Timeline for Distribution of gravitational force on a non-rotating oblate spheroid

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Nov 14, 2014 at 15:09	comment	added	fibonatic		The reason why celestial bodies are approximately oblate spheroids originates from their rotational velocity, which in their rotating reference frame induces a centripetal force. When you look at the effective surface acceleration you could also include this centripetal force, which would make it even more clear that you will have lower gravity at the equator.
Nov 5, 2014 at 9:32	comment	added	John Rennie		@imakesmalltalk: you could probably find a derivation somewhere in Googlespace. In principle it's simple as you just split the spheroid up into suitable elements and integrate. In practice you'd be surprised how often basically simple functions turn out to have horribly complicated integrals. That's what happens here.
Nov 5, 2014 at 8:52	comment	added	user49111		+1 Thanks for your answer John. I've heard this many times, that $F_G$ is stronger at the poles. Its no brainer that as $r$ is reduced, $F$ increases (as $F_G \propto \frac{1}{r^2}$). But what I wanted to know is that how exactly does gravity remains stronger at the pole? My intuition too says that its obvious. But how can it be shown?
Nov 5, 2014 at 8:17	history	answered	John Rennie	CC BY-SA 3.0