# Homework Help: DIffering Weight at equator and poles

1. Jun 10, 2013

### gake

1. The problem statement, all variables and given/known data

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.

I understand how shape affects it, but I am puzzled about the effect of the spin.

2. Jun 10, 2013

### lewando

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?

3. Jun 10, 2013

### gake

gggg3

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)

Last edited: Jun 10, 2013
4. Jun 10, 2013

### lewando

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).

5. Jun 11, 2013

### haruspex

6. Jun 11, 2013

Thank you :)