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Gravity's change rate and the Centrifugal force

  1. May 9, 2015 #1
    Hi everyone, this is my first thread i hope i do it right.​
    I will get right into it by saying, how can i prove that :
    w=w0*(1±4πƒv/g)
    Where "w" is the weight of an object moving along the ecuador, v is it's velocity and g is the gravitational acceleration.
    I have tried to calculate the centrifugal/gravity acceleration ration by deviding at=v2/r over g=Gm/r2, and i got that it equals 2πƒv/g instead of 4πƒv/g i have no idea where to go further than this other than put all this in a binomial equation and come up with an exponent that will give the (1±4πƒv/g) parameter.
    Thank you in advance.
     
  2. jcsd
  3. May 9, 2015 #2

    mfb

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    What is f and can you show your calculations?
    Centrifugal force is proportional to velocity squared, not linear velocity.
     
  4. May 9, 2015 #3
    Excuse my vocabulary i believe i meant tangetial acceleration
    here is the calculations i did, its so poor and lacks logic i just followed guts feelings
     

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  5. May 9, 2015 #4

    mfb

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    Acceleration relative to the ground does not go into that formula at all.

    Yes, it lacks logic, and there is no explanation either, which makes it hard to impossible to understand what you did.
     
  6. May 10, 2015 #5
    i thought g(r) = G m(r)/r2
    what i did is to show that, that 4πƒv/g is the tangential acceleration/gravitational acceleration ratio and the weight of any object moving along the ecuador with a velocity of v should change with respect of that formula, i think i should work more on this i can't find anything else on the internet and the exercise is in german.
     
    Last edited: May 10, 2015
  7. May 10, 2015 #6

    mfb

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    It is not.

    This is an English forum, but if you also make the original German exercise available it does not harm I guess (I am German).
    But more important: you should explain what you are doing, what the symbols mean and so on.
     
  8. May 10, 2015 #7
    enlighten me please ive been stuck here for 3 days
     

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  9. May 10, 2015 #8

    A.T.

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    That's a question about the Coriolis force, not about the centrifugal force.

    v is the velocity relative to the rotating frame of the Earth.
    w0 is the weight at the equator when v=0
    f is the rotation frequency of the earth
     
  10. May 10, 2015 #9

    mfb

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    @A.T.: The coriolis force is zero.

    But we are considering small velocities relative to the rotation of earth here.
    @S_Aghzafen: if earth would not be rotating, what would be the force for an object at rest? What would be the effective force for an object moving at speed v?
    If you compare an object at speed v0 from the rotation of earth and v0+v for small v, what do you get?
     
  11. May 10, 2015 #10

    A.T.

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    Not in the frame of the Earth. It's upwards when the ship goes east, and downwards when it goes west.
     
  12. May 10, 2015 #11
    Indeed it is not about the centrifugal force ive made a typing mistake
    What is the Coriolis force ?

    it looks like it changes the weight a little bit by a very small amount
     
  13. May 10, 2015 #12

    A.T.

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    Actually it depends on the reference frame you choose. In the Earth frame the effect is attributed to the Coriolis force, in other frames you can attribute it to the centrifugal force.
     
  14. May 10, 2015 #13

    mfb

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    Oh right, sorry, was thinking of north/south somehow. Yes, the approach via the Coriolis force is easier.
     
  15. May 10, 2015 #14
    In my opinion this question is about centrifugal force.
    First solve these questions:
    Find the weight of an object at the equator at rest with respect to the rotating earth.
    Express this in its mass, the earth rotation and g.
    Find the definition of w0.
     
  16. May 10, 2015 #15

    A.T.

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    w0 is the weight of the ship at the equator at rest with respect to the rotating earth.
    w is the as the weight of the ship moving along the equator.
    If we stick to the frame of the Earth, the difference is the Coriolis force.
     
  17. May 10, 2015 #16
    this is what i've done until now by approaching via the Coriolis effect (i had no idea that such fictious forces exist)
    since the Coriolis force will cause the weight of a moving object to change depending on its direction and latitude, i assume that :
    w0 - w' = Fc : where w0 is the weight of an object with v=0, w' the weight of a moving object along the ecuador and Fc is the Coriolis force where Fc=±2m*ω×ν since the moving object is along the ecuador the angle between v and ω must be 90° therefore ω×ν=ω*ν*sin(90°) where ω is also a cross product of v×r/r2 since the angle between these last mentioned therms is also 90° the final value of ω×ν=v2/r
    now we get :
    w0 - w' = ±2m*v2/r
    mg ± 2m*v2/r = w'
    mg*(1 ± (2v2)/(g*r)) = w'
    what comes next is just math where i consider v = 2πƒr and that's all
    i get
    w' = mg*(1 ± 4πƒv/g) : where mg = w0
    what do you think about this ?
     
  18. May 10, 2015 #17

    A.T.

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    ω is the angular velocity of the earth bound frame, while v is the velocity of the ship relative to that frame.
    Here again 2πƒ is the angular velocity of the earth bound frame, while v is the velocity of the ship relative to that frame.
     
  19. May 10, 2015 #18
    The OP did not mention a ship ;-).
    According to the formulas on http://en.wikipedia.org/wiki/Coriolis_effect#Rotating_sphere
    for v along the equator (v=v_e) and phi=0, there is only an upward force.
    Thus the general case reduces to a centrifugal force here.
    Lets keep it as simple as it needs to be.
     
  20. May 10, 2015 #19

    A.T.

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    The original text does, though I don't know why it matters to you what the object is.

    Yes, for eastern travel, as I already said.

    No it doesn't. This the Coriolis term. The centrifugal term is separate from this, and is the same for the moving and resting ship in the Earth frame. So it cannot account the changed weight in that frame.
     
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