What do you mean by "mass needs charge?" There are massive particles without charge, such as the Z and Higgs.

My thoughts is that this is a physics question rather than one of philosophy. So why didn't you ask this in Phys.SE?

"all particles would fall to each other without opposing forces" Without any other forces they would all move around in complicated orbits determined by the gravity of all the other particles, they wouldn't just collect in one place (gravity is a time-symmetric theory, meaning if you take a movie of particles moving only under the influence of gravity and play it backwards, the resulting trajectories should still be obeying the same dynamical laws)

@Hypnosifl But what if there was no charge from the start? If two particles fall towards Earth, there is only a tidal force. How can there be another kind of force? How do they get tangential velocity (which they have nowadays but not back then). Massive bodies like Earth can't form).

@Sandejo But the Higgs and Z do possess charge. So does the neutrino. I think you mean electric charge only?

@MoziburUllah I think this gets closed on physics. I don't take the risk...

"If two particles fall towards Earth, there is only a tidal force." Are you talking about Newtonian gravity, or general relativity? In Newtonian gravity there is a gravitational force on point particles, while tidal force relates specifically to different forces on different parts of an extended body. General relativity explains gravity in terms of geodesics in curved spacetime but its predictions match Newtonian physics in certain limits, if you aren't dealing with large concentrations of mass leading to event horizons, or relativistic velocities, things should be qualitatively similar.

Also note that in both Newtonian physics and GR, if you assume no non-gravitational forces then masses falling towards each other should just pass through one another (no repulsive force to cause them to 'crash') and then move in opposite directions symmetrically until the gravitational pull slows them down and they start to fall towards one another again. "How do they get tangential velocity (which they have nowadays but not back then)" What do you mean? There's no reason you can't have tangential velocity in your initial conditions.

@Hypnosifl How can particles go through another? Tidal forces are non-local. They drive particles apart. If two particles fall from different heights their distance grows linearly (they are not accelerated wrt each other. Except for strong tidal forces. Only if you are accelerated upward (because of the EM force) you see the particles accelerate. The Earth can't even exist in the first place. Neither can an accelerating rocket.

They can go through one another if there is no repulsive force that prevents them from occupying the same space (in quantum physics there is also the pauli exclusion principle which would prevent this for fermion particles, but not for boson particles). As for tidal forces, that's what I meant by "extended body"--any collection of particles (or continuous mass distribution) gets stretched by tidal forces, in Newtonian physics this is just a consequence of different local gravitational forces on different individual particles, which cause them to accelerate differently, even without EM forces.

Not sure why you think acceleration requires EM forces, the gravitational force experienced by a point mass m that's at radius r from another mass M would be given by the equation F = GMm/r^2, and the point mass' acceleration is given by a = F/m, so the point mass' acceleration due to the gravity from mass M would be a = GM/r^2. This is not a tidal force, it's just acceleration due to the gravitational field from M at a single point in space.

@Hypnosifl The force F in your formula is the EM force. There is no force in GR. The mass M exists because EM forces keep M in shape. A mass m falling to M seems accelerated wrt to the surface of M, which is the surface that actually accelerates upward (without gaining velocity). An accelerating rocket (because of EM forces) makes it look as if objects are accelerated in the rocket

Higgs and Z do not possess any charge; they are their own antiparticles. Note that I did not mention neutrinos.

"The force F in your formula is the EM force. There is no force in GR." My equation was for Newtonian gravity, but as I mentioned, the predictions of GR come arbitrarily close to those of Newtonian physics in certain limits. "The mass M exists because EM forces keep M in shape" In Newtonian gravity (or the Newtonian limit of GR), shape is irrelevant for a spherically symmetric body, the external field felt by other particles is identical to that which would be created by a point particle with mass M situated at the body's center of mass.

@Hypnosifl But a spherical distribution of particles with mass only can form a solid ball. If you fall freely in it, you only notice particles being stationary wrt to you or with constant velocity (ignoring tidal effects). You can't even imagine such a distribution as this means being stationary wrt to it, for which an EM force is required. In short, you need EM acceleration to notice accelerated motion.

A spherical distribution need not be solid, you could still have a spherical distribution even if there were no non-gravitational forces and the particles could pass through one another--they could all be in circular orbits around their common center of mass, or continually oscillating on radial paths going through the center like an object falling through a tunnel through the center of the Earth.

Anyway as I said you don't actually need a spherical distribution, you could replace it with a single point mass M, it would cause the other mass m to accelerate in exactly the same way.

Also even if it was a solid ball with all the particles at rest relative to the center, I don't understand your comment "If you fall freely in it, you only notice particles being stationary wrt to you or with constant velocity (ignoring tidal effects)"--the Newtonian gravity equation that I quoted indicates you would be accelerating relative to any point inside the ball that you passed by it, even if "you" were a point particle. Why do you think a falling observer would see them as stationary or with constant velocity?

@Hypnosifl Because all particles fall along with you. You would only perceive changes in velocity if one experiences an acceleration because one of the three basic forces. Gravity is not a force (hence the difficulty to quantize it, i.e., the presence of black holes).

Are you talking about the case of the solid ball (in which case the surrounding particles would not be falling, held in place by non-gravitational forces), or of one of the other cases I mentioned where there is no need for forces other than gravity? Also when you talk about what the observer sees, are you talking about only local measurements, or are you including long-distance measurements? If spacetime is close to flat (the Newtonian limit) so you can use something close to an inertial frame for long-distance measurements, you could see yourself accelerating relative to distant particles.

@Hypnosifl Like I stated, I'm talking locally. If you fall with particle around you, you will see, apart from tidal effects, no change in velocity. So only accelerating by other forces will make you see the particles accelerate. Like an object falling to Earth, or on a scale doesn't feel the force of gravity. It's the scale pushing upwards, not the gravity pulling downwards.

You did say something about what would be observed locally in your explanation of the question, but you never said that the question itself was restricted to local observations, and some aspects of the question seemed to refer to non-local observations, like an object falling towards the Earth "which apparently diminishes the distance between the mass and the surface of the Earth" (a local comparison could only be made at the exact instant the position of the falling mass coincided with the surface of the Earth). If you want to restrict the question this way, could you edit to make more clear?

@Hypnosifl I think the end of my question is clear about that.

The end of your question indicates that "local linear effects" are at least one of the issues you are concerned about, but nowhere do you state that this is the only issue you are asking about, and as I said the part about an object falling towards the surface of the Earth at least seems to indicate something else. I understand what you mean now, but I'm just saying if you want to increase your chance of getting relevant answers, it couldn't hurt to spell it out more clearly, preferably near the beginning or in the title.

@Hypnosifl Wouldn't you notice locally that particles fall towards a CoM only if there was a force to counter the falling? Which is different from charges falling towards other charges.

@Felicia - Yes, if you mean "locally" in the idealized sense of an infinitesimally small region of spacetime around you, and you also use classical general relativity (not quantum gravity) combined with a model of EM which treats it as a field on spacetime rather than an effect of curved spacetime itself (EM can be explained in terms of curved spacetime, see the Kaluza-Klein theory, which IIRC is at least partly incorporated into string th.)

But treating EM + gravity in terms of a field defined on curved spacetime is presumably just one approximation to however a quantum theory of everything would deal with the problem, I think most physicists working on ToEs would consider it unlikely that such a theory would preserve this fundamental dichotomy between "true forces" and "effects of spacetime curvature".

Ultimately though this seems like purely a physics question rather than a philosophical one, if you asked it on the physics stack exchange and people thought you were talking about electric gravity or something, it's probably because you didn't make it clear enough that you were just talking about observations in the sorts of "local inertial frames" used to define the equivalence principle.