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Okay,I am having a little problem in understanding the effect of latitude/rotation of Earth on g.

In this figure, the test mass,m at place P experiences Centrifugal Force, Fcf acting outwards.

enter image description here

As we can see, the force,Fc=mrω2 has 2 rectangular components:-

a. Fcf = mrω2cosλ: Opposite to Weight, W=mg
b. Fcf = mrω2sinλ: Normal to Weight, W=mg

Now, my First doubt:-

Individually, what is the role of both the components, especially the second one, Fcf = mrω2sinλ?

Second doubt(Important One):-

In Observer Frame of Reference, the figure and derivation is completely true because Centrifugal Force(Pseudo Force) only exists in observer's point of view.

Now for an observer outside Earth, this will not be the case because for that person, Centrifugal Force is not existing but the particle at point P will experience Centripetal Force.

So how will we conclude the same/similar results(as given in the image) without introducing Centrifugal Force(which is a Fictitious Force) but to consider Centripetal Force(which is a Real One) in this scenario? Please give the figure for this case.

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  • $\begingroup$ Do you still have a problem with the concept if the test mass is at the equator? $\endgroup$
    – Floris
    Commented Jul 13, 2017 at 19:23

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Individually, what is the role of both the components, especially the second one, Fcf = mrω2sinλ

It means your book is doing something fishy. Your book treats the Earth as a sphere, with acceleration due to gravity pointing directly toward the center of the Earth. The outward normal force would be radially away from the center of the Earth. The component normal to the surface of the Earth means the oceans would flow toward the equator.

The solution to this is that the Earth is an oblate spheroid rather than a sphere. Except at the equator and the poles, gravitational acceleration at a point on the surface of the Earth does not point directly downward. Instead, it has a component normal to the surface that cancels the component of the fictitious outward centrifugal normal to the surface of the Earth.


So how will we conclude the same/similar results (as given in the image) without introducing centrifugal force(which is a fictitious force) but to consider centripetal force (which is a real one) in this scenario?

There are two real forces acting on an object at rest on the surface of the rotating Earth, the not-quite downward gravitational force and the upward normal force. These don't quite cancel one another. The net force points directly toward the Earth's rotational axis -- i.e., exactly opposite to the fictitious centrifugal force. This non-zero net force explains the fact that the object is undergoing uniform circular motion, making one revolution per day.

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  • $\begingroup$ Sorry David, I am unable to understand your answer. It would become better if you give a good figure so I can clear up my doubt. $\endgroup$
    – lakhi
    Commented Jul 14, 2017 at 13:27
  • $\begingroup$ @lakhi -- I don't do figures. You'll have to ask someone else. One way to think about this: The Earth has an equatorial bulge. Without this bulge (i.e., if the Earth was a perfect sphere), the Earth's oceans would form a thin but very deep band centered about the equator, and the poles would be essentially airless. The Earth is instead an oblate spheroid. $\endgroup$
    – David Hammen
    Commented Jul 15, 2017 at 21:21

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