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I know of an explanation that the Earth is a inertial frame of reference since it is rotating about its own axis, and since this is happening there is a centrifugal "force" or effect which counters the gravitational one. I know this is a fictitious force and I don't really find myself truly understanding anything if I have to use fictitious forces so I'm not satisfied with this explanation.

Also, I know of an explanation that observes this from a non-inertial frame of reference, saying that since gravity is the centripetal force in effect a part of it has to go on rotating a mass on a certain radius form its axis, therefore the "weight" part of the gravity slightly decreases. How is this happening?

If we assume a constant gravitational force on a constant mass on the surface of the Earth, then on the poles it is evident that the weight of the mass would be a consequence of purely the gravitational pull. However, on the equator the gravitational force itself is still the same, how is then the net acceleration decreased thus also the weight?

Qmechanic
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bonehead
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1 Answers1

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Well, if you consider Earth as a perfect sphere and assuming that at sea level, gravity acceleration will be the same for all points in this perfect sphere, then weight will not change when you go to poles or equator. And that makes perfect sense, since the spherical Earth will be symmetric. However, Earth is NOT a perfect sphere, then gravity won't be equal on different points. A quick mathematical explanation:

On Poles, the Centripetal Force is null, then Weight is maximum on poles.

$W=F_g$

$g_p=GM/R^2$

On Equator, the centripetal force will be maximum, then Weight will be minimum:

$W=F_g - F_c$

$g_e=GM/R^2 - \omega R^2$

$g_e=g_p - \omega R^2$

By the way, earth's format is considered as an oblate spheroid.

Lp_cam
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