[Physics] Solidifying understanding of centrifugal force at the equator vs poles

centrifugal forcecentripetal-forcenewtonian-gravitynewtonian-mechanicsreference frames

I'm just trying to get my head around the physics here and understand exactly what's going on, because there's a lot of conflicting information on the internet.

Viewed from an inertial reference frame in space i.e. looking at the earth, if a person is standing at a pole there are only two forces acting on them:

The normal force, and gravity.

So $mg – N_{p} = 0$

However, at the equator, the person is accelerating, and so:

$mg – N_{eq} = \frac{mv^2}{r}$

Which means that the normal force (the weight the person will measure at the equator) is reduced:

$N_{eq} = mg (1 – \frac{v^2}{rg})$

So the reduction in weight is not due to the centrifugal force.

However, if viewed from an observer in the rotating frame i.e. on the equator, they experience only the normal force $N$ and gravity $mg$.

So in theory, $N_{eq} – mg = 0$

However, this person knows (although they can't detect it) that they're rotating, so to reconcile that they add a fictitious force opposite to the direction of the centripetal force but equal in magnitude to it, called the centrifugal force:

$N_{eq} – mg + \frac{mv^2}{r} = 0$

which gives

$N_{eq} = mg (1 – \frac{v^2}{rg})$

Is my thinking correct here?

EDIT:

A further question I have is, why is there no coriolis force in this case? The reference frame is rotating, after all.

Best Answer

There are three misconceptions that I can see in your reasoning.

  1. The poles are the only places on Earth where you are not accelerating due to the Earth's rotation, so you have that backwards.

  2. You seem to think that the normal force has to be the same at the poles and the equator, which isn't true. Finding the normal force at the poles does not give you the normal force at the equator.

  3. The forces involved are vectors, not scalars, and they are not collinear (except for the special case of the equator). The gravity and normal force are approximately collinear with the Earth's radius everywhere, but the centripetal (or centrifugal) force is not; it points towards (away from) the axis of rotation. So you need to do some vector math/trigonometry to get the actual values.

You seem to be struggling with the distinction between the centripetal force and the centrifugal force. It seems like you have the right idea, but its hard to tell due to the other issues. Let me try to explain what those are.

The centrifugal force is a "fictitious force" (meaning there is no object that exerts this force) that appears to arise in a rotating coordinate system; considering the centrifugal force in a rotating coordinate system maintains the usefulness of Newton's 2nd law.

The centripetal force is the required amount & direction of force that the net force on an object must satisfy in order for the object to move in a circle of a certain radius at a certain speed.

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