DIffering Weight at equator and poles

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A person weighs less at the equator than at the poles primarily due to both the Earth's spin and its shape. The spin of the Earth creates centripetal acceleration, which is greater at the equator, resulting in a weaker normal force. If the Earth were to spin significantly faster, the centripetal force could match gravitational force, leading to a state of effective weightlessness at the equator. At the poles, centripetal acceleration is nonexistent, making weight consistent with gravitational force. Understanding these dynamics clarifies why weight varies based on location on Earth.
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Homework Statement



23) A person weighs less at the equator than at the poles. The main reason for this has to do with
A) the spin of the Earth.
B) the shape of the Earth.
C) both the spin of the Earth and the shape of the Earth.
D) the influence of the sun, moon, and all the planets.
E) the law of action and reaction.
Answer: C



I understand how shape affects it, but I am puzzled about the effect of the spin.
 
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If the Earth was spinning really, really fast, would you have a better chance of being launched into space standing on the equator or on one of the poles?
 
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lewando said:
If the Earth was spinning really, really fast, would you have a better chance of being launched into space standing on the equator or on one of the poles?

Intuitively, I wish to say the equator, but I'm not certain why.

Is centripital acceleration greater at the equator, resulting in a weaker normal force? (As far as my question goes)
 
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Your intuition is working well. Centripetal acceleration is greater at the equator than the pole(s). It is nonexistent at the poles. It is a lot less than gravitational acceleration but still significant.

One way to think about it is this: in order for you to stay attached to the Earth at the equator (assume for the moment there is no gravity), a centripetal force needs to act upon you to keep you in place. Otherwise you would travel along the line of your tangential velocity and become quickly "unattached". Turning gravity back on, gravitational force more than provides the required centripetal force to keep you in place. If we increase the Earth's rotation enough so that the centripetal force required to keep you on the surface exactly matches the force due to gravity, you become effectively weightless, just barely earthbound. Stop the Earth from spinning and your weight at the equator will be no different from the poles (ignoring shape effects).
 
Thank you :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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