Is there really no gravitational force at the center of a massive object?

[DaveC426913]

At the centre of a very massive object, the gravitational force is zero. Is that net zero or is it gross zero?

What I mean is: If two people are pulling on my limbs with equal force in opposite directions, my net movement is zero, but I am under great stress. The net forces on me sum to zero, but the gross forces are that of two people pulling on me.

Is it the same with gravity? If I were at the center of the Earth, is there a test I could do to measure the absolute gravitational pull on me?

 

[Doc Al]

Imagine the massive object is perfectly symmetric and that there is a spherical cavity at its center. What's the gravitational field within that cavity? Zero. You would experience no gravitational pull whatsoever within that cavity.

 

[DaveC426913]

Imagine the massive object is perfectly symmetric and that there is a spherical cavity at its center. What's the gravitational field within that cavity? Zero. You would experience no gravitational pull whatsoever within that cavity.

Yes. This is exactly what I've stated in my OP.:grumpy:

My question is: does gravity "balance out" to zero, or is there a way of determining that I am in a strong field that merely "nets" zero?

For example: If I drew a "rubber sheet" diagram of the Earth's grav field, the spot in the centre would have zero curvature to be sure, but that spot would stiill be in a depression (relative to interplanetary space). Does the "depth" of the depression in the rubber sheet analogy have a counterpart in GR? Can I detect that I am in a depression, or can I only detect the curvature? (I know the rubber sheet analogy is flawed, but it's suiting my purpose to explain my question)

 

[Doc Al]

Yes. This is exactly what I've stated in my OP.:grumpy:

What about "no gravitational field whatsoever" was unclear?

My question is: does gravity "balance out" to zero, or is there a way of determining that I am in a strong field that merely "nets" zero?

I'm not sure what "a strong field that nets zero" means. The field isn't strong and yet somehow "cancels out"--it's nonexistent.

I understand your analogy of two people pulling your limbs, where there's no net force on you but still great tension. The gravity case is not like that: no one's pulling on you.

For example: If I drew a "rubber sheet" diagram of the Earth's grav field, the spot in the centre would have zero curvature to be sure, but that spot would stiill be in a depression (relative to interplanetary space). Does the "depth" of the depression in the rubber sheet analogy have a counterpart in GR? Can I detect that I am in a depression, or can I only detect the curvature? (I know the rubber sheet analogy is flawed, but it's suiting my purpose to explain my question)

If you drew a diagram of the Earth's gravitational field, the spot in the center would read zero, not have a slope of zero. (Perhaps you're thinking of a diagram of gravitational potential?)

 

[Dale]

My question is: does gravity "balance out" to zero, or is there a way of determining that I am in a strong field that merely "nets" zero?

For example: If I drew a "rubber sheet" diagram of the Earth's grav field, the spot in the centre would have zero curvature to be sure, but that spot would stiill be in a depression (relative to interplanetary space). Does the "depth" of the depression in the rubber sheet analogy have a counterpart in GR? Can I detect that I am in a depression, or can I only detect the curvature? (I know the rubber sheet analogy is flawed, but it's suiting my purpose to explain my question)

You would not have any gravitational acceleration and no tidal forces or other stresses, but you would still have time dilation.

 

[DaveC426913]

So, if you were at the centre of a neutron star (not that this is at all practically possible, I'm just trying to scale up the magnitude of the forces...)
So - if you were at the centre of a neutron star, in an open space a mere mile across, you would not be able to detect the effects of so much mass so near? The gravitational field would literally be non-existent?

I mean, you could do some tests, such as measuring yoiur subjective time against a clock out in flat space, but otherwise, you would experience no untoward effects?

 

[Danger]

A hole a mile across in the middle of a neutron star wouldn't leave much neutron star. They're not more than a couple of miles across to start with.
As for comparing your situation with an 'external' clock, I suspect that time dilation effects would screw that up.

 

[belliott4488]

If you drilled an infinitessimally thin tunnel through the center of a massive body, and then measured the force along the length of this tunnel, you'd find that the force varied linearly with the distance from the center (as you probably know). This means that the potential would have a parabolic shape (just as with the simple harmonic oscillator). This is the correct way to think of the rubber sheet analogy; the rubber sheet would be flat right at the center of the parabola, thus no forces from any direction.

That might not really address your question, though. I suspect what you're asking gets to the superposition principle, which says that the field at any point is equal to the sum of the contributions from all sources affecting that point. It's not so much that all the sources exert forces that cancel out on the test object (even though I've heard it stated that way); it's really that the field itself is canceled out, so that it is not there at all.

 

[sadhu]

suppose you are at center of the Earth there will be very little or no tension on you
because th part of your body exactly at the center will experience no force but perts around it will experience force net balance of all forces at the very point ,but since the forces vary very little with small changes in position due to massive size of Earth hence net force acting on your limbs will be negligible.

the of rope pulling your hand is true but only if you understand that forces are not discrete in gra. case...

 

[DaveC426913]

suppose you are at center of the Earth there will be very little or no tension on you
because th part of your body exactly at the center will experience no force but perts around it will experience force net balance of all forces at the very point ,but since the forces vary very little with small changes in position due to massive size of Earth hence net force acting on your limbs will be negligible.

This is not true.

You are suggesting that the only place you experience no gravitational field is at the exact centre of the hollow - that your arms and legs, being a small distance from the centre will experience a small force.

No.

If you float around inside a hollow sphere at the centre of the Earth or the centre of a neutron star, you will not experience a force due to gravity anywhere in that hollow sphere. you could float all the way over and hug the wall and you would still not experience any force due to gravity.

 

[DaveC426913]

It's not so much that all the sources exert forces that cancel out on the test object ... it's really that the field itself is canceled out, so that it is not there at all.

This is what I'm asking, yes.

OK, so I guess the overwhelming answer I'm getting is that the field really does disappear.

 

[DyslexicHobo]

This is not true.

You are suggesting that the only place you experience no gravitational field is at the exact centre of the hollow - that your arms and legs, being a small distance from the centre will experience a small force.

No.

If you float around inside a hollow sphere at the centre of the Earth or the centre of a neutron star, you will not experience a force due to gravity anywhere in that hollow sphere. you could float all the way over and hug the wall and you would still not experience any force due to gravity.

How do you figure this? I could be wrong, but since your head would be close to the top of the sphere (we'll just call the top the part that's closest to your head :) ), the distance between your head and those particles would be smaller than the distance between your head and the bottom of the sphere, so the gravitational force between your head and the top would be greater than the force between your head and the bottom. If I'm interpreting what you're saying correctly, the whole storyline of Halo would be destroyed! The inner parts of all the halos would be free-floating without gravity.

 

[neutrino]

If I'm interpreting what you're saying correctly, the whole storyline of Halo would be destroyed! The inner parts of all the halos would be free-floating without gravity.

Dave is talking about Gauss' Law as applied to gravitational fields.

 

[Doc Al]

So, if you were at the centre of a neutron star (not that this is at all practically possible, I'm just trying to scale up the magnitude of the forces...)
So - if you were at the centre of a neutron star, in an open space a mere mile across, you would not be able to detect the effects of so much mass so near? The gravitational field would literally be non-existent?

I mean, you could do some tests, such as measuring yoiur subjective time against a clock out in flat space, but otherwise, you would experience no untoward effects?

Right--that's what I would say.

OK, so I guess the overwhelming answer I'm getting is that the field really does disappear.

Right again.

How do you figure this? I could be wrong, but since your head would be close to the top of the sphere (we'll just call the top the part that's closest to your head :) ), the distance between your head and those particles would be smaller than the distance between your head and the bottom of the sphere, so the gravitational force between your head and the top would be greater than the force between your head and the bottom.

Look up Newton's Shell theorems! The gravitational field within a uniform spherical shell is zero everywhere.

 

[sadhu]

This is not true.

You are suggesting that the only place you experience no gravitational field is at the exact centre of the hollow - that your arms and legs, being a small distance from the centre will experience a small force.

No.

If you float around inside a hollow sphere at the centre of the Earth or the centre of a neutron star, you will not experience a force due to gravity anywhere in that hollow sphere. you could float all the way over and hug the wall and you would still not experience any force due to gravity.

right you are ,if you consider Earth as hollow sphere(Newtons shell theorem)
but what i mentioned was taking Earth as a solid sphere
if you consider Earth a hollow sphere then their will be absolutely no tension on any part of your body

 

[Dale]

So, if you were at the centre of a neutron star (not that this is at all practically possible, I'm just trying to scale up the magnitude of the forces...)
So - if you were at the centre of a neutron star, in an open space a mere mile across, you would not be able to detect the effects of so much mass so near? The gravitational field would literally be non-existent?

I mean, you could do some tests, such as measuring yoiur subjective time against a clock out in flat space, but otherwise, you would experience no untoward effects?

In Newtonian gravity there would be no effect whatsoever. I can demonstrate that mathematically fairly easily.

In GR I believe that there would be no curvature anywhere inside the shell, so there would be no tidal forces etc. and only time dilation. But, I don't have the math to back that up at all.

 

[Janus]

How do you figure this? I could be wrong, but since your head would be close to the top of the sphere (we'll just call the top the part that's closest to your head :) ), the distance between your head and those particles would be smaller than the distance between your head and the bottom of the sphere, so the gravitational force between your head and the top would be greater than the force between your head and the bottom. If I'm interpreting what you're saying correctly, the whole storyline of Halo would be destroyed! The inner parts of all the halos would be free-floating without gravity.

Try visualizing it this way:

Draw two concentric circles to represent the "shell". Draw an object just inside of the inner circle to represent your head close to the top of the hollow. Now draw a line through this object perpendicular to the radius. Every part of the shell "above" this line works together to produce a pull "upward" on your head and every part of the shell "below" this line works together to produce a pull "downward". Now while your head is closer to the part of the shell "above" it, there is a lot more of the shell "below" it. The extra mass in one direction exactly cancels out the decreased distance in the other.
This is true no matter where you put your object in the sphere.

As far as Halo goes, if the game assumes otherwise (and I don't know, as I've never played it.) then they have got the physics wrong. But what do you expect? It's just a video game.

 

[DaveC426913]

How do you figure this? I could be wrong, but since your head would be close to the top of the sphere (we'll just call the top the part that's closest to your head :) ), the distance between your head and those particles would be smaller than the distance between your head and the bottom of the sphere, so the gravitational force between your head and the top would be greater than the force between your head and the bottom.

That may seem logical but it is false. Indeed, there is no gravity inside a hollow sphere.

If I'm interpreting what you're saying correctly, the whole storyline of Halo would be destroyed! The inner parts of all the halos would be free-floating without gravity.

Correct me if I'm wrong, but HALO is based on a ring - a la Larry Niven's Ringworld. The gravity on a ringworld is not due to mass, it is artifical gravity from rotation, just like any other rotating space station writ large. The Ringworld rotates around its star at 770km/s.

 

[belliott4488]

How do you figure this? I could be wrong, but since your head would be close to the top of the sphere (we'll just call the top the part that's closest to your head :) ), the distance between your head and those particles would be smaller than the distance between your head and the bottom of the sphere, so the gravitational force between your head and the top would be greater than the force between your head and the bottom. If I'm interpreting what you're saying correctly, the whole storyline of Halo would be destroyed! The inner parts of all the halos would be free-floating without gravity.

hm ... I posted a response to this earlier today, but it seems to have vanished. Here's the gist of it:

While the mass is one direction is clearly at a larger distance (on average) than the mass in the opposite direction there's also more of it. One of the wonderful things about inverse-square laws is the force from uniformly distributed sources is independent of the distance. That's because if you look at the mass (for the case of gravity) in a given direction over some solid angle, the amount of mass goes up as the square of the distance (think of the spherical area subtended by this angle). The force from the mass goes down as the inverse square of the distance, however, so the two dependencies cancel each other, and you end up with a constant force in all directions.

This is the same argument Newton used to show that the universe could not be infinite. Since the intensity of starlight drops as the inverse square of distance, but the number of stars in a given solid angle goes up as the square of the distance, an infinite universe (uniformly filled with stars) would produce a blindingly bright sky. As it happens, there are other reasons why the sky is dark, but the argument is still sound.

 

[DyslexicHobo]

Aha! That is very cool! I never thought about it that way, thanks for the enlightenment. Here I was thinking I was being helpful. :P It's always nice to learn something new.

 

[belliott4488]

Aha! That is very cool! I never thought about it that way, thanks for the enlightenment. Here I was thinking I was being helpful. :P It's always nice to learn something new.

Glad to help, and glad that it made sense - I wasn't sure if I was explaining things clearly enough, especially the bit about the number of stars (or amount of mass) in a given solid angle increasing with r^2.

 

[DyslexicHobo]

Glad to help, and glad that it made sense - I wasn't sure if I was explaining things clearly enough, especially the bit about the number of stars (or amount of mass) in a given solid angle increasing with r^2.

I *think* I can visualize what's happening, but I'm not quite sure. I do realize that as a particle approaches the inside wall of a sphere, there is more mass on the opposite side pulling it towards the center; the part of the sphere that the particle is close to is exerting more of a force than the rest of the sphere... a lot of a little, and a little of a lot cancel out. :P

I can't even explain the way I think I know it, so I'm not sure if I'm understanding it completely right.

 

[DaveC426913]

I *think* I can visualize what's happening, but I'm not quite sure. I do realize that as a particle approaches the inside wall of a sphere, there is more mass on the opposite side pulling it towards the center; the part of the sphere that the particle is close to is exerting more of a force than the rest of the sphere... a lot of a little, and a little of a lot cancel out. :P

I can't even explain the way I think I know it, so I'm not sure if I'm understanding it completely right.

Yep. I'ts hard to visualize, but you've got it.

 

[belliott4488]

I *think* I can visualize what's happening, but I'm not quite sure. I do realize that as a particle approaches the inside wall of a sphere, there is more mass on the opposite side pulling it towards the center; the part of the sphere that the particle is close to is exerting more of a force than the rest of the sphere... a lot of a little, and a little of a lot cancel out. :P

I can't even explain the way I think I know it, so I'm not sure if I'm understanding it completely right.

Yes, you definitely have the general idea. A rigorous mathematical proof would require evaluating an integral to add up all the contributions to the force from all the bits of mass in the spherical shell; that would show that they all cancel perfectly. Short of that, however, you can just take in on faith that the math works out and confirms your intuition. Keep in mind, however, that this is very much dependent on the force following an inverse square law - otherwise it wouldn't balance out so conveniently - I don't think that part is so intuitive.

 

[capnahab]

When you are at the center of the mass does net or gross make any difference?

 

[Doc Al]

When you are at the center of the mass does net or gross make any difference?

Right now, the net force on you is zero (or close enough for discussion purposes). If a 40 ton boulder were placed on top of you, the net force on you would still be zero.

Notice any difference?

 

[capnahab]

Would this be called "death by crushing?" But if you are at the theoretical center of mass does it make any difference?

 

[DaveC426913]

Would this be called "death by crushing?" But if you are at the theoretical center of mass does it make any difference?

Hang on. It's not a direct analogy. You don't experience any gravitational forces at the centre of a mass (unlike at the centre of a rock and a hard place ).

Doc Al was simply pointing out that, in the general case, when talking about net forces versus gross forces, there is often a substantial difference. The upshot is that my opening question about gravity is a valid question - even if the final answer is still a "no".

 

[DaveC426913]

So, because
1] I like to beat things until they're not just dead, but parts start falling off and
2] I like to do graphics,

I've graphicized the original question.

In the graph, distance d is 100% an artifact of the illustrative representation, and has absolutely no real-world counterpart. There is no direct measurement I could do, even in principle (such as, say a gravitometer) that that would give me distance d. Correct?

 

[disregardthat]

I am wondering about something concerning gravitys effect on mass. Consider a body under a gravitational force. I have the impression that it "pulls" every atom with the same force (I am not taking change of gravitational force for the parts of the body further away from the gravitational source in consideration) in such a way that you won't feel anything (no compression or stretching). Is this correct?

 

[dst]

If you approached the OP from general relativity, you would find the hollow section to be a uniformly curved blob such that all the curvature added = 0 curvature and hence no "field"? How would time dilation be a factor here?

 

[Dale]

If you approached the OP from general relativity, you would find the hollow section to be a uniformly curved blob such that all the curvature added = 0 curvature and hence no "field"? How would time dilation be a factor here?

You would have no time dilation inside the shell, but you would have time dilation relative to points outside the shell. A photon emitted inside the shell would not be redshifted as it traversed the inside, but after it got outside the shell it would be redshifted as it went up to a distant observer. Sorry if I wasn't clear.

 

[DaveC426913]

You would have no time dilation inside the shell, but you would have time dilation relative to points outside the shell. A photon emitted inside the shell would not be redshifted as it traversed the inside, but after it got outside the shell it would be redshifted as it went up to a distant observer. Sorry if I wasn't clear.

I'm not sure that the notion of time dilation makes sense without comparing to another frame of reference.

I think I see what you mean though.

You would have no time dilation [measuring two points that are both] inside the shell, but you would have time dilation relative to points [inside the shell relative to points] outside the shell. A photon emitted inside the shell would not be redshifted as it traversed the inside, but after it got outside the shell it would be redshifted as it went up to a distant observer. Sorry if I wasn't clear.

 

[belliott4488]

I am wondering about something concerning gravitys effect on mass. Consider a body under a gravitational force. I have the impression that it "pulls" every atom with the same force (I am not taking change of gravitational force for the parts of the body further away from the gravitational source in consideration) in such a way that you won't feel anything (no compression or stretching). Is this correct?

This is a different question and really ought to be in its own thread.

The short answer, however, is that you are asking about what are called "tidal forces". They are generally negligible unless the physical extent of the object in question, i.e its size, is large with respect to the gradient of the field.

 

[belliott4488]

So, because
1] I like to beat things until they're not just dead, but parts start falling off and
2] I like to do graphics,

I've graphicized the original question.

In the graph, distance d is 100% an artifact of the illustrative representation, and has absolutely no real-world counterpart. There is no direct measurement I could do, even in principle (such as, say a gravitometer) that that would give me distance d. Correct?

Please explain. What is this graph supposed to represent? It doesn't look to me like the gravitational potential for either a solid sphere with a small hole drilled through it, or like a spherical shell, which are the two cases that we've discussed so far.