Exam_3_practice - Version PREVIEW Exam 3 Practice...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Version PREVIEW – Exam 3 Practice – chelikowsky – (59005) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Conceptual 05 09 001 (part 1 of 2) 10.0 points Calculate the weight of a solid rocket booster of the space shuttle with mass 5 . 7 × 10 5 kg on Earth. The Earth has a mass of 6 × 10 24 kg and a radius of 6 . 4 × 10 6 m. 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 5 . 586 × 10 6 N. Explanation: Let : m = 6 × 10 24 kg and g = 6 . 67 × 10 11 N · m 2 / kg 2 . W = m g = (5 . 7 × 10 5 kg) (9 . 8 m / s 2 ) = 5 . 586 × 10 6 N . 002 (part 2 of 2) 10.0 points What would this weight be on Mars, with mass 0 . 11 m E and radius 0 . 53 R E ? The value of the universal gravitational constant is 6 . 67 × 10 11 N · m 2 / kg 2 . 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 2 . 18089 × 10 6 N. Explanation: Let : M E = 6 × 10 24 kg , R E = 6 . 4 × 10 6 m , M M = 0 . 11 m E , R M = 0 . 53 R E , and G = 6 . 67 × 10 11 N · m 2 / kg 2 . On Mars g = G M M R 2 M = 6 . 67 × 10 11 N · m 2 / kg 2 [0 . 53 (6 . 4 × 10 6 m)] 2 × [0 . 11 (6 × 10 24 kg)] = 3 . 82612 m / s 2 . W W = m g m g W = g g W = 3 . 82612 m / s 2 9 . 8 m / s 2 ( 5 . 586 × 10 6 N ) = 2 . 18089 × 10 6 N . Finding G Like Cavendish 003 10.0 points An apparatus like the one Cavendish used to find G has a large lead ball that is 5 . 2 kg in mass and a small one that is 0 . 056 kg. Their centers are separated by 0 . 046 m. Find the force of attraction between them. The value of the universal gravitational con- stant is 6 . 67259 × 10 11 N · m 2 / kg 2 . 1. @@@ 2. @@@ 3. @@@ 4. @@@
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version PREVIEW – Exam 3 Practice – chelikowsky – (59005) 2 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 9 . 18269 × 10 9 N. Explanation: Let : d = 0 . 046 m , G = 6 . 67259 × 10 11 N · m 2 / kg 2 , m 1 = 5 . 2 kg , and m 2 = 0 . 056 kg . By Newton’s universal law of gravitation, F = G m 1 m 2 d 2 = (6 . 67259 × 10 11 N · m 2 / kg 2 ) (5 . 2 kg) (0 . 056 kg) 0 . 046 m 2 = 9 . 18269 × 10 9 N . Suspended Over a Planet 004 10.0 points A “synchronous” satellite is put in orbit about a planet to always remain above the same point on the planet’s equator. The planet rotates once every 10 . 1 h, has a mass of 2 . 1 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given: G = 6 . 67 × 10 11 N m 2 / kg 2 . Calculate how far above the planet’s surface the satellite must be. 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 9 . 74957 × 10 7 m. Explanation: Basic Concepts: Solution: According to Kepler’s third law T 2 = 4 π 2 G M r 3 where r is the radius of the satellite’s orbit. Thus, solving for r r = bracketleftbigg G M T 2 4 π 2 bracketrightbigg 1 3 = bracketleftbigg (6 . 67 × 10 11 N m 2 / kg 2 ) 4 (3 . 1415926) 2 × (2 . 1 × 10 27 kg) (36360 s) 2 bracketrightbigg 1 3 = 1 . 67396 × 10 8 m . Now, the altitude h of the satellite (measured from the surface of Jupiter) is h = r R = (1 . 67396 × 10 8 m) (6 . 99 × 10 7 m) = 9 . 74957 × 10 7 m . Orbiting Small Moon 005 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 A small Moon of a planet has an orbital period of 2 . 7 days and an orbital radius of 5 . 13 × 10 5 km.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern