# Intuitive reason for no gravity in a hollow sphere

1. Dec 16, 2004

### Bartholomew

There is no net gravity in a hollow sphere. You can get this result using calculus but that's not really understanding it. There must be some special property of the hollow sphere which is unlike most, or all, other objects, because most or all other objects do not have zero net gravity in their interiors.

Is there any other object which has no net gravity in its interior? Or even in regions of its interior?

I was thinking about the intuitive reason and you can think of it like this: in the middle of the sphere there is clearly no net gravity. If you move towards one side of the sphere then that side pulls on you more strongly and the other side pulls on you more weakly, but also the side you move towards gets smaller. This can be understood in terms of the circle on the sphere's surface formed by the intersection of the sphere surface and the plane perpendicular to the sphere's radius at your location. All of the sphere's surface in front of this circle pulls you forward, and all of the sphere's surface behind this circle pulls you back, and as you move forward the surface in front of the circle shrinks and the surface behind the circle increases.

So you have two influences as you move to the edge--the increasing closeness tends to pull you towards the edge of the sphere and the changing mass ratio tends to pull you back towards the middle. What is still not intuitive to me is why these must cancel each other exactly. Any ideas?

2. Dec 16, 2004

### krab

Say you have an observation point r1 from one side and r2 from the other. Draw 2 lines through your observation point. The angle they make with each other can be used to imagine a whole cone of lines on either side of the point. The area at the base of the cones is proportional to r1^2 and r2^2. According to the law of gravity, the piece of the sphere at the cone base pulling one way is proportional to this area divided by distance squared. So it is independent of distance. Therefore the pull one way is the same as, and balanced by, the pull the other way.

3. Dec 16, 2004

### Pieter Kuiper

You need to divide "front" and "back" in many small facets. Each of these will cancel for an inverse square force, so also the total.

There are figures here: http://ebtx.com/ntx/newton.htm

4. Dec 16, 2004

### DaveC426913

"...most or all other objects do not have zero net gravity in their interiors..."

How do you know this?

(I suspect it is intuitive.)

If you're looking for something halfway between intuition and mathematical formulae, try sketching it out. N planets equidistant around a point. Place yourself anywhere inside, measure the attractive forces. You should be able to use numbers that make the math easy. (BTW, I don't know if you get the same result in only 2 dimensions.)

5. Dec 16, 2004

### DaveC426913

Building a Bus Depot: Calculus without Math

Actually, there's a way cooler way. This is something I've read about that I think is absolutely fascinating. It's calculus without math.

Here's an example:

You have three neighbouring towns in a triangle, and a budget for only one bus depot. What is the best geographical location between the 3 towns to minimize travel for travellers from all three towns?

Draw a map for the 3 towns, properly spaced, on a board. Drill holes where the 3 towns are. Drop a string through each hole, tie a weight to the end of each string. Tie the 3 strings together at one point with a single knot.

Now, jiggle the map around. The knot will drift to the most conservative point. That's where you put your bus depot.

Here's where it gets cooler.

Say the three towns are all different sizes - pop. 5000, 3000 and 1000. The mayors insist that the bus depot be proportional to the population. The largest town should have the bus depot closest, to serve the most people.

OK, easy. Change the suspended weights! Make the weight equivalent to the population. Use weights of 5oz., 2oz. and 1oz. Jiggle the board again, and the knot will settle much closer to the town of 5000.

This can be extended with arbitrary complexity. It is simply a matter of whether or not you can engineer a faithful mechanical model of the mathematical property.

So, your task is to make a model that represents the hollow sphere. It *might* work in only two dimensions, i.e. you might be able to model a flat disk, but I'm not 100% sure of this.

Anyway, the tricky part will be to configure the strings so that their *pull* is inversely proportional to the *square* of the distance. (It pulls twice as hard when it's at half the distance from the hole) That will be tricky!

If our intuition about the gravity in hollow spheres is correct, you should find that the knot is equally comfortable at *any* point within the circle of holes.

Last edited: Dec 16, 2004
6. Dec 16, 2004

### Bartholomew

Krab said:
Hmm... This sounds good but the area at the base of the cones is only proportional to r1^2 and r2^2 when the cones hit the sphere squarely. When they hit it at an angle then they "wrap around" the surface and the area is greater than if they hit squarely.

I wonder where the cones hit at the sharpest angle for the greatest area. Maybe it's that circle of division between the front and the back of the sphere surface!

Aha, Kuiper, I checked out your link and that explains everything. Except, how do you know that the two wedges there are proportional? I got it partially... you know that when two chords cross each other the products of the lengths of the chords above the crossing point and below the crossing point are equal to one another so that the two line segments on each side are proportional, and the angle between them is equal in this case, so that's enough to make the two wedges similar triangles (if they were triangles). But how do you go from proportional chords (the chord being the third, as-yet undetermined side in the two "triangles") to proportional arc lengths?

So, Dave, I guess that would mean a ring would not have zero gravity in its interior except at the center, because the corresponding areas at the "cones" (wedges in this case... no doubt there's a geometric name for those but I've forgotten it) would only increase linearly in mass with distance. Good luck trying to rig an empirical setup for the spherical case then!

7. Dec 16, 2004

### Pieter Kuiper

Sir Isaac just went to the limit of infinitesimally small facets. In his figure (at the bottom of http://ebtx.com/ntx/newton.htm) the HI and KL described as being very small.

8. Dec 16, 2004

### T@P

wasnt the whole gravity in a sphere thing tested using electro magnetism?

9. Dec 17, 2004

### ObsessiveMathsFreak

What if a black hole was hollow? Would there be no gravity on objects in its hollow?

10. Dec 17, 2004

### DaveC426913

calculus w/o Math: solution

Yeah, that's what I was thinking.

The tricky part isn't expanding it from 2 to 3 dimensions, the tricky part is rigging the inverse square force. How do you get a piece of string (or spring) to pull twice as hard when it's half as long?

But I figured it out (long drive home last night):
3D 'gravity inside a hollow sphere' empirical setup

The strings run out through the holes and then down to weights resting on a curved surface. The surface is constructed such that the slope doubles proportionally to the distance down the slope.

Notes:
- This setup is obviously stretching the limits of practical 'calculus without math'. It stretches it both with the 3-dimensionality and with the simulation of a geometrically increasing force.
- In practice, the circular nature of the funnel is a needlessly complex design. You would get the same result with a rectangle that curves on only one axis - like a roof shingle - if you positioned it far enough from the sphere to ensure any hypotenuse-induced effects were negated.
- What I don't know is whether the 'granularity' of the experiment is sufficient. i.e: my sphere has only twelve masses. It is conceivable that the net zero effect only happens when the granularity of the masses gets very large.

Last edited: Dec 17, 2004
11. Dec 17, 2004

### rayjohn01

I think the original question is interesting from a general viewpoint -- I believe this type of problem was the reason why Newton did not publish his Principia earlier , first it pre-supposes that gravity has an inverse square law , I do not think this is simply intuitive , one) because there is no reason to suppose that gravity acts in straight lines ( at the time )
two) that gravity from one point mass should not be screened by another , and that the effects on a third should add vectorially -- Newton had to work all those things out before he could model the earth or the moon as point masses.
I believe that there are many problems which do not have simple intuitive solutions because intuition implies some sort of prior knowledge , not just that, but on many occasions the maths ( a formal way of thinking ) results in solutions which were never anticipated -- were not intuitive and a TOTAL surpise -- see the 'positron' by Dirac .
Quantum mechanics -- the two slit experiment -- and the constancy of light speed .
If we have no experience then intuition can in fact be very misleading.
Ray.

12. Dec 17, 2004

### Bartholomew

Of course, intuition can often be wrong; it must be carefully molded before it is reliable, yes, often according to information which was obtained in other ways. I dislike the term "counter-intuitive" for this reason. Everything is "counter-intuitive" until you understand it; it just means your intuition is not good enough yet. Once you understand it, it becomes intuitive. So, I'm not talking about intuition as any kind of a priori knowledge.

I don't know much about the positron, but I don't believe there is any good idea which cannot be made intuitive with sufficient understanding.

Dave, how did you make that picture? I'm just wondering.

I don't think the setup is quite right. In your diagram, the gravity increases according to the square of the central ball's distance from the center as the central ball goes towards one of the holes. You want the force of the string on the central ball to increase to a very large figure, theoretically infinite, as the central ball goes towards one of the holes.

But if you somehow fixed that force problem, it does not look like the granularity would be much of a problem--a dodecahedron is close to a sphere. That might be enough to make the central ball slide around a little. In particular it would tend to be pulled through any hole it got close to--a slight attraction towards the sphere wall, though I'm not sure if that attraction would exist except at the edges. That's because if you imagine it as a dodecahedron and imagine yourself getting close to the center of one of the faces, the distance changes as usual but the mass ratio I mentioned does not change as fast (none of the face disappears).

13. Dec 17, 2004

### DaveC426913

Entirely manually drawn in Adobe Illustrator. (I'm going to be presumptuous and be flattered by the fact that you're not sure how I made it. Maybe you thought it was computer generated... )

My diagram is schematic. In reality, you'll design the slope carefully to provide a quadrupling of pull when the distance is halved.

Say the sphere is 12 inches in radius. When the ball is 12 inches from a hole (i.e in the centre), the pull on all strings is (say) 1 Newton. Move the ball to within 6 inches of a particular hole, and the pull from that string should be 4 Newtons. It does fall apart as you get too close to the holes, because you can't make the weights with no upper limit on their mass. However, that's why you make the holes big! At some point, you say, 'this experiment fails when distances get down to X inches'. It can still be made to prove the point though.

No, that would imply that, as you approach a planet, gravity increases to infinity, which it clearly does not.

14. Dec 17, 2004

### Bartholomew

As you approach a point mass (Newtonian), gravity increases to infinity. You can't get too close to the point mass that could model the gravity of a planet because the planet is in the way.

My mistake about the slope, you're right, it would work. You'd also have to factor the changing angles of the strings into how you calculate the slope--if the string is more vertical then I think the ball would tend to settle lower because the tension is not as parallel to the slope. For a practical demonstration you should probably have some kind of little pulleys to reduce friction on the holes. Also the weights coming from the top of the sphere should be slightly heavier, or on a slightly sharper slope, to counter the force of real gravity on the central knot.

Yes, I was wondering whether you had used some kind of CAD to draw it. It looked very "clean." I haven't used Adobe Illustrator.

Last edited: Dec 17, 2004
15. Dec 18, 2004

### sekta

why wouldnt there be a mid point to the black hole? there has got to be a dead space inside to counter the amount of pull. true, we may never know that BUT...

and i know im not on the same level as you guys. i have been reading your posts you guys are some smart people! :yuck:

16. Dec 18, 2004

### DaveC426913

"You'd also have to factor the changing angles of the strings into how you calculate the slope--if the string is more vertical then I think the ball would tend to settle lower because the tension is not as parallel to the slope."

Yup. Though there are other ways to minimize this. More pulleys will minimize the angles (frictionless of course), or putting the funnel much farther from the sphere.

BTW, you can reverse the funnel, making it a dome (i.e. upsidedown cereal bowl), with the weights going down the outside. This gives you more freedom in designing angles and distances.

"...the weights coming from the top of the sphere should be slightly heavier, or on a slightly sharper slope, to counter the force of real gravity on the central knot...."

Well, the knot is merely a knot in the string. The string is effectively weightless compared to the weights.

CorelDraw? Same thing.

17. Dec 18, 2004

### DaveC426913

"why wouldn't there be a mid point to the black hole? there has got to be a dead space inside to counter the amount of pull. true, we may never know that BUT... "

I don't think he was asking about a "regular" spherical black hole. Yes, at the dead centre of a regular black hole, the net gavity would be zero, just like at the dead centre of the Earth, the net gravity is zero. (ignore the practicality of what actually[/] happens at the centre of a black hole, since I believe otherwise the question is meaningless.)

"What about black holes? What if a black hole was hollow? Would there be no gravity on objects in its hollow? "

I don't know if it's physically possible for a black hole to be formed in the shape of a hollow hollow sphere. A black hole is a phenomenon, not merely a construct of matter.

However, if we propose an object that is shaped like a hollow sphere, and has the gravitational equivalent of a black hole ... then...

You would experience a zero *net* force at any point inside the sphere. That doesn't mean you don't feel gravity, it just means the gravity is pulling you in every direction at once to the same amount. In truth, in a black hole, the gravity is so great you would be ripped apart limb from limb.

But yes, no matter where you positioned yourself within the hollow (not just at the centre), you would feel the same pull in every direction.

18. Dec 18, 2004

### Bartholomew

Well, a black hole is not governed by Newtonian physics anyway... but if it were, and it could somehow be hollow, the pull in all directions would be felt by every particle of your body, down to as small a level as you care to go. So you wouldn't feel anything at all, it would be just a zero gravity situation. For you to be torn apart the force on one part of your body would have to be different from the force on another part of your body.

19. Dec 18, 2004

### DaveC426913

No...

Wait, yes.

You're right! I was going to say the gravitational gradient near a black hole is so strong it can have an effect even over as short distances as the length of your body, which is you'd get torn apart (one arm would get pulled one way while the other arm would get pulled the other way).

But, as we've just concluded, all points within the sphere will have a net zero gravity.

Which means, yes, if you could build a black hole into a hollow sphere, you could float around inside it, nary the wiser!

Wow. That's weird. (Or would be, if BHs obeyed Newtonian mechanics.)

20. Dec 19, 2004

### Kittel Knight

You must be kidding!
Take a look:

This guy has no idea what a black hole is! :yuck:

Do you really believe this black hole woldn't attract you , just cause you both would be inside a hollow sphere?! :yuck:

Sorry , I couldn't resist! :rofl:

21. Dec 20, 2004

### DaveC426913

Amusing, you, claiming someone else doesn't know what they're talking about.

Exotically-shaped black holes are a common subject. Google 'Kerr black holes' for example. Or maybe put down the game controller and pick up a 'B' 'O' 'O' 'K'.

Last edited: Dec 20, 2004
22. Dec 21, 2004

### Kittel Knight

:yuck:

PLZ, don't lend me your books! :yuck:

23. Dec 25, 2004

### DaveC426913

Fair enough, but hey, don't take my word for it.

Did you Google Kerr black holes?

24. Dec 27, 2004

### donjennix

KRAB got to one of the essential points about the "hollow sphere" problem when he showed the relation between the distance the test particle is from the sides of the sphere versus the attraction on the particle due to the effective mass. Basically the lack of attraction inside a hollow (uniform) sphere is a result of and a test of the inverse square law. The law has been extensively tested for columb's inverse square rule for charges - down to errors of 10^(-16) or something similar. If the gravitational attraction is not governed by the inverse square law then the effect will be absent (if the law is inverse 2.0001 power, it would be quite hard to detect, of course).

Talking about black holes in this context is kind of odd. As another contrubuter pointed out, the inverse-square law is Newton's law. The characteristics of black holes cannot be deduced from Newton's physics and thus trying to make a conjecture of how one would behave in the Newtonian universe is not sensible. I don't know how to work out the field equations, but I bet that if the walls of the uniform sphere began to approach relativistic mass that many odd consequence would happen - one would probably be "frame dragging".

Last edited: Dec 27, 2004
25. Dec 27, 2004

### Integral

Staff Emeritus
The link to the Kerr Black hole referred to space inside of the EVENT HORIZON, this is not the same as "inside" of a black hole. The mass of black hole is concentrated inside of a singularity. The Singularity is caused by the INFINITE density of the black hole. It is NONSENSICAL to speak of the inside of a black hole. For an object to have an inside, it must have a measurably large volume, if a black hole had a measurable volume it would not have an infinite density. Please stick to the topic.