hw3 tsoi - homework 07 – GADHIA TEJAS – Due Mar 7 2007...

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Unformatted text preview: homework 07 – GADHIA, TEJAS – Due: Mar 7 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has mass M and speed v . G is the universal gravitational constant. D v v M M Which of the following is a correct relation- ship among these quantities? 1. v 2 = 4 GM D 2. v 2 = 2 GM 2 D 3. v 2 = M GD 4. v 2 = GM D 5. v 2 = GM 2 D correct 6. v 2 = 4 GM 2 D 7. v 2 = 2 GM D 8. v 2 = GM D 2 Explanation: From circular orbital movement, the cen- tripetal acceleration is a = v 2 D 2 . Using the Newton’s second law of motion, we know the acceleration is a = F M , where F is the force between two stars and is totally supplied by the universal force. So we obtain 2 v 2 D = a = F M = GM D 2 = ⇒ v 2 = GM 2 D . Question 2 part 1 of 2 10 points Compare the gravitational force on a 23 . 5 kg mass at the surface of the Earth ( R E = 6 . 4 × 10 6 m, M E = 6 × 10 24 kg) with that on the surface of the Moon ( M M = 1 81 . 3 M E , R M = 0 . 27 R E ). What is it on the Earth? Correct answer: 229 . 607 N (tolerance ± 1 %). Explanation: Let : m = 23 . 5 kg , and R E = 6 . 4 × 10 6 m . F = GmM E R 2 E = (6 . 67 × 10 − 11 N · m 2 / kg 2 ) (23 . 5 kg) (6 . 4 × 10 6 m) 2 × (6 × 10 24 kg) = 229 . 607 N . Question 3 part 2 of 2 10 points What is it on the Moon? Correct answer: 38 . 7407 N (tolerance ± 1 %). Explanation: F = GmM M R 2 M = (6 . 67 × 10 − 11 N · m 2 / kg 2 ) (23 . 5 kg) (0 . 27 × 6 . 4 × 10 6 m) 2 × (6 × 10 24 kg / 81 . 3) = 38 . 7407 N . homework 07 – GADHIA, TEJAS – Due: Mar 7 2007, 4:00 am 2 Question 4 part 1 of 2 10 points Given: G = 6 . 6726 × 10 − 11 N m 2 / kg 2 . Three masses are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). y ( m )- 5- 3- 1 0 1 2 3 4 5 x ( m )- 5- 4- 3- 2- 1 1 2 3 4 5 1 kg 7 kg 3 kg What is the magnitude of the resulting force on the 1 kg mass at the origin? Correct answer: 2 . 7125 × 10 − 11 N (tolerance ± 1 %). Explanation: Let : m o = 1 kg , ( x o , y o ) = (0 m , 0 m) , m a = 7 kg , ( x a , y a ) = (1 m ,- 4 m) , m b = 3 kg , ( x b , y b ) = (4 m , 2 m) . Newton’s Universal Gravitational Law for m o and m a is F ao = G m o m a radicalbig ( x a- x o ) 2 + ( y a- y o ) 2 = (6 . 6726 × 10 − 11 N m 2 / kg 2 ) · (1 kg) (7 kg) radicalbig (1 m) 2 + (- 4 m) 2 = 2 . 74754 × 10 − 11 N . tan θ a = y a x a θ a = arctan parenleftbigg y a x a parenrightbigg = arctan parenleftbigg- 4 m 1 m parenrightbigg = 284 . 036 ◦ . F a x = F a cos θ a = ( 2 . 74754 × 10 − 11 N ) cos 284 . 036 ◦ = 6 . 66377 × 10 − 12 N . F a y = F a sin θ a = ( 2 . 74754 × 10 − 11 N ) sin 284 . 036 ◦ =- 2 . 66551 × 10 − 11 N ....
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hw3 tsoi - homework 07 – GADHIA TEJAS – Due Mar 7 2007...

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