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Help with circular motion of zero gravity

  • Thread starter koopa347
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  • #1
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Homework Statement


How long would the day have to be in order that for a person standing at the equator of the earth would see a bathroom scale that she is standing on read zero?
This is all the info that is given.

Homework Equations


Fc=m(v^2)/r


The Attempt at a Solution


I setup my force equations so the sum of f=Fc=Fn-Fg. I then found out the radius of the earth, and this is where i got stuck , i know that the radius of the earth is 6377.34km, but how do i figure out v and m; if v has to be greater than earths current centripetal force?
 

Answers and Replies

  • #2
cepheid
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Welcome to PF

You're spinning so fast that gravity is just barely providing enough force to keep you in your circular "orbit." In other words, the centripetal force required for this orbit is equal to your weight (which is why you feel no normal force, which as you've correctly stated, is the difference between the two).

If the centripetal force is equal to the weight, and both are proportional to m, then m cancels from both sides of the equation (the mass of the object in the orbit doesn't matter). Hence you're finding the speed at which the centripetal acceleration equals g. You know r. You know g. You just have to solve for v.
 
  • #3
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What do you mean by both are proportional to m? Could you type out an equation that shows this step? Do you just mean
(m)(9.8)=mv^2/r
so those m's cancel, so we know now that we're left with 9.8=v^2/r?
 
  • #4
cepheid
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Science Advisor
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What do you mean by both are proportional to m? Could you type out an equation that shows this step? Do you just mean
(m)(9.8)=mv^2/r
so those m's cancel, so we know now that we're left with 9.8=v^2/r?
Yes, that's exactly what I meant. To be proportional to "x" means to be equal to "x" multiplied by some factor.
 

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