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• O • Is University of South Alabama XY WileyPLUS Ahmad * G Chegg Study Guided Solution X D Travis Scott - Butterfly Ette 4 x + ca Secure | https//edugen.wileyplus.com/edugen/student/mainfruni Y NEW I Apps R WileyPLUs a Differential calculus LS University of South. Apple C iCloud Yahoo b Bing w Wikipedia F Facebook , y Twitter in LinkedIn D The Weather Channel O Yelp >> other Bookmarks wleyPLUS: MywlePLUSI Help I contact Us I Log Out WileyPLUS haiday, Fundamentals of Physics10e Calculus-based Physics I & II (PH 201-202) Home Read, Study & Practice Assignment Gradebook ORION Downloadable eTextbook Assignment > Open Assignment PULL SCREEN PRINTER VERSION 1 BACK NTS ASSIGNMENT RESOURCES 201-6 Chapter 04, Problem 056 ontest as Boston An Earth satellite moves in a circular orbit 810 km above Earth's surface with a period of 100.9 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite? 04, Problem 5G Chapter 04. Problem (a) Number Units Units Inster is promitem (b) Number Chapter 05, Problem 006 Review Score Review Results by Study objective LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO NIN-LECTURE ITH HELP Question Attempts: 0 of 3 used SAVE FOR LATER SUBMIT ANSWER licensa Agreement privacy Policy ID 2000-201Zohn Wiley & Sons, Inc. All Rights Reserved. A Division of ohn Wiley & Sons, Inc. Version 4.24.1.20 https://edugen-wileyplus.com/ediugen/shared/assignment/test/agist uni?id-asnmt 204900Explanation / Answer
Solution:
a) The height of the satellite above the earth's surface = R + h = (6400+810)x103 m
Speed of the satellite = v = 2 pi (R+h) / T = 2 pi [(6400 +810) x 103 ] m / ((6054) s
= 7482.9 m/s = 7500 m/s (2 significant digits)
b) Centripetal acceleration = v2 /R = (7482.9)2 / (6400+810)x103
= 7.77 m/s2 = 7.8 m/s2 .
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