Newton's shell theorem - all mass is concentrated at its centre

[laser]

Homework Statement

Conceptual

Relevant Equations

F proportional to d^2, d is the distance between two point particles

Firstly, I know how to derive Newton's shell theorem but was looking on how to explain it conceptually to myself without going into all of the maths. The part of the theorem I am concerned about is that the force acts as if all mass is concentrated at its centre.

Consider a solid sphere on the floor, with a point particle above the sphere (not in the sphere)

Intuitively, the top of the sphere should exert a greater force than the bottom of the sphere, and the average should not be at its centre, because of F proportional to 1/R^2.

For example, take the mass dM closest to the point particle, and farthest from the point particle. Imagine the distance between the point particle and the infinitesimal mass closest is 5m, and the radius of the sphere is 2m. That means that all the the distance from the centre to the point particle is 7m. Using Newton's law of gravitation, we should get F=GMm/49.

Taking the point on the sphere closest to the point particle, we get dF = GdM*m/25, and taking the point farthest, we get dF=GdM*m/81. Averaging these forces should (intuitively) give the force as if those two dM are from the centre of the sphere.

Now obviously my understanding is wrong because Newton is a God, so any help would be appreciated!

 

[Orodruin]

You are only considering the parts on the sphere that lie on the line passing the center of the sphere and connects to the particle. This ignores the three-dimensional nature of the sphere.

 

[laser]

You are only considering the parts on the sphere that lie on the line passing the center of the sphere and connects to the particle. This ignores the three-dimensional nature of the sphere.

True, but apply the same logic to the other dM's. Firstly, there are no forces in the x and y directions as they cancel, only the z direction. The top hemisphere of the sphere exerts a much greater force than the bottom hemisphere. Do you agree that the average of the force caused by the top hemisphere and the force caused by the bottom hemisphere should equal the force caused by both as if all the mass were concentrated at the centre of the sphere?

Also - applying the logic to the other dM's. Break up the sphere into two hemispheres, top and bottom. For every dM on the top hemisphere it has a pair at the bottom hemisphere. Following my logic for my previous example, the force appears to be acting from a point a little higher than the centre of the sphere.

 

[Orodruin]

Do you agree that the average of the force caused by the top hemisphere and the force caused by the bottom hemisphere should equal the force caused by both as if all the mass were concentrated at the centre of the sphere?

No. Each of the hemispheres will give a smaller force than the total so obviously the average will be smaller.

 

[laser]

No. Each of the hemispheres will give a smaller force than the total so obviously the average will be smaller.

Maybe you are misunderstanding me - what I meant to say was the average of the force caused by the top hemisphere (with mass M) and the force caused by the bottom hemisphere (also with mass M) vs the force caused by mass (M) at the centre of the sphere.

 

[Orodruin]

So essentially you are asking if the total force from both hemispheres would be the same as the force from a point mass in the center with the summed mass.

This is trivially true. Not only is it true, but it is also independent of you cutting the sphere at the ”equator”.

 

[PeroK]

Maybe you are misunderstanding me - what I meant to say was the average of the force caused by the top hemisphere (with mass M) and the force caused by the bottom hemisphere (also with mass M) vs the force caused by mass (M) at the centre of the sphere.

It's called the shell theorem because it's usually proved for a spherical shell. The case of a solid sphere follows as a corollary, as a solid sphere is a set of concentric spherical shells.

I know there is an intuitive argument for why the force inside the shell is zero. I'm not sure I've seen one for the force outside the shell. That's the power of mathematics, I guess. It can prove things that cannot be thought out reliably by simple or intuitive arguments.

 

[laser]

So essentially you are asking if the total force from both hemispheres would be the same as the force from a point mass in the center with the summed mass.

This is trivially true. Not only is it true, but it is also independent of you cutting the sphere at the ”equator”.

Oh - That was basically my question, because I don't find it trivial!

Let me pose an analogy for you: In the moment of inertia world, the moment of inertia about a point 9m away is 81M, and the moment of inertia about a point 11m away is 121M. The average of this is 101M, which is not the same as if M was concentrated in the middle of 9m and 11m.

 

[laser]

It's called the shell theorem because it's usually proved for a spherical shell. The case of a solid sphere follows as a corollary, as a solid sphere is a set of concentric spherical shells.

I know there is an intuitive argument for why the force inside the shell is zero. I'm not sure I've seen one for the force outside the shell. That's the power of mathematics, I guess. It can prove things that cannot be thought out reliably by simple or intuitive arguments.

Yeah, I get that - but I am unable to see the reasoning for the shell! I can visualise the argument for the force inside being zero fine enough, but the more I think about the force outside the shell, the more I'm convinced that it's not correct. Of course, the maths proves otherwise!

 

[Bandersnatch]

True, but apply the same logic to the other dM's. Firstly, there are no forces in the x and y directions as they cancel, only the z direction. The top hemisphere of the sphere exerts a much greater force than the bottom hemisphere.

The horizontal forces do cancel. But the horizontal components constitute a smaller proportion of the full force vector coming from a mass element near the top than for a corresponding mass element near the bottom, since the angles the vectors make with the horizontal are more acute. I.e. you have stronger attraction from the nearby mass elements by the virtue of the inverse squares, but at the same time more of it is 'wasted' on cancelling each other out.

 

[nasu]

Intuitively, the top of the sphere should exert a greater force than the bottom of the sphere, and the average should not be at its centre, because of F proportional to 1/R^2.

This part is true but the conclusion is not relevant. The shell theorem is about the total force exerted by the two pieces and not about the average of the forces exerted by them. In general, the sum of two forces is not the same as the average of the two forces.

 

[laser]

This part is true but the conclusion is not relevant. The shell theorem is about the total force exerted by the two pieces and not about the average of the forces exerted by them. In general, the sum of two forces is not the same as the average of the two forces.

Do you mean that the sum of two forces equivalent to double the average of two forces? I can see where you're coming from, because force is a vector, and you must add the vectors vectorially (if that is a word!), not just their magnitude!

 

[laser]

The horizontal forces do cancel. But the horizontal components constitute a smaller proportion of the full force vector coming from a mass element near the top than for a corresponding mass element near the bottom, since the angles the vectors make with the horizontal are more acute. I.e. you have stronger attraction from the nearby mass elements by the virtue of the inverse squares, but at the same time more of it is 'wasted' on cancelling each other out.

Thanks, that kinda convinced me. However, can't you apply the same logic to a vertical disc? However, the force doesn't act like it comes from the centre here.

 

[PeroK]

This is why Newton had to invent calculus! It's not simple logic.

 

[Hill]

Newton himself, in the Principia, has it proven geometrically.

The use of infinitesimals and limiting processes in geometrical constructions are simple and elegant and avoid the need for any integrations. They well illustrate Newton's method of proving many of the propositions in the Principia.

 

[Orodruin]

Thanks, that kinda convinced me. However, can't you apply the same logic to a vertical disc? However, the force doesn't act like it comes from the centre here.

I mean, as I said already in the beginning of this thread, your issue is just considering points on the straight line of action of the net force. This is never going to work out well. Yes, closer points will give more force, but there are fewer of them. Even forgetting that some of the force is lost to cancelling horizontal components, there simply are less points at a distance smaller than the distance to the center than there are points further away.

 

[laser]

there simply are less points at a distance smaller than the distance to the center than there are points further away.

ah, I thought they were equal! But nah lol, the equator is indeed fat :D