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-1

It is not only about up and down, there are deeper dimensions to this. Like infront and behind. The Moon seems to be "behind" us, as in expressions like: "Going BACK to the Moon". Although it does go all around and never gets anywhere, much like ones own bottom, actually. Even Ptolemy didn't argue with that fact. Might this be a clue to this geometric ...

3

Most of the 60 moons in the Saturn system are far away from the rings and very small, so their effect on the rings is negligible. But larger ones that are closer in (Enceladus) do have a rather significant effect on the rings, but as the gravitational pull of these moons is radially outward, it is hardly visible. On the other hand, small moons inside the ...

0

I think that one thing confusing you, is that up/down is a 2 dimensions concept (just like your image by the way), but the Universe is in 3 physical dimensions (let's not talk about time here). So when you say « going up from Earth is extracting from it's gravity », it can be in many direction. If two persons are in North Pole and South Pole of Earth, and ...

0

Your confusion is that you are treating gravity from all those objects as a bunch of discrete 'pulls' - but it doesn't manifest itself like that. Gravity from all bodies in the universe effects you as one force (I'm simplifying and excluding getting close to black holes etc where you have dramatically changing gravitational potential over the length of your ...

2

I think what you have established here is just that $\rho$ tends to increase with mass. The density of planets isn't constant. Let $\rho = \rho_0 (M/M_{earth})^{\alpha}$, so that $M = (4/3)\pi R^{3} \rho_0 (M/M_{earth})^{\alpha}$ Then $$g = \frac{GM}{R^2} = \frac{4\pi G}{3} R \rho$$ Replace $R$ with $(3M/4\pi \rho)^{1/3}$ so that $$g = \frac{4\pi G}{3} ... 1 For planets of constant mean density you have:$$ M=\rho \times 4\pi r^3 $$and the surface value of g is:$$ g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r  So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ ...

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