Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
5
Circular Motion; Gravitation
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5-1 to 5-3: Uniform Circular Motion; Highway Curves
5-4: Nonuniform Circular Motion
5-6 and 5-7: Law of Universal Gravitation
5-8: Satellites; Weightlessness
5-9: Kepler's Laws

Problem 11
A
v=0.403rev/sv=0.403rev/s
Giancoli 6th Edition, Chapter 5, Problem 11 solution video poster
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VIDEO TRANSCRIPT

This astronaut is experiencing a centripetal force equal to seven point eight five times their weight, so: centripetal is seven point eight five times ‘m’ times ‘g’. The centripetal force also is ‘m’ times ‘v’ squared over ‘r’ and we can now solve for ‘v’. The ‘m’s cancel, multiply both sides by ‘r’ and we get ‘v’ equals the square rout of seven point eight five ‘r’ times ‘g’, and substituting in numbers, we have the square root of seven point eight five times nine point eight times twelve point zero metres, this makes thirty point four metres per second. So that’s one of our answers, then we have to convert that into revolutions per second, so that’s ‘v’ equals thirty point four metres per seconds times one revolution for every two pi ‘r’ metres, so one revolution the astronaut goes the circumference of a circle two pi ‘r’, so that’s thirty point four metres per second times one revolution for every two times pi times twelve. This works out to zero point four zero three revolutions per second.

COMMENTS
By kyle.joseph.landry on Tue, 2/2/2021 - 5:54 PM

Hello,

My intuition tells me that you would divide the distance (circumference) by the speed. If the circumference is 75.4m and I am travelling at 30.4m/s, my answer for time per revolution would be 2.5 seconds. Am I wrong here?

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