ps2solnsF06

ps2solnsF06 - %Engineering 6 Fall 2006 Problem 2.1...

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%Engineering 6, Fall 2006, Problem 2.1 Solution %(Planetary calculation) %Turn off extra line breaks in display format compact ; %Clear variables and command window clear; clc; %Define variables for the diameters of Earth and Jupiter % (in meters) and their rotational periods (in hours) Ediam = 7926; Jdiam = 88736; Eperiod = 23.934; Jperiod = 9.85; %Calculate the speed of the planet's surface % at the equator, using the following equations: % Planet circumference = pi*diameter % Speed of surface = circumference/period (i.e., the % surface travels a distance equal to the circumference % in a time equal to the period of rotation) Espeed = pi*Ediam/Eperiod; Jspeed = pi*Jdiam/Jperiod; fprintf( 'Speed of the surface of the Earth at its\nequator is %.1f miles per hour.\n\n' , Espeed); fprintf( 'Speed of the surface of Jupiter at its\nequator is %.1f miles per hour.\n\n' , Jspeed); Displayed results: Speed of the surface of the Earth at its equator is 1040.4 miles per hour. Speed of the surface of Jupiter at its equator is 28301.8 miles per hour.
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%Engineering 6, Fall 2006, Problem 2.2 Solution
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ps2solnsF06 - %Engineering 6 Fall 2006 Problem 2.1...

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