For a person standing at the equator, if he sees an object in free fall, he will see that the object accelerates downward at the rate $$a = g - \omega^2R$$ where $R$ is the radius of the planet and $\omega$ is the spin angular velocity of the planet. I can't really provide a diagram right now but lets assume that the north-south axis is in the vertical direction. At latitude $\phi$ then an object would go about a circle of radius $R\cos\phi$ with angular velocity $\omega$ and so the "horizontal" (horizontal to the north-south axis, not horizontal to a person on the surface at latitude $\phi$) component of acceleration is $$a_{\text{hor}} = g\cos\phi - \omega^2R\cos\phi$$ while the vertical component remains $$a_{\text{ver}} = -g\sin\phi$$ and so the net force seems to be off centre. So I guess my question is I know that weight seems to increase as we move towards the poles but does the force also slightly go off center when we move away from the equator or did I make some mistake in my analysis?
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asked Jun 25, 2020 at 10:10
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2$\begingroup$ Related: physics.stackexchange.com/q/429441/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jun 25, 2020 at 11:35
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Look at this image of the earth.
(image from Gravity for Geodesy I: Foundations)
- The gravitational force (red arrow) points to the center of the earth.
- The centrifugal force (yellow arrow) points away from the axis of rotation.
- The total force (green arrow) is the vector sum of the two above. In general (except at the poles or at the equator) it does not point to the center of the earth, as you already pointed out in your question. The earth will deform in such a way that the surface is perpendicular to the total force, hence resulting in the ellipsoid form.
answered Jun 25, 2020 at 11:24