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?


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


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.

  • $\begingroup$ With regards to 1), I might be being particularly dense, but where else on earth would you not accelerate? 2) I do not think that, I did not say that. The whole point of the question is that the normal force is different at the poles and at the equator. 3) The force due to gravity is radial, so its vector nature doesn't matter in this case. The normal force opposes that radial force, so I fail to see why the vector nature is important here? $\endgroup$ – user1654183 Feb 15 '16 at 0:20
  • $\begingroup$ 1) Imagine the Earth rotating. When you are standing at the pole, what path are you taking through space relative to your observer? Hint: You are not moving at all 2) Ok, I guess I misunderstood 3) The force due to gravity and the normal force are radial, but the centripetal acceleration is not radial, at least not spherically radial - it is cylindrically radial (i.e. it points towards the axis of rotation, not the center of the Earth, like the others). Thus, the vector nature of the forces is relevant and important. $\endgroup$ – Brionius Feb 15 '16 at 0:29
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    $\begingroup$ Here is an illustration of the directions of these forces that I found by googling. $\endgroup$ – Brionius Feb 15 '16 at 0:32
  • $\begingroup$ 1) That's exactly what I said. At the poles, the person is not rotating. It seems you have misunderstood nearly everything I wrote down, and I'm trying to tell whether it's me that had the problem here or you. 3) At the equator, the centripetal acceleration is radial. Hence again, the vector nature of the force does not matter. $\endgroup$ – user1654183 Feb 15 '16 at 0:35
  • $\begingroup$ I'm quoting from your question here - However, at the poles, the person is accelerating. That simply isn't true. I agree with you (as I wrote in the answer) that the special case of the equator is a valid place to treat the forces as scalars. $\endgroup$ – Brionius Feb 15 '16 at 0:45

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