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Physics: Weight at Equator vs North Pole

Discussion in 'MCAT Study Question Q&A' started by MidwestMED1, May 16, 2014.

  1. MidwestMED1

    MidwestMED1 2+ Year Member

    Mar 13, 2013
    This was in Berkeley Review Ch2 Passage 1, but an overview of the question with explanation can be found here as well.

    Essentially the question posits if one weighs more at the equator vs north pole. The explanation is yes ie

    North Pole: there is no centripetal acceleration so N- mg = m(0) ------> N=mg

    Equator: there is centripetal acceleration so N - mg = -mv^2/r ----> N = mg - mv^2/r

    My confusion: If gravity is not a "downward" force, but rather a force towards the center of the earth, then what makes the north pole fundamentally different from the equator?

    Which of my understandings/assumptions is incorrect:
    The radius to the center of the earth is the same in either location.
    If the earth orbits the moon in uniform circular motion then wouldn't a centripetal force tangent to velocity always to be required to keep the orbit. I don't understand why there wouldn't be a centripetal force at the north pole if there is a uniform orbit (is there?)
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  3. Teleologist

    Teleologist 2+ Year Member

    Jul 7, 2013
    on your 6
    Where do you think gravity is acting from?

    Also, is the moon above or below the Earth? Is the Earth's force of gravity pulling up on the moon or tugging down on the moon?

    More or less the same, but that's irrelevant.

    I think you mean the moon orbits the Earth, Galileo.

    The Earth spins on its tilted axis.

    If you pushed a door right at the hinge to open the door, would there be a lever arm? Would you generate torque?
    Last edited: May 16, 2014
  4. DrknoSDN


    Feb 21, 2014
    Most importantly the centripetal force vector from gravity is going to be perpendicular to velocity pointing towards the center of the earth.
    Gravity is providing the acceleration required to "constantly turn"

    This problem is much better with something easier to visualize.
    Spinning a wet basketball on your finger for example. What happens to the water?

    At the poles nothing significant (ignoring gravity), but at the "equator" the water is flying off the ball because the molecular adhesive forces are being overcome by centripetal (centrifugal) force.
    Replace molecular adhesive forces with gravity and you have the same scenario. Gravity is clearly the force that causes things on earth's surface to stay put, but centripetal force diminishes that a little at the equator.

    In another thread it was showed that with an equatorial rotation speed of 1670 kilometers/hour (463 m/s), and a radius of 6,378,000 meters you have:
    Ac=v^2/r = (463^2)/6.378E6 = 0.034 m/s^2 opposite gravity. (so you have a slightly lower weight)... very slightly.

    This problem isn't dealing with satellites orbiting the earth. A satellite (moon** or man made) can have a polar orbit just as easily as an equatorial orbit because to orbit your centripetal acceleration is equal to the forces of gravity.
    This problem is saying that at the poles you have zero centripetal acceleration but at the equator you have a little.

    **Over eon's stellar tidal forces would push a polar moon orbit to become equatorial. (beyond scope of question) o_O
    Last edited: May 16, 2014
  5. MidwestMED1

    MidwestMED1 2+ Year Member

    Mar 13, 2013
    Thanks, the lever arm and basketball analogies really helped solidify things.
  6. Teleologist

    Teleologist 2+ Year Member

    Jul 7, 2013
    on your 6
    Right, the point is that with no moment arm, there would be no torque; net torque is after all equal to Force * length of moment arm. Same with the rotation of the Earth around its axis.
  7. neurodoc

    neurodoc Neurologist 10+ Year Member

    Sep 1, 2003
    SF Bay Area
    You would weigh more at the poles than at the equator for two reasons : 1) the reduction in the gravitational acceleration at the equator due to the earth's rotation; and 2) the fact that the earth is an oblate sphere (flattened at the poles) which means that the gravitational acceleration is less at the equator due to the increased distance from the center of the earth at the equator. The Cornell site (see OP) explains this well. Both effects are significant and work in the "same direction" to make objects at the poles heavier than those at the equator.
    syoung and DrknoSDN like this.

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