Homework 7 Solutions - homework 07 ALIBHAI, ZAHID Due: Mar...

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Unformatted text preview: homework 07 ALIBHAI, ZAHID Due: Mar 7 2007, 4:00 am 1 Question 1 part 1 of 1 10 points In introductory physics laboratories, a typ- ical Cavendish balance for measuring the gravitational constant G uses lead spheres of masses 2 . 34 kg and 13 . 9 g whose centers are separated by 5 . 54 cm. Calculate the gravitational force between these spheres, treating each as a point mass located at the center of the sphere. Use G = 6 . 67259 10 11 N m 2 / kg 2 . Correct answer: 7 . 0714 10 10 N (tolerance 1 %). Explanation: Basic Concepts: Force of gravity: F = G m 1 m 2 r 2 The force of gravity is F = G m 1 m 2 r 2 = 6 . 67259 10 11 N m 2 / kg 2 2 . 34 kg . 0139 kg (0 . 0554 m) 2 = 7 . 0714 10 10 N Clearly, this force is very tiny. The Cavendish balance is set up very delicately to notice this force. Question 2 part 1 of 1 10 points Given: G = 6 . 673 10 11 N m 2 kg 2 Mars has a mass of about 6 . 97 10 23 kg, and its moon Phobos has a mass of about 9 . 44 10 15 kg. If the magnitude of the gravitational force between the two bodies is 4.69 10 15 N, how far apart are Mars and Phobos? Correct answer: 9 . 67557 10 6 m (tolerance 1 %). Explanation: Basic Concept: F g = G m 1 m 2 r 2 Given: m 1 = 6 . 97 10 23 kg m 2 = 9 . 44 10 15 kg F g = 4 . 69 10 15 N G = 6 . 673 10 11 N m 2 kg 2 Solution: r 2 = Gm 1 m 2 F g = parenleftbigg 6 . 673 10 11 N m 2 kg 2 parenrightbigg ( 6 . 97 10 23 kg )( 9 . 44 10 15 kg ) 4 . 69 10 15 N = 9 . 36166 10 13 m 2 Thus r = radicalbig 9 . 36166 10 13 m 2 = 9 . 67557 10 6 m Question 3 part 1 of 1 10 points Planet X has three times the diameter and six times the mass of the earth. What is the ratio of gravitational accelera- tion at the surface of planet X to the gravita- tional acceleration at the surface of the Earth, g x g e ? 1. g x g e = 5 49 2. g x g e = 2 3 correct 3. g x g e = 9 64 4. g x g e = 9 25 5. g x g e = 2 25 6. g x g e = 6 49 homework 07 ALIBHAI, ZAHID Due: Mar 7 2007, 4:00 am 2 7. g x g e = 7 16 8. g x g e = 1 32 9. g x g e = 5 16 10. g x g e = 2 81 Explanation: Let : M X = 6 M e , R X = 3 R e . Because mg e = GM e m R 2 e , where M e is the mass of the earth and R e is the radius. So that g e = GM e R 2 e . Similar equation applies to the planet X g x = GM x R 2 x . The ratio between the surface gravitational accelerations is g x g e = M x R 2 e M e R 2 x = 6 M e R 2 e M e (3 R e ) 2 = 2 3 . Question 4 part 1 of 1 10 points Two iron spheres, of mass m and 2 m , respectively, and equally spaced points r apart are shown in the figure. A m B C D 2 m E r r r r r r At which location would the net gravita- tional force on an object due to these two spheres be a maximum?...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Homework 7 Solutions - homework 07 ALIBHAI, ZAHID Due: Mar...

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