Centrifugal Force Affecting Weight?

[dmission]

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Got a question physics review question that makes no sense to me, would love some help understanding it. The question says:

• How does the centrifugal force and the distance from the center affect the weight of an object along the equator (which is farther from the earth's center)?

• The correct answer is that "the increased distance and the centrifugal force both act to decrease the weight of the object"

Now, from what I remember, centrifugal force is really the lack of a centripetal force. Now it makes sense to me how the weight would be less due to the increased distance from the center (which is the case along the equator, apparently); this is just F=GMM/R^2. However, I've got 0 idea as to why centrifugal force would also reduce the weight. It seems to me like it'd play no major role here, since there was no rotation around the earth.

Thanks!

 

[MT Headed]

Sometimes with MCAT concept questions it pays to be silly and/or extreme. Let's say the earth started spinning really fast. People on the north pole might get a little dizzy. People on the equator would get flung into outer space because they have negative weight. People at latitude 30 would step on the bathroom scale and be pleased with their sudden weight loss.

Yeah there's equations with Wa and Wg and stuff, but the MCAT is a timed test and I'll use every tool at my disposal to get through it quickly.

 

[dmission]

Thanks for the reply. How exactly does the centrifugal force lower the weight, though?

 

[Labminion]

According to wikipedia:

"A reactive centrifugal force is the reaction force to a centripetal force. A mass undergoing curved motion, such as circular motion, constantly accelerates toward the axis of rotation. This centripetal acceleration is provided by a centripetal force, which is exerted on the mass by some other object. In accordance with Newton's Third Law of Motion, the mass exerts an equal and opposite force on the object. This is the reactive centrifugal force. It is directed away from the center of rotation, and is exerted by the rotating mass on the object that originates the centripetal acceleration."

So, a centrifugal force is a reaction force to the centripetal force.

The person standing at the equator is experiencing a centripetal force, because the earth rotates from west to east. So, the net force on the person at the equator is:

Fc = Fgravity-Nforce(exerted on person by earth) = ma= m(v^2/r)
Fc = mg-N=m(v^2/r)

The person's apparent weight is the normal force. This is what a scale would read if a person at the equator were standing on it. So, the normal force (apparent weight) at the equator is less than it would be at the poles, because N = mg-m(v^2/r). You appear to weigh less due to the centripetal force, as indicated by subtracted the "m(v^2/r)" term from Fgravity.

Does that make sense?