This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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?...
View
Full
Document
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.
 Spring '08
 Turner
 Mass, Work

Click to edit the document details