Is the Shell Theorem Valid for Both Solid and Hollow Spheres?

[bobie]

Hi,
I found this derivation at wiki: http://en.wikipedia.org/wiki/Shell_theorem

Can someone tell me if the theorem is valid only for a solid sphere , or
the pull of a hollow sphere is exacly the same it would be if all the masse where at the center wrt to a mass at any distance D from the centre?
does it apply also when D = R?
If it does can you explain how (apart from the math derivation) if ,say, an electron is on the circumference it gets the maximum pull, and it suffices that it moves 1 mm toward the centre, that pull vanishes all of a sudden? How can such minimal difference in distance cause such a huge result?
Thanks

 

[HallsofIvy]

As long as the density is a function of r only, which includes the case of a hollow sphere, yes, the gravitational pull at any point outside the sphere is exactly the same as if all the mass were concentrated at the center.

 

[bobie]

yes, the gravitational pull at any point outside the sphere is exactly the same as if all the mass were concentrated at the center.

Thanks, Hallsofivy.
I was referring to that and also to the following:

If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

What puzzles me is the borderline situation, when the recipient mass is touching the hollow sphere, like a particle on earth:
suppose it is a hollow sphere with a hole in it, when the particle leans on the brink it gets the full pull, if it moves a little it falls in the hole just for an instant but, as soon as it crosses the tiny crust, if gets no more pull and gets suspended underneath the surface. If the conclusion of the theorem is this, isn't it odd? its like crossing a magic screen, shouldn't change be gradual?doesn't the equation break down at the surface?

I posted this in the math forum, because you can scrutinize it numerically, if the radius r = 10 Km, do you think the particle can stop when it is 1cm unterneath the hole?
Should I post this in the physics forum?
Thanks

 

[Matterwave]

In the case of a 2-D sphere of mass (literally 2-D sphere), the mass density is infinite at the surface of the sphere, since you have a finite mass compressed into literally 0 volume. This is why you get a discontinuity in the force as you move from outside the sphere to inside the sphere.

All physical "hollow spheres" will have some thickness, and therefore when you move from outside the sphere to inside the sphere, you move through the thick part of the sphere. Do the math here, and you will find that the force gradually weakens until you get to 0 force inside the inner radius of the sphere.

But one should not that 0 force does not mean 0 velocity, it means 0 acceleration. So a particle dropping through a spherical shell (2-D, with a small hole in it) will move at constant velocity after it enters the shell rather than "stop". There would be no discontinuity in the motion. No infinite acceleration (although there would be an infinite jerk lol, but again we already established that a true 2-D sphere of matter is not physical).

 

[bobie]

In the case of a 2-D sphere of mass (literally 2-D sphere).

thanks Matterwave,
why are you calling it a 2-D ...? It is a regular 3-D sphere.

Just imagine the planet earth, with a crust of 1 cm or whatever you like, the radius of 10^8 cm and a hole.

Where do we get infinite mass density? shot or sand would move through the hole for a fraction of a second and then suddenly stay suspended a few cm underneath the hole?

 

[Matterwave]

I'm referring to a 2-D sphere. I.e. a sphere with no thickness.

Read the second part of my post?

 

[bobie]

I'm referring to a 2-D sphere. I.e. a sphere with no thickness.
Read the second part of my post?

I am referring to a concrete example.
I read it, just to make the discussion simpler, then imagine we lower a marble/grain of sand through the hole (by a string) a few cm under the crust and leave it there.

It will get no speed , feel no pull and should stay there as suspended in deep space.

That is what the theorem says, or did I misinterpret it?

 

[Matterwave]

Then see this part about my post:

All physical "hollow spheres" will have some thickness, and therefore when you move from outside the sphere to inside the sphere, you move through the thick part of the sphere. Do the math here, and you will find that the force gradually weakens until you get to 0 force inside the inner radius of the sphere.

But one should note that 0 force does not mean 0 velocity, it means 0 acceleration. So a particle dropping through a spherical shell (2-D, with a small hole in it) will move at constant velocity after it enters the shell rather than "stop". There would be no discontinuity in the motion. No infinite acceleration.

As for your new question. Then yes, if the Earth were hollow, and you put a grain of sand inside the inner radius, it would stay suspended there and not accelerate. But inside the Earth is not deep space...

 

[bobie]

yes, if the Earth were hollow, and you put a grain of sand inside the inner radius, it would stay suspended there and not accelerate.

So I got it right, thanks, Matterwave.

But that was my primitive concern: it is rather odd that at a distance of only 2cm the pull changes abruptly to nought, where does the equation break?
At 1 cm under the Earth crust the marble would still get the downward pull by 99.99999999..% of the Earth mass and only 0.0000000000000...1% of upward pull from the crust above it, it seems most unlikely that the two pulls compensate each other.
Where is the catch?

 

[Matterwave]

So I got it right, thanks, Matterwave.

But that was my primitive concern: it is rather odd that at a distance of only 2cm the pull changes from 983 cm/s² to nought, where does the equation break?
At 1 cm under the Earth crust the marble would still get the downward pull by 99.99999999..% of the Earth mass and only 0.0000000000000...1% of upward pull from the crust above it, it seems most unlikely that the two pulls compensate each other.
Where is the catch?

Like I mentioned in my previous post, the acceleration goes from 983 cm/s^2 outside the hollow shell, to 0 cm/s^2 inside the hollow shell by decreasing inside the thickness of the shell. Only if your shell had no thickness (the 2-D sphere) would you get a discontinuous change in the acceleration. You said you wanted to consider a physical sphere that has some thickness.

To answer your second question, you simply have to apply the math. This is a very well known, and very well proved result. I don't think you will find it satisfactory if I just reproduce the proof inside the wikipedia article. You can maybe read it carefully to see if you can see its reasoning.

The intuition is that if you are close to the inner edge of the sphere, although you are pulled by the mass close by stronger (per unit mass), there is more mass pulling you the other way. These two effects, for a problem which has spherical symmetry, will cancel.

 

[bobie]

t if you are close to the inner edge of the sphere, although you are pulled by the mass close by stronger (per unit mass), there is more mass pulling you the other way.

Yes, I got that, but it seemed to me at a first glance the the two forces are unbalanced.

I'll make a few pen-and-pencil calcs and get back to you. Please tell me: if the humble discrete-summation results contrast with abstract formulas and integration , which prevail?

Thanks, and congrats for your new chevrons,

 

[Matterwave]

The proof is given to you explicitly in the wikipedia article you yourself provided.

I have not tried to apply the shell theorem for the discrete case. But after some thought, it seems to me that the theorem would only hold true given that the distribution is evenly spread along an imaginary sphere (no clumps obviously), and only true in an axis where the distribution is symmetric about that axis. In other words, what does it mean for a discrete distribution to be spherically symmetric? Spherical symmetry requires a continuum, because only a continuum can create a sphere, which is infinitely smooth. A discrete distribution can only be symmetric about a finite number of axes (at least, a finite distribution of charges, I'll leave it to a mathematician to correct me if a countably infinite number of discrete charges can have an countably or uncountably infinite number of symmetry axes). As you have more and more points of course, you approximate a sphere better and better and so for a regular sized object, the result is very nearly true.

Generally speaking, you are allowed to move from the discrete case to the continuous case if the discrete points of concern are very densely packed compared to the distance scales of interest.

 

[bobie]

I have not tried to apply the shell theorem for the discrete case.
I'll leave it to a mathematician to correct me if a countably infinite number of discrete charges can have an countably or uncountably infinite number of symmetry axes).

I have tried to do my homework, Matterwave, but figures don't match.If I got it right you are allowing for a certain discordance between abstract theory and concrete world. But that would produce a minimal, insignificant difference in results.

I have tried to consider a concrete sphere (similar to the Earth r = 10^8 cm) but with ideal properties of symmetry and density. The real problem is that when you integrate the equation (which must be a hyperbola) in reality you cannot go as far as zero, but must stop at the level of molecules (10^-7cm)or atoms (10^-8cm).

In our example, considering that each unit mass (1 cm² of area) has pull =1 (→G = 10^16), the total pull a unit mass in the center would get a pull of: 4π*10¹⁶, and a mass at any distance inside the shell should get (2pi *)10¹⁶ in both opposite direction
If we consider a unit mass m (1 mm wide) at 1 cm distance inside the Earth surface, if you stop at 10^-8 cm, you get a considerably smaller value , if you integrate from 1 cm to 0 you get a greater one, but never a balanced result.

Could you or anyone show me the way to formulate and integrate an adequate equation for our example?

I called x the distance from the shell (0 to 1 cm), the distance D between m and the points on the shell is D = √((1-x)²+10^8x-x²) and the radial pull (cos λ) = (1-x) /D
the equation is therefore G*(1-x)/ D³

Is there anything wrong in this procedure?

 

[verty]

Bobie, you are asking about the attraction close to atoms, but do we even know what gravity is like close to atoms? For example, within an electron's distance from the nucleus, what is the gravity? Perhaps there is no gravity that close, right? Quantum physics may have an answer to that question but it may have many answers or no answers.

 

[bobie]

Bobie, you are asking about the attraction close to atoms,.

No , I was not.
I am considering a tiny ball of metal m (1mm wide) at distance 1 cm from a hollow sphere like the Earth (r = 10⁸).
At that distance (h= x) the spherical cup has the radius of its base a = 44.72 m (√10^8*1-1²). The ring at the base (x = 1) has no radial pull on m, as cos λ = 0
for the other rings (x = .99...) the pull is G*(1-x)/√(a²+(1-x)²)³, right?

When I say we have to stop at x = 10^-8 cm (the pull is the normal ,regular pull an electron gets in a H atom, you do not need QM at all) I mean we have to stop because a gets < 1 and that means there is no more matter, the surface/volume/mass of the sphere does not decrease as far as zero. A real sphere does not end in an abstract dimensionless point but with an atom.

Can you check if my equation is correct?

 

[verty]

No , I was not.
I am considering a tiny ball of metal m (1mm wide) at distance 1 cm from a hollow sphere like the Earth (r = 10⁸).
At that distance (h= x) the spherical cup has the radius of its base a = 44.72 m (√10^8*1-1²). The ring at the base (x = 1) has no radial pull on m, as cos λ = 0
for the other rings (x = .99...) the pull is G*(1-x)/√(a²+(1-x)²)³, right?

When I say we have to stop at x = 10^-8 cm (the pull is the normal ,regular pull an electron gets in a H atom, you do not need QM at all) I mean we have to stop because a gets < 1 and that means there is no more matter, the surface/volume/mass of the sphere does not decrease as far as zero. A real sphere does not end in an abstract dimensionless point but with an atom.

Can you check if my equation is correct?

Should $\pi$ not appear in your formula? The attraction of a ring should include $2 \pi R$, I think.

 

[bobie]

Should $\pi$ not appear in your formula? The attraction of a ring should include $2 \pi R$, I think.

The integration gives the pull of 1 unit mass (cm²) on each ring, then you multiply the result that by 2 pi to get the full value.
(I work in a simple way, as I am not familiar with Wolfram and calculus. When I integrate with a constant , sometimes wolfram gives me a different resul if I put it in a differen placeof the equation, so I omit it altogether to be sure of the result. I also use integers instead of decimals ).

But you need not, you can expect your result to be exacly 10¹⁶, right?
Can you get a balanced result, in accordance with the theorem?

btw, congrats for your brand new medal, verty