Please check my calculation. Very confused. Gravitation.?

2016-12-27 12:42 am
My book said: As the Earth is rotating about its axis, objects on the earth's surface are performing circular motion. Therefore, objects on the earth are accelerating and they need a centripetal force. The weight contributes a small portion to the centripetal force needed. Therefore, the measured g is smaller. The decrease in g is largest at the equator. At higher latitude, the radius of the circle is smaller while w is the same. Therefore, the centripetal force needed is smaller. The measured g is larger at higher latitude.

However from my calculation,
let the latitude be θ,
W=weight,
N=normal force,
R=radius of the earth

Wcosθ - Ncosθ = mw^2 Rcosθ

W-N=mw^2(R)..........(1)

Wsinθ=Nsinθ

W=N......(2)

Which is totally the same as considering an object on the equator.
Why am I wrong, please help

回答 (1)

2016-12-27 11:39 am
✔ 最佳答案
Your first equation is already wrong. Be aware that W and N are not exactly in opposite directions.

The earth attractive force W on an object (of mass m, say) is directed towards the centre of the earth (i.e. along the radius of the earth R). This is not the "weight" that we measured, say using a spring balance, on earth surface because part of W is shared by providing the centripetal force (Fc) on the rotating earth.

The centripetal force (Fc) is in a direction parallel to the equatorial plane of the earth, whereas the earth attractive force on an object (W) is directed towards the equator. Thus Fc and W are not in the same direction. The VECTOR DIFFERENCE of these two forces will give you the magnitude and direction of the "weight" (mg) that we experienced on earth surface. The "normal reaction (N)" is just in opposite direction to the "experienced weight (mg)". Hence, W and N are not exactly in opposite directions.

You could refer to the following diagrams for the directions of the involved forces:

https://qph.ec.quoracdn.net/main-qimg-010eb95027a9e7c0863ed103dbf5be7a?convert_to_webp=true

http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap4-5.gif

The second diagram shows that the "normal reaction" is in opposite direction to the "experienced weight", which, strictly speaking, is not perpendicular to the earth surface, i.e. not "normal" at all.


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