ) (9 . 27 m / s 2 ) 6 . 67259 × 10 11 N m 2 / kg 2 = 1 . 36242 × 10 33 kg . Question 4, chap 9, sect 99. part 1 of 1 10 points

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homework 07 – VENNES, ROSS – 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 Earth’s surface until its weight is 0 . 62 times its weight on the Earth’s surface? Correct answer: 1719 . 91 km (tolerance ± 1 %). Explanation: Basic concepts By Newton’s Universal Law of Gravitation, W 1 r 2 Let : n 0 . 62 . Solution: The radius of the Earth is R Earth = 6 . 37 × 10 6 m . The rocket will be at a distance of h + R Earth from the center of the Earth when it weighs n W . Thus n W W = 1 ( h + R Earth ) 2 1 R 2 Earth = R 2 Earth ( h + R Earth ) 2 , so R 2 Earth = n ( h + R Earth ) 2 R 2 Earth n = ( h + R Earth ) 2 R Earth n = h + R Earth , so h = R Earth n R Earth = R Earth parenleftbigg 1 n 1 . 0
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hw7 solutions - homework 07 VENNES ROSS Due 1:00 am...

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