# Why rotating reference frames are not inertial? [duplicate]

Let's say I'm standing on the equator, and that there is no other reference point in the sky.

If the planet is rotating, then I measure my weight to be lower than if it is not.

But given that I have no reference frames to "prove" than I am rotating, why can't I just claim that my weight should be the same since I'm standing still and the universe is rotating around me?

(After all, this kind of argument does work for translation... so what's different about rotation?)

• "But given that I have no reference frames to "prove" than I am rotating,..." yes, you do: see Foucault pendulum – AccidentalFourierTransform Apr 23 '16 at 10:22
• To tell that you are rotating, you dont have to have a reference point, a Foucault pendulum could do the job – Courage Apr 23 '16 at 10:26
• I think that unfortunately the Foucault pendulum does not "work", ie show a rotation, at the Equator? – Farcher Apr 23 '16 at 10:30
• Yes, I was about to say that. A Foucault pendulum shows the biggest effect at the poles and zero effect at the equator. Instead I suggest you throw an object north and observe the Coriolis force. – John Rennie Apr 23 '16 at 10:30
• Not only that, but it completely misses the point of my question, but unfortunately for me I can't find the words to explain it better. However, I can say that if a Foucault pendulum is your answer, then I think you haven't understood the crux of my confusion... in fact I was already familiar with the Foucault pendulum before I asked this, and it did not answer my question. I'm confused why the world behaves the way it does, not asking how it behaves. – user541686 Apr 23 '16 at 10:34