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Circular motion/Gravitation

  1. Aug 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Earth is a spherical object, which completes a full rotation every 24 hours. How long should the day on Earth be so that an object at the equator is able to float freely above the ground?

    2. Relevant equations
    v=2πr/T
    Fr=mv^2/r
    v=velocity
    r= radius
    T = time for comleting 1 revolution in seconds
    Fr= centripetal force/force in a radial direction

    3. The attempt at a solution
    1 revolution in 24 hours so T = 24x3600sec = 86400

    Fr=mv^2/r
    Fr at the earth is equal to mg
    mg=mv^2/r
    mass cancels out
    g=v^2/r Eq. 1

    v=2πr/T
    vT/2π = r Eq. 2

    Sub Eq. 2 into 1

    g=v^2/(vT/2π)
    re arrange the equation to
    v=gT/2π
    v= 9.8*86400/2π
    v=134 760m/s

    Thats where I'm at so far....I'm just not sure if that's correct...and if so where to go from there.
     
  2. jcsd
  3. Aug 12, 2016 #2

    Doc Al

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    This is where you're messing up. That value for T is what the period is now. You want to solve for the period when the earth is sped up so that the object floats.

    So, do just what you did except now solve for T.
     
  4. Aug 12, 2016 #3
    So for the object to float

    Fr should equal 0 correct?
     
  5. Aug 12, 2016 #4

    Doc Al

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    No. Your thinking before was correct: Fr = mg.

    This is good. Keep going. (Just don't assume a value for T.)
     
  6. Aug 12, 2016 #5
    Ahh I'm confused.

    So if I take v=gT/2π and solve for T i get T=2πr/v Eq 2....what do I do with that? I don't have any of the values to help me solve for T.

    You said g=v^2/r Eq. 1 was correct....My assumption would have been that Fr would have to equal 0 in order for the object to float freely above the ground
     
  7. Aug 12, 2016 #6

    Doc Al

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    No. You want the centripetal acceleration to equal the acceleration due to gravity. (Or, in other words, the centripetal force to equal the weight.) Which is what you did to get that equation.

    Substitute that expression for v in your first equation. Then you can solve for T. (You'll need the radius of the earth. Look it up!)
     
  8. Aug 12, 2016 #7

    Okay....

    g=v^2/r
    r = 6371 km = 6371000m

    v=sqrt g*r
    v= 7901m/s

    v=2πr/T

    T=2πr/v
    T=2π*6371000/7901
    T=5066 seconds
    T= 84.4 mins

    So a day on earth should be 84.4 mins so that an object can float freely above the equator.

    One thing I'm still confused about:
    Fr=mv^2/r
    Fr=mg - I came to that because as an object on earth the only thing creating a centripetal force would be gravity therefore Fr = mg. For the object to float I still thing Fr should equal 0 so that there is no pull on the object towards to earth....What am I missing here?
     
  9. Aug 12, 2016 #8

    haruspex

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    You are mixing up two frames of reference.
    In an inertial frame, the object is accelerating towards the centre of the Earth. This is centripetal acceleration, v2/r, and requires a net force to produce it. If it is floating (orbiting) then the only force is mg, mg=mv2/r.
    In the frame of reference of the object, there is no acceleration, so the net force is zero. But now there is the "fictitious" force, centrifugal, to balance the force of gravity: mg=mv2/r.
     
  10. Aug 12, 2016 #9

    Doc Al

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    What you're missing is that you must have a non-zero centripetal force in order for the object to move in a circle. For it to "float" the speed must be just right so that the required centripetal force equals the weight. If the speed is too slow (like when the period is 24 hours!) the object doesn't float--it sits on the ground. And if it's too fast, it goes flying off. Make sense?
     
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