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Sep 20, 2024 at 22:36 vote accept KDP
Mar 16, 2024 at 7:05 comment added KDP "At first I was surprised by the fact that such a complicated external potential gives rise to simple linear relations for the surface force components." I was too until I realised that the oblate is the equilibrium shape when centrifugal force matches gravitational force and the centrifugal has a linear relationship with r ($F_{centrifugal} = m \omega^2 r$) so the surface gravity must too.
Mar 15, 2024 at 22:11 history edited Ghoster CC BY-SA 4.0
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Mar 15, 2024 at 21:55 comment added Ghoster $\vec F= -\vec\nabla U$. The gradient of a potential function is always orthogonal to the equipotential surfaces.
Mar 15, 2024 at 21:52 comment added KDP Nice idea to find a simpler solution for the surface gravity by using the interior potential. Wish I had thought of that! So the force is orthogonal to the equi-potential surface ($(a'e')^2 = (3/5) (ae)^2$) (Eq26) and not to the equi-force surface ($(a'e')^2 = (9/10) (ae)^2$) (Eq27)?
Mar 15, 2024 at 21:21 history edited Ghoster CC BY-SA 4.0
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Mar 15, 2024 at 21:11 history edited Ghoster CC BY-SA 4.0
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Mar 15, 2024 at 8:15 comment added KDP Nice work @Ghoster +1 I checked your simplified equations and they are spot on. I think together, we are the first to come up with an exact solution for the surface gravity of a homogenous density oblate spheroid of arbitrary ellipticity.
Mar 15, 2024 at 7:08 history edited Ghoster CC BY-SA 4.0
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Mar 15, 2024 at 5:46 history edited Ghoster CC BY-SA 4.0
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Mar 15, 2024 at 5:35 history answered Ghoster CC BY-SA 4.0