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Okay, that makes sense.

You said "the short answer, less gravity" and "This answer assumes that the density is the same throughout the planet depth." So it looked like you were saying "less gravity assumes that the density is the same throughout the planet depth." The boundary between increasing gravity and decreasing gravity (i.e acceleration as a function of r) is constant gravity: [math]\displaystyle{a(r) = \frac{Gm(r)}{r^2} = g}[/math] [math]Gm = g r^2[/math] [math]G dm = 2 g r dr[/math] [math]\displaystyle{\rho(r) = \frac{dm}{dV} = \frac{dm}{ 4 \pi r^2 dr}}[/math] [math]\displaystyle{\rho(r) = \frac{dm/dr }{ 4\pi r^2 } = \frac{2gr}{4\pi r^2 G}}[/math] So [math]\displaystyle{\rho(r) = \frac{k}{r}}[/math], where [math]\displaystyle{k = \frac{g }{ 2 \pi G}}[/math]. In other words, the acceleration will be g all the way down as long as the density is inversely proportional to the distance from the center. It will decrease with r (ie. increase with depth) if the density depends on r more sensitively than 1/r, and will increase with r (decrease with depth) if it depends less.

Genady is correct the density and distribution of the mass makes a difference. There is nothing wrong with your maths, but note you initial conditions. However there is more to this than meets the eye since gravity, whether considered as force or as an acceleration is a vector, not a scalar. The distribition and density of the mass affects the direction of gravity. So the geometric centre and the centr of gravity no longer conincide. It is interesting to note the the discovery of the massiveness of the Himalaya was discovered by Everest from deductions on observations variations of alignment of plumb bobs on the plains beneath. Yet another effect has not been mentioned and is oft forgot. The reason why gravity is greater at the poles than the equator , due to centrigugal effects of our spinning planet. Once again this also has an effect on direction.

This is right. So, why I thought the clarification is needed? Just consider the extreme case where the whole mass is near the planet center and the rest of the planet is massless. As the body goes from the surface toward the center it just gets closer to the same mass. The gravitational force increases. In the other extreme case where the whole mass is in the surface shell and the planet is empty inside, the body will become weightless as soon as it crosses the surface. Ergo, the answer depends on the density distribution with the depth. Also, take a case where the entire mass is in the shell halfway down. As the body gets closer to that depth the gravity increases, and when it crosses that shell the gravity drops to 0. Ergo, the answer depends on the depth.

No, you had it right the first time. The derivation of the shell theorem that exchemist just mentioned applies to each (spherically symmetric) shell individually, regardless of its density. For constant density, the gravitational field is linear in the radius: [math]\displaystyle{a(r) = \frac{G m(r)}{r^2}}[/math] [math]\displaystyle{a(r) = \frac{G \rho (\frac{4}{3} \pi r^3)}{r^2}}[/math] [math]a(r) = G \rho (\frac{4}{3} \pi r)[/math] [math]a(r) = G \rho (\frac{4}{3} \pi R) \displaystyle{ \left(\frac{r}{R}\right)}[/math] [math]\displaystyle{a(r) = \frac{GM}{R^2}\left(\frac{r}{R}\right)}[/math] [math]\displaystyle{a(r) = g\left(\frac{r}{R}\right)}[/math]

Look up Newton’s shell theorem. That explains why the mass at a radius greater than the location of the red dot has no gravitational effect on it.

An active ball joint mechanism (ABENICS) enhanced by interactions of spherical gears: https://ieeexplore.ieee.org/ielx7/8860/9556521/9415699/supp1-3070124.mp4?arnumber=9415699

I like it, which means I'm tempted by its seductive promise of dispelling illusions. And I'm open to such ontological flights, though I sometimes fly into the whirling blades of what, then, is a field, be it a field of force (a vector field) or a field of potential energy (a scalar field). If everything that "is" is just a field perturbation, or a gradient, or a vector, what is being perturbed, waved, pointed, attenuated, quantized and so on? What the heck are fields anyway? (Rhetorical question, don't answer) I'm saying, mainly, that any metaphysics is possibly hopeless, and that maybe physics tends to attach a particle term to any quantized jiggle in the jello of reality because anything else feels like chaos and madness. Even David Bohm couldn't get rid of solid particles. Still, I'd love to look into any speculative take on a world of just waves.

You got that right: it's math, not physics. So why bother with it? It's just math you made up, and has no applicability to physics. You choose the method that applies to the problem. It comes from understanding physics and the experience of having solved similar problems many, many times. You can evaluate and see what conservation laws can be applied, and what other equations can be applied. If you think you can solve a problem with mass*energy, go ahead and show that it works: derive the formula from known physics, and show that gives the correct answer. Tell us when it applies and when it doesn't, so others can test it (and make sure that whatever example you give isn't correct by accident) That's not a good enough citation. momentum is not kinetic energy. They are distinct concepts. You can check that KE is lost because you can calculate the KE for each object, and the values do not match. As in the example I gave.

And then the science enthusiast has to deal with trolls who are already established as "Resident Experts" on this board...

swansont You are totally wrong. First, it's just present science BELIEF that axial tilt and the length of the day are not constant, and had different values in the past. The Menorah Matrix PROVES that it is not coincidental and it was always the way it is because a TON of other mathematical patterns confirms it. Picada Please, read the Menorah Matrix before state that "The Menorah Number System is not based on Base10". You are wrong. Period.