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The title above was a question on an exam that was marked wrong for me. I answered that if the Earth rotated slower (i.e. longer days), my apparent weight would increase. I based this on the observations that weights of objects of equal masses increase the higher in latitude you go. At the poles, the apparent weight of things is slightly higher than at the equator. I tried to argue my case with analogies, but it didn't work and the counter-argument was that $F_C = m * r * ω^2$ and that if the tangential velocity decreased, the centripetal force would also decrease. Since the centripetal force pulls inward, this decrease in centripetal force would lessen our apparent weight.

Who's correct here? If I am, how can I prove that this argument above is flawed and that an increase in the Earth's rotational speed will decrease an object's weight? I also want to circumvent the possible argument that "well, centrifugal force is a fictitious force, so it doesn't count."

Qmechanic
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a3dur4n
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    The rotation of the Earth gives rise to two relevant effects. One is the centripetal acceleration part that you are focused upon. But this is vastly washed out by the equatorial bulging, causing the equatorial radius to be a bit bigger than the polar radius. – naturallyInconsistent May 19 '23 at 00:38
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    In the exam question, does the Earth spin slower without changing its shape? If the Earth's spin were slower, then the flattening would be smaller. – PM 2Ring May 19 '23 at 00:41
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    FWIW, we have some great detailed answers regarding the equatorial bulge, eg https://physics.stackexchange.com/q/8074/123208 & https://physics.stackexchange.com/q/69562/123208 – PM 2Ring May 19 '23 at 00:50

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You would absolutely weight more.

The centripetal force isn't a "fake force" it's a label for a force or part of a force. On Earth, part of mg is what makes up the centripetal force. Saying centripetal force is another force pulling you towards the center is a classic case of having your cake and eating it too.

Your apparent weight on a scale is $mg - \frac{mv^2}{r}$. Not much more to it than that.

Señor O
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The counter argument seems to be based on a flawed understanding of the centripetal force. Let's call $W$ your weight and $N$ the normal force that the scale applies on you. The net force $W$ - $N$ is the centripetal force on you that allows you to rotate along with the Earth. If the tangential speed goes down, the centripetal force goes down, so $N$ must go up, i.e. the weight you read on the scale is higher.

If you are not on the equator, there is a minor complication that the centripetal force doesn't point exactly downward but toward the spin axis. This isn't an essential consideration for the argument above, but technically the relevant (downward) component of the centripetal force is $m\omega^2r\cos\theta$, where $\theta$ is your latitude and $r=R\cos\theta$, $R$ being the Earth's radius.

"The centrifugal force is fictitious so it doesn't count" is flawed in its own right, but maybe that's a discussion for another time.

Puk
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It's easiest to think about this problem in an inertial reference frame. If you're standing still on the surface of the earth, you're undergoing uniform circular motion. The centripetal force, which keeps you in uniform circular motion, is the force of gravity minus the normal force. If the earth slows down, the centripetal force does indeed decrease. However, since gravity is presumably constant, this means that the normal force must increase. The normal force is your apparent weight, so your apparent weight increases. You will feel the effects of gravity more strongly because, in your noninertial reference frame, there's less of a centrifugal force counteracting it.

Technically Natural
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