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Grade 12 Physics  Circular Motion  Universal Gravitationby synchrostarr
Tags: circular, circular motion, grade, grade 12, grade 12 math, grade 12 physics, motion, physics 
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#1
Oct2013, 10:00 AM

P: 1

1. The problem statement, all variables and given/known data
This question is from the Nelson Grade 12 Physics textbook. The force of attraction between masses m_{1} and m_{2} is 26N in magnitude. What will the magnitude of the force become if m_{2} is tripled, and the distance between m_{2} and m_{1} is halved? 2. Relevant equations F_{G}=(Gm_{1}m_{2})/r^{2} G=6.67x10^{11} 3. The attempt at a solution I tried to look at both equations, the original and the altered equation so I had: 26N=(Gm_{1}m_{2})/r^{2} and then for the changed equation I had" F_{G}=(Gm_{1}3m_{2})/(1/2)(r^{2}) But I was unable to figure out what to do from there. If it is at all helpful, the answers in the back of my textbook said that the correct answer is 3.1x10^{2}N. 


#2
Oct2013, 10:37 AM

Mentor
P: 11,625

You're on the right track. Note that if you halve the distance then the "1/2" applies to r, not to r^{2}. After you insert the changes into the formula, see if you can't factor them out so you end up with something like: F_{new} = C x (Gm_{1}m_{2})/r^{2} 


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