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1) It's a well-known result that the net force inside a hollow, uniform spherical shell is zero at all points.

However, for a spherical shell with finite mass inside of it, we say that the mass inside pulls the shell inwards on all sides equally, thereby creating mechanical pressure that the shell must bear. By Newton's third law, this should imply that there is an outward force (albeit one that sums up to zero at every point) being applied on the object that is inside as well.

But how will a spherical non-rigid body at the center change its shape with time due to an equal amount of force being applied from all sides? Say, if I kept a semi-solid object at the center, would it stretch all the way till the walls and stick to it? What if it was off-center?

2) Will a rigid body at the center not feel pressure from being pulled from all sides? In stars and planetary bodies approximated as uniform spheres (for example, Irodov 1.216 pg45: https://imgur.com/pMCiWqW), while calculating the pressure at an arbitrary radius inside the sphere, we consider elemental shells and integrate the pressure experienced by all of them by the spheres inside. However, we neglect the forces of the outer shell(s). If the inner body pulls the outer shell, why neglect the other way around?

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    $\begingroup$ I don't follow your argument that there is a force on the central body. The inwards gravitational force on the shell is resisted by equal and opposite compressive forces in the material of the shell. There is no force on any body inside the shell. $\endgroup$ – John Rennie Dec 4 '19 at 11:20
  • $\begingroup$ @JohnRennie The resistive forces are not relevant here. I'm talking about the reaction force. Gravitation is mutual, so if the inner body is responsible for pressure on an outer shell, that should mean that there is a reaction force on the inner object, albeit the object can't accelerate because it is being pulled equally in all directions (thereby making the NET force zero). Are you telling me that the inner body creates no pressure on the outer shell? $\endgroup$ – Arnav Das Dec 4 '19 at 11:34
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    $\begingroup$ This is your second time asking the same question based on the same misconception. So far, five separate people have tried to get you to reconsider that misconception. Perhaps you should ask a question directly targeted on how forces combine to get good (and hopefully persuasive) answers on that? $\endgroup$ – Bob Jacobsen Dec 4 '19 at 14:08
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/517590/2451 $\endgroup$ – Qmechanic Dec 4 '19 at 15:09
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1) You answer it yourself, the total reaction force is zero at every point, regardless of position, so a non-rigid object will not deform

2) If the sphere is rigid, the shell will not move inwards, but if it is made of individual particles, it will start (or try) moving inwards, which will create a pressure on inward shells, but this pressure will not be gravitational, is a contact force, and it has its own reaction. This issue is absent if the object inside the sphere is not in contact with the shell, because the net gravitational force is zero are there are no contact forces between the shell and the inside object

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    $\begingroup$ Can you elaborate on (1)? The net force is zero, but that doesn't mean the object won't deform. If you pull a non-rigid body equally from both sides, the net force on it is zero, but it still deforms. $\endgroup$ – Arnav Das Dec 4 '19 at 12:17
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    $\begingroup$ but the net force is zero at every point, you are not pulling it in different directions $\endgroup$ – Wolphram jonny Dec 4 '19 at 12:53
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    $\begingroup$ The net force is zero only because the gravitational force exerted by all elemental masses in the ring cancel out, id est they pull on the object from all directions equally. That's the only way you can get the net force to be zero. What else could it be? The gravitational force doesn't mysteriously vanish just because of how the masses are oriented. This is trivial. $\endgroup$ – Arnav Das Dec 4 '19 at 12:56
  • $\begingroup$ yes, I agree with that, but if you have an extended object and, at every point, the net force pulling it is zero (it does not matter if it is the resultant or many forces), then it will not move. I believe you are imaging something like a a force pair on a rigid body (acting at different points) whose resultant is zero but the torque is not. This case is different, here the resultant at every point is zero, F=ma, so it will not move $\endgroup$ – Wolphram jonny Dec 4 '19 at 12:59
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    $\begingroup$ The resultant force on it is zero, so its COM won't accelerate. That obviously doesn't mean that it won't deform. Essentially, if a human were to be put in the center of a Dyson sphere, would they feel no forces at all, or would they be painfully stretched in all directions? $\endgroup$ – Arnav Das Dec 4 '19 at 13:02

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