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Assuming matter is more or less uniformly distributed in the Universe, what would be the net gravitational force acting on a stationary (from the reference frame of the Universe as a whole) lump of matter, from the rest of the Universe in these cases?

  1. The Universe is infinite,
  2. The Universe is finite and spherical, but the object is not at its centre

I am asking because neither of my naïve results seem to make sense:

  1. Infinite Universe: ∞ forces of opposite directions, each cancelling out?
  2. Finite Universe: significant amount of all matter accelerating towards the centre of mass
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  • $\begingroup$ What do you mean by not bound? $\endgroup$ – AHB Jan 3 '17 at 7:05
  • $\begingroup$ Let's say the planet has been ejected from a solar system, it is not orbiting a star $\endgroup$ – ᆼᆺᆼ Jan 3 '17 at 19:21
  • $\begingroup$ You mean we shouldn't consider gravity actions on an orphan planet? $\endgroup$ – AHB Jan 4 '17 at 6:12
  • $\begingroup$ Never mind, removed any mention of a "planet" $\endgroup$ – ᆼᆺᆼ Jan 4 '17 at 13:53
  • $\begingroup$ Do you mean in the framework of Newtonian mechanics? $\endgroup$ – Qmechanic Jan 4 '17 at 18:27
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In 1st case, assuming an infinitesimal part of the lump, We consider the universe as the superposition of 1. The rest of the lump and 2. The other masses which we considered them all in a single mass density $\rho$.

The forces of the uniform infinite matter cancel on any infinitesimal part by symmetry. What remains is the force of the rest of the lump.

After integrating over the volume of the lump, The second part also becomes zero as a system can't exert force on itself.

About The second configuration, aside form the fact that the lump can't exert force on itself as we stated above, we must be calculate a thing.

A particle in a uniform sphere of mass density $\rho$, if off-center, WILL feel force. As you say, we assume the lump's in the center (of course it's unclear because if it possesses dimension (it's not a point), It matters which point of it is in the center!).

The gravitational field inside a uniform sphere given by:

$$\mathbf{g}=-4\pi\rho G r\mathbf{\hat r}$$

So the total force due to the big sphere on the lump will be:

$$\mathbf{F}=\int_\text{the lump}\mathbf{g} dm=-4\pi\rho G \int_\text{the lump}\mathbf{r}dm=-4\pi\rho G M \mathbf{r_\text{center of mass}}$$

The last integral is by definition equal to $M \mathbf{r_{center of mass}}$. So,

Whether or not, the lump in the second configuration will feel force depends on the result of Where the center of mass is located. If the center of mass of the lump is in the center of the big sphere, the lump will not feel force.

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First of all your question assumes there is a universal proper inertial frame which may or may not be true. With both case scenarios the answer is it depends. Gravitational attraction is not linearly related to the distance between mass/energy in question. Because of that relationship the homogeneous distribution of matter at a larger scales is overpowered by the less homogeneous distribution at smaller scales.

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  • $\begingroup$ I guess the question is then, how can definitely finite (at smaller scales) overpower potentially infinite? Even if greatly diminished, ∞ / x = ∞, right? $\endgroup$ – ᆼᆺᆼ May 8 '19 at 6:45

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