I understand that Gauss' Law (or manual integration) yields 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 it pulls the shell inwards on all sides equally. By Newton's third law, this should imply that there is an outward force being applied on the object that is inside as well.

But how will a spherical non-rigid body at the centre 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 centre, would it stretch all the way till the walls and stick to it? What if it was off-centre? [Edit: People seem to not realise what I'm asking. I know the net force is zero, but that doesn't mean that there are no forces to begin with. A non-rigid body changes its shape when anti-parallel, equal forces are applied on it.]

Will a rigid body at the centre not feel pressure from being pulled from all sides? In stars and planetary bodies approximated as uniform spheres, while calculating pressure at an arbitrary radius inside the sphere, why do we consider only the forces inside the elemental shell under consideration?

(For question, see Irodov 1.216 pg45: https://imgur.com/pMCiWqW ; this question is not directly relevant to my query, but it's what incited it)

  • $\begingroup$ Why would an interior body move at all? There are no forces on it? $\endgroup$ – Bob Jacobsen Dec 4 '19 at 4:36
  • $\begingroup$ Newton said in the shell case there shall be no force. $\endgroup$ – user192234 Dec 4 '19 at 4:37
  • $\begingroup$ @BobJacobsen The net force is zero, and that's due to the fact that all forces cancel out. There's an obvious difference between zero force and no net force. Pull a perfectly rigid object with 5N from opposite sides and nothing happens, but pull a non-rigid object and it stretches. That's why the interior body's orientation would change. It wouldn't move as a whole. $\endgroup$ – Arnav Das Dec 4 '19 at 4:41
  • $\begingroup$ @user192234 It doesn't matter what Newton said. His math works out and the NET force comes out to be zero. That doesn't mean that there are no forces at all. $\endgroup$ – Arnav Das Dec 4 '19 at 4:42
  • $\begingroup$ The net force is zero inside the volume because all the forces are zero. The E field and gravitational field are zero everywhere inside. There are no different forces. $\endgroup$ – Bob Jacobsen Dec 4 '19 at 4:44

When you apply two antiparallel equal forces to a non-rigid body, it changes its shape, but what does it mean? It means that different parts of the body start to move with respect to each other, while keeping the center of mass still (because the net force is zero).

Let's consider a tiny chunk of matter from the non-rigid body. Before the deformation it was still. During the deformation it moves. This means that the net force on the tiny chunk isn't zero, even if the net force on the whole body is.

That's what happens when you deform a non-rigid body by forces that sum up to zero: the net force on the whole body is zero, but if you ideally divide it into tiny chunks and consider the net forces on each one of them, you'll discover that they are not zero, while their sum is. If the net force on each chunk was zero, they simply wouldn't move and there would not be any deformation.

Now let's go back to the hollow sphere. Let's consider a non-rigid body inside it. Let's imagine to divide it into tiny chunks of matter. What is the force acting on one chunk? Well, the gravity field vector is zero at every single point in space, then the net gravitational force on each chunk is zero, no matter how small the chunks are. Therefore gravity isn't forcing any kind of deformation in the body. No pressure and no shear stress.


This question conflates two effects.

There are no gravitational forces inside a uniform spherical shell due to that shell. None, period: they add to zero on an extended body because they’ll all zero on every point in that body.

But the shell has to be held up against the gravitational effect of material inside it, which is in turn pulling down on material outside it.

A 1km thick shell in the earth at 1001 km radius doesn’t exert any gravitational force on the shells at 0 to 1000 km. But 0-1000km does attract 1001km. The 1001km shell will fall unless mechanical forces, in this case compression pressure, holds it up.

In the electrical case of uniformly charged shells, the force on the outer shell is usually repulsive, and it’s the strength of the material that keeps the shells together.

  • $\begingroup$ If there's mass around an object, surely there should be some forces acting on the body (despite the fact that they cancel out)? If a human was put inside a Dyson sphere, would they die a painful death or would they feel no forces whatsoever? $\endgroup$ – Arnav Das Dec 4 '19 at 4:57
  • $\begingroup$ If the forces cancel at a point (I.e. any point, hence all point), there is no force there. There’s no analogy with e.g. pulling the ends of a string these are forces cancelling at a point by making a net zero field there. A body at that point just feels the zero field making zero force. $\endgroup$ – Bob Jacobsen Dec 4 '19 at 5:01
  • $\begingroup$ Why is there no force there? There's mass all around that point. It should be attracted in such a way that all forces cancel out. Why would all the forces exerted by an object cease to exist entirely just because of how an object is shaped? $\endgroup$ – Arnav Das Dec 4 '19 at 5:05
  • $\begingroup$ You seem to think a point charge near two others feels two separate forced instead of just their sum. Ok, if that’s true, which way does it accelerate? Two ways? $\endgroup$ – Bob Jacobsen Dec 4 '19 at 5:15
  • $\begingroup$ Neither direction. Because the forces pull it equally in both directions. How would the point charge 'feel' the summation of the forces? It has to feel both, separately. $\endgroup$ – Arnav Das Dec 4 '19 at 5:17

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