Genady

On the scale of several hundred megaparsec and more the visible universe is homogeneous and isotropic. It means that on that scale it is the same in all directions.

As the bodies outside our visible universe make pretty much spherically symmetric shell, their gravitational effect inside the shell is identically zero.

Even without water issues, the growing roots of Ficus tree in my yard have killed over the years surrounding coco palms, one by one.

It is curious that the same mechanism gives both the attractive effect of dark matter and the repulsive effect of dark energy.

The hypotheses above may be complementary rather than contrasting because they seem to relate to different aspects of the "four why's of animal behavior" of Nikolaas (Niko) Tinbergen, i.e., causation, development, adaptation, and phylogeny.

What does it mean? The model tells us how it happens. Some future model might derive this model, i.e., the equations, from more generic principles. The previous model, aka Newtonian gravity, included a notion of mass acting on the ship. This is an approximation, which is not included in the current best model.

The spacetime geometry obeys certain differential equations. The solution of these equations depends on initial and boundary conditions. The mentioned mass is a part of these conditions.

I don't know what is unclear. Clarify. Restate the unintentionally struck question, please.

The physics is clear then.

This is true for homogeneously and isotropically distributed sources. What about other "smeared" sources? The Weyl tensor is not necessarily zero, but the curvature is not away from the source.

What happens if the sources are everywhere?

Yep, it was obvious indeed.

Perhaps it's obvious but I don't see where \(g_{rr}\) in the denominator comes from, here: \[a^r = -\dfrac{c^2}{2} \dfrac{1}{g_{rr}\ g_{tt}} \dfrac{dg_{tt}}{dr}\]

What is \(T\) in \[a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}\]

Because Earth or other body you mention are far away and your ship cannot be affected directly by them without "an action at a distance." What it can be affected by directly is the geometry of spacetime at that same point where and when it is "here" and "now". I still don't know what you mean in that question. Does what?

The following story from Zee, A. Einstein Gravity in a Nutshell (p. 52) is in the same line: Long ago, an undergrad who later became a distinguished condensed matter physicist came to me after a class on group theory and asked me, “What exactly is a tensor?” I told him that a tensor is something that transforms like a tensor. When I ran into him many years later, he regaled me with the following story. At his graduation, his father, perhaps still smarting from the hefty sum he had paid to the prestigious private university his son attended, asked him what was the most memorable piece of knowledge he acquired during his four years in college. He replied, “A tensor is something that transforms like a tensor.” 😉 Here is one, from MTW's Gravitation. All variables are real numbers. The most important one, \(T_{00}\), is energy density:

Any geometry with zero Ricci tensor and non-zero Riemann tensor.

It can be a non-flat solution for some sets.

Only perhaps if it's preexisting / primordial. Got it. Sure*. In case the mass we have is zero, the geometry is not necessarily flat. * Depends, not uniquely determined.

I don't know what you mean. Does mass cause geometry to exist?

But why Earth gets near there? Because it follows a geodesic according to the spacetime geometry. So, I would "blame" the spacetime geometry again. But you would blame some other mass-energy changes for that geometry. Then I would blame geometry for changes which occur to that mass-energy. Etc. The point is that geometry and mass-energy "conspire" in such a way that they together obey the Einstein field equation of GR.

Because the same changes in geometry can occur in other circumstances, e.g., different body or bodies, with different parameters / locations/ movements, but your response will be the same.

1 and 2 (they are different expressions of the same, but 2 is more precise.) You have to do it because what happens at your location at that time. I.e., the spacetime geometry at that spacetime event.

All solutions obey this equation. The mass is not at the same location where the Schwarzschild metric is. I don't see why it would matter. There are simply more independent variables in the curvature tensor than there are independent equations in the Einstein field equation.

Aren't there many examples, at least in principle? In particular: "Q: The information that gets lost when we go from the Riemann tensor to the Ricci tensor does not affect the energy-momentum tensor nor Einstein’s equations. What is the meaning of this lost information then? A: It means that for a given source configuration, there can be many solutions to Einstein’s equations. They all have the same right-hand side, namely \(T^{\mu \nu}\). But they simply have different physical properties. For example, the simplest case is to ask: what if this energy-momentum stuff is zero? If it is zero, does it mean that there is no gravitation, no interesting geometry at all? No. It allows gravitational waves." Susskind, Cabannes. General Relativity: The Theoretical Minimum. Not according to this: homework and exercises - Non-zero components of the Riemann tensor for the Schwarzschild metric - Physics Stack Exchange