Something I've always wondered

[Firepanda]

If you dug out a hollow sphere in the centre of the earth, ignoring heat or pressure or any chance that the hole would collaspe in on itself, then would I just float there if I was in it? Or would i be stretched out in every direction?

 

[atyy]

Just float.

 

[Ilivian]

Float and stretched :D

 

[TurtleMeister]

Float and stretched :D

Actually, I think he would just float no matter where he was located within the sphere and no matter how large the sphere.

 

[coverme]

your body is not perfectly sphere(not a sphere in any sense), and the gravity is zero at a central point only(a point you know, really unimaginable), but your body lies over a point(considerabaly large) I think your body receives unequal force and your body will be broken disaster.

 

[HallsofIvy]

No. Turtle Meister is correct. If you cut a spherical hole in the center of the Earth the mass outside the hole cancels at any point inside that hole. As long as the hole was large enough to hold your entire body, you would feel no gravitational pull on any part of your body.

Consider a spherical surface inside the matter. From any point inside the hole, imagine a cone from that point to the spherical surface in both directions. The gravitational force on that point due to the part of the spherical surface inside that cone is proportional to 1/r², where r is the distance from the point to the spherical surface but the amount of surface inside the cone is proportional to r². The two cancel out.

 

[Dadface]

No. Turtle Meister is correct. If you cut a spherical hole in the center of the Earth the mass outside the hole cancels at any point inside that hole. As long as the hole was large enough to hold your entire body, you would feel no gravitational pull on any part of your body.

Consider a spherical surface inside the matter. From any point inside the hole, imagine a cone from that point to the spherical surface in both directions. The gravitational force on that point due to the part of the spherical surface inside that cone is proportional to 1/r², where r is the distance from the point to the spherical surface but the amount of surface inside the cone is proportional to r². The two cancel out.

I'm not so sure that is correct.At any point in the hole the net gravitational pull is zero because the attraction in anyone particular direction is balanced by an equal pull in the opposite direction the result being a stretching force as suggested above.

 

[atyy]

Just float.

Ok, I correct myself. Considering only the force from the sphere, you'd just float. But since your finger will attract your toe gravitationally, there will be some stretching or compression there. Of course, your finger will attract your toe even if you are in completely empty space.

 

[Nick89]

I'm not so sure that is correct.At any point in the hole the net gravitational pull is zero because the attraction in anyone particular direction is balanced by an equal pull in the opposite direction the result being a stretching force as suggested above.

If there was a 'stretching force' that would mean there would have to be a net force on your body. Since the net gravitational force everywhere inside the sphere is zero, there is no such force, and there is no stretching.

 

[Bob S]

Since the gravitational force varies linearly with distance from the center of the Earth for r<R (radius of Earth), you will feel essentially no gravity. If you were in a truly inertial reference frame, stabilized by gyroscopes, you will find your enclosure rotating around you.

 

[Dadface]

If there was a 'stretching force' that would mean there would have to be a net force on your body. Since the net gravitational force everywhere inside the sphere is zero, there is no such force, and there is no stretching.

But you are pulled equally in all directions and this is why the net force is zero.

 

[Nick89]

But you are pulled equally in all directions and this is why the net force is zero.

Yes. Every particle in your body is pulled on equally in all directions; or equivalently: there is no pull at all.

You seem to be mixing this up with being pulled by your feet and your hands for example. If you are being pulled by your hands and feet, in opposite directions, then yes, you will feel that. However, that is not the same! The net force in that example is still zero, but only if you take your whole body as 'the system'. If you are in the center of the Earth in a hollow sphere, every particle in your body feels no net force. So there is no pull. It's a different scenario!

 

[Dadface]

Yes. Every particle in your body is pulled on equally in all directions; or equivalently: there is no pull at all.

You seem to be mixing this up with being pulled by your feet and your hands for example. If you are being pulled by your hands and feet, in opposite directions, then yes, you will feel that. However, that is not the same! The net force in that example is still zero, but only if you take your whole body as 'the system'. If you are in the center of the Earth in a hollow sphere, every particle in your body feels no net force. So there is no pull. It's a different scenario!

Imagine an object with a single force on it .The object will accelerate in the direction of that force.Now imagine the same object with the same original force on it but now with a second force of equal magnitude but acting in the opposite direction .As an example the object could be a length of wire in equilibrium supporting a weight, the wire itself being supported at the top.The net force is zero but the wire will have stretched its molecules having been pulled further apart..
Of course the object I have referred to is macroscopic but you want to go into the microscopic i.e into particles so let's go there.Firstly we know that collections of molecules can be separated by forces.Next, if we take a single molecule we know that it can be stretched eg in an electric field it can be polarised.Getting smaller still can protons and electrons etc be stretched or distorted in some way?Remember that a macroscopic object is made of real separate microscopic particles as is the wire in my example above,it is not made of point particles.
It's a great question though Nick and I may try to think about it some more.I may come back and change my mind.

 

[Nick89]

The fact that particles may or may not be able to be stretched is beside the point.

In the example of the rope you use, you did not say on which part of the wire the forces are acting. If one force is acting on one end of the wire, and the other force on the other end, then yes it will be stretched. But, in this case, the forces act on the same point of the wire! So the wire is not stretched.

Of course, there are forces acting on one end of the wire, and other forces on the other end. But if you look at one point on the wire (it doesn't even matter if we take that point to be an atom, or a proton, or a hypothetical position in space), the net force on that point will be ZERO.

This can easily be proven using Newton's laws of gravitation, and is valid for any point inside a hollow sphere. So, every point on your wire will have NO net force acting on it, and subsequently, it will not be stretched.
EDIT
Here's a picture that will hopefully clarify it for you:

In (a), there is no net force on the chain / wire. But the chain is still stretched, because the two equal but opposite forces don't act on the same position in the chain.

In (b), there is still no net force on the chain / wire. But now, the chain is not stretched. Each particle / point in space feels two equal but opposite forces. There is no net force on any particle / point in space.In a hollow sphere, you have situation (b) (except that the forces acting on your are uniformly around you in a sphere, instead of just two forces horizontally; but that's a bit hard to draw ;) ).

 

[Doc Al]

Nick89 is correct. In order for an object within the hollow to be stretched due to gravity (aka, experience a tidal force), there would need to be a variation of gravitational field strength (force per unit mass) with position within the hollow. But the gravitational field within the hollow is uniformly zero. Not only is there no tidal force, there's no gravitational force at all.

 

[Dadface]

Nice one Nick, thank you very much for clarifying things.

And thank you Doc Al

 

[TeTeC]

I agree with Nick89. See for example Hecht's textbook on introductory physics. There is a chapter about gravity, and this "theorem" is demonstrated. I'm not giving a page number because I have a french version.

This is also true in General Relativity.

 

[Nick89]

Of course, this is all neglecting any outside forces such as gravitational force from the sun and other planets. Usually they can be safely ignored because they are so small, but in the absence of the Earth's gravitational pull I'm not so sure...

 

[Cantab Morgan]

This can easily be proven using Newton's laws of gravitation, and is valid for any point inside a hollow sphere. So, every point on your wire will have NO net force acting on it, and subsequently, it will not be stretched.

But perhaps not quite so easily proven. If I recall correctly, Newton himself delayed publishing the result for many years, because he was not satisfied that his proof (requiring a volume integral) was good enough.

 

[Buckleymanor]

Nick89 is correct. In order for an object within the hollow to be stretched due to gravity (aka, experience a tidal force), there would need to be a variation of gravitational field strength (force per unit mass) with position within the hollow. But the gravitational field within the hollow is uniformly zero. Not only is there no tidal force, there's no gravitational force at all.

The original OP was a hollow inside the Earth.
The Earths crust and it's oceans are not uniform so there would be variations of the gravitational field strength.
Not to mention that the Earth rotates so you would be flung to the side of the hollow by variations in the field strength and centrapetal effects.

 

[Doc Al]

The original OP was a hollow inside the Earth.
The Earths crust and it's oceans are not uniform so there would be variations of the gravitational field strength.

Well, it's certainly true that Newton's shell theorems only apply for a spherically symmetric mass distribution.

 

[coverme]

earth a sphere (this is clear) now make a hollow concentric sphere.,your body is inside sphere.all the gravitational forces are mean to terminate at the centre of this hollow sphere (vector sense) , and acceleration due to gravity will be zero at this point (at r=o)because all the force cancels but our body is bigger than the point isnot it? and our body is not rigid as the Earth part,so the force must have some effect in the body part, so i assumed it to forced to be rolled to become sphere, is it fine??

 

[Doc Al]

earth a sphere (this is clear) now make a hollow concentric sphere.,your body is inside sphere.all the gravitational forces are mean to terminate at the centre of this hollow sphere (vector sense) , and acceleration due to gravity will be zero at this point (at r=o)because all the force cancels but our body is bigger than the point isnot it? and our body is not rigid as the Earth part,so the force must have some effect in the body part, so i assumed it to forced to be rolled to become sphere, is it fine??

The point is that the gravitational field is zero everywhere within the hollow of a spherically symmetric shell of mass, not just at the exact center. Look up Newton's Shell theorems.

 

[coverme]

could you tell me at which reference site or book this shell ntheorem lies

 

[Doc Al]

could you tell me at which reference site or book this shell ntheorem lies

You should be able to find it discussed in most classical mechanics books (or calculus books, for that matter). Here's a site: http://en.wikipedia.org/wiki/Shell_theorem"

 

[Klockan3]

Of course, this is all neglecting any outside forces such as gravitational force from the sun and other planets. Usually they can be safely ignored because they are so small, but in the absence of the Earth's gravitational pull I'm not so sure...

No, this wouldn't make a difference if you do not consider the tidal effects, the deal is that the sun pulls as much on the Earth as on you. Of course the tidal effect would pull you away or towards the sun/moon though(Mostly towards the moon since it is closer, even though the gravitic pull is greater from the sun), and the tidal effects are really really weak so you wouldn't notice them since you are so close to the centre of Earth anyway.

Not to mention that the Earth rotates so you would be flung to the side of the hollow by variations in the field strength and centrapetal effects.

Can you clarify this? The centripetal effect is not valid in this scenario at all unless you consider that it makes the Earth's mass distribution form as an ellipsoid instead of a sphere, and even then that effect is really minimal, the Earth is extremely close to being a sphere.

The point is that the gravitational field is zero everywhere within the hollow of a spherically symmetric shell of mass, not just at the exact center. Look up Newton's Shell theorems.

Well, if you utilize the fact that gravity fields are divergence free in vacuum we can just put a spherical gaussian surface inside the earth, and then due to symmetry all sides of this surface must have equal gravity flux while at the same time the divergence inside is 0 which means that the theorem is proven. It is fairly simple, and the deal with it being divergence free comes automatically from dissipating proportionally to the square of the distance.

 

[Doc Al]

Well, if you utilize the fact that gravity fields are divergence free in vacuum we can just put a spherical gaussian surface inside the earth, and then due to symmetry all sides of this surface must have equal gravity flux while at the same time the divergence inside is 0 which means that the theorem is proven. It is fairly simple, and the deal with it being divergence free comes automatically from dissipating proportionally to the square of the distance.

Absolutely. Using Gauss's law for gravity is the easy and elegant way to prove the shell theorems. But brute force integration is still instructive. (And may be more accessible to students just learning about Newton's law of gravity.)

 

[Buckleymanor]

Can you clarify this? The centripetal effect is not valid in this scenario at all unless you consider that it makes the Earth's mass distribution form as an ellipsoid instead of a sphere, and even then that effect is really minimal, the Earth is extremely close to being a sphere.

Minimal or not. objects weigh 0.5% more at the poles because of a combination of the Earths oblateness and centrapetal effects.
So if you were to hollow out a symetrical sphere in the middle of the Earth a body placed in the sphere would be gravitationaly attracted more towards the equator by a similar amount.
Due to the greater mass surounding at the equator and speed of rotation.

 

[granpa]

the suns tidal forces would be tiny near the center of the Earth and exactly zero at the exact center.

 

[Klockan3]

Minimal or not. objects weigh 0.5% more at the poles because of a combination of the Earths oblateness and centrapetal effects.
So if you were to hollow out a symetrical sphere in the middle of the Earth a body placed in the sphere would be gravitationaly attracted more towards the equator by a similar amount.
Due to the greater mass surounding at the equator and speed of rotation.

No, it would not be a similar amount, that is not an effect by the gravitation but the centripetal, the change in gravitational field will be extremely minimal due to that effect.

 

[coverme]

You should be able to find it discussed in most classical mechanics books (or calculus books, for that matter). Here's a site: http://en.wikipedia.org/wiki/Shell_theorem"

Thank you very much