Does something weigh less 1km under the ground?

[K^2]

Still wrong. Gravitational field inside a perfect spherical shell is exactly zero. Everywhere. There are no tidal forces, no stretching, no bloating. It's precisely zero everywhere.

 

[pallidin]

This is ridiculously simple.
I remand further comment to the mods.

 

[K^2]

This is ridiculously simple.
I remand further comment to the mods.

In the mean time, look up Gauss Theorem.

 

[pallidin]

The Gauss Theorem in NO WAY validates your position at all.
Understanding the actual dynamics of a water balloon in that type of scenario can help, as it portends to reality.

 

[K^2]

I do understand the dynamics of a water balloon in a zero force field. You don't seem to.

You seem to think that there can be a net force acting on balloon under zero net force.

 

[pallidin]

I do understand the dynamics of a water balloon in a zero force field. You don't seem to.

You seem to think that there can be a net force acting on balloon under zero net force.

Ok, now your getting it.

 

[K^2]

Alright. So you are saying that there is a non-zero net force when there is a zero net force?

 

[pallidin]

No, no! Vector cancellation, mathematically, does cancel the force, BUT, this is not what happens in reality for your described circumstance.

 

[Pythagorean]

To put it in proportional terms, if you had a model of the planet the size of a wrecking ball and gave it one coat of paint, the thickness of the paint would represent from the highest mountain to the deepest ocean. All of life on Earth exists in an extremely thin layer on the surface of the planet.

 

[K^2]

No, no! Vector cancellation, mathematically, does cancel the force, BUT, this is not what happens in reality for your described circumstance.

So... Universal law of gravity does not describe reality?

 

[Curl]

No, no! Vector cancellation, mathematically, does cancel the force, BUT, this is not what happens in reality for your described circumstance.

Dude please, I'll try be polite but FFS please open your eyes. The balloon example is ridiculous. Mathematically, the net force on each INFINITESIMAL part of a balloon (on each molecule, if you will) is precisely zero. You are dividing up your balloon as if some parts are being pulled away, but its simply not true. You can divide up the water into as many volume elements as you want, and each one will still have zero force. That is why there is no bloating.

But forget that, the point is that since the Earth is denser into core, going deep inside will INCREASE the gravitational force on a person because you are CLOSER to the dense part of the Earth (iron and nickel I hear?)

So you are giving up the field from the outer shell, however you are getting closer to the dense core and there will be an inflection point where yes, you'll start getting lighter. However that inflection point is deeper than 800m.

 

[DrGreg]

Imagine that you are a water-filled balloon in the center of a hollow earth.
The water IN THAT BALLOON WILL BE PULLED away(slightly) from it's center and optimize on the spherical edge inside the envelope of the balloon.

This would be true if the net force at the very centre of the hollow Earth were zero, but non-zero (upwards) at all other points inside the cavity. But it's not true. The net force is zero everywhere within the hollow cavity. The molecules of water can't tell the difference between multiple forces summing to zero and no force at all. (To all intents and purposes there is no difference.)

 

[tmst08]

... the point is that since the Earth is denser into core, going deep inside will INCREASE the gravitational force on a person because you are CLOSER to the dense part of the Earth (iron and nickel I hear?)
...

Thanks, Curl. I believe you have supplied the answer to a question that has occurred to me on occasion for many years, as did BH? near the top of this thread.

I like this place.

-Tom

 

[K^2]

Keep in mind that this increase doesn't last for very long. It does start to drop after some depth, and goes to zero at the center.

 

[DaveC426913]

The molecules of water can't tell the difference between multiple forces summing to zero and no force at all. (To all intents and purposes there is no difference.)

I participated in a discussion about this exact thing here on PF a long time ago.

As far as I understand, there is no difference between zero net gravity and zero gravity.

If you ramped up the density of the hollow shell to the point where, externally, the object would experience millions of g's on the surface, the gravity inside the sphere would still be zero. No (local) experiment you did could tell you whether you are in zero g or in some sort of "balanced million g field". A difference that cannot be detected is not a difference.

 

[D H]

Lighter? Yes. Considerably lighter? No. The effect is *tiny*. Let's say your father's friend had to carry 50 pounds of equipment. Going 1/2 mile (~800m) underground would make that equipment feel about 1/10 of an ounce lighter.

Well, nuts! I made a sign error there.

The material weighs more underground, assuming typical surface rock. The effect is negligible at a depth of 800 meters.

Keep in mind that this increase doesn't last for very long. It does start to drop after some depth, and goes to zero at the center.

Yes and no. It does go to zero at the center, but the graph of gravitational acceleration inside the Earth is a bit strange. Gravitational acceleration increases with depth down to 500 km or so. Gravitation then decreases down to a depth of 1600 km or so, where it starts increasing again. It continues to increase down to 2900 km or so below the surface, where g reaches it global maximum of about 10.48 m/s².

The peaks at about 500 km and 2900 km are due to physical transitions inside the Earth. The core-mantle boundary at 2900 km represents a huge change in density. A less abrupt transition occurs within the transition zone between the lower and upper mantle. The rock of the upper mantle start to become unstable at about 400 km below the surface and progressively transition to ever denser forms with increasing depth. On the other hand there is nothing particularly special about the 1600 km depth where the local minimum occurs.

Gravitational acceleration increases with depth (decreases with height) if the local density is less than 2/3 of the average density of all of the stuff at and below the point in question but decreases with depth (increases with height) if the local density is more than 2/3 the average density. It is this average versus local density that accounts for the peaks at about 500 km and 2900 km and the local minimum at about 1600 km.

 

[cragar]

I wonder how the Gravitational field at the center of the Earth would affect an atomic clock as far as gravitational time dilation . And if we are further underground , would the air be denser and may have a slight buoyant force

 

[K^2]

There is a metric that describes gravity inside a spherical body. Using that metric, you'd be able to derive time dilation. I think it should be maximum at the center, but I'm not sure.

DH, I didn't realize there was more than one peak, but it does make sense. Thanks.

 

[cragar]

would my clock tick faster at the center of the Earth ,
and can my atomic clock tell the difference between no gravitational field , and a field cancellation of vectors to produce a zero field .

 

[DaveC426913]

would my clock tick faster at the center of the Earth ,
and can my atomic clock tell the difference between no gravitational field , and a field cancellation of vectors to produce a zero field .

This was discussed in https://www.physicsforums.com/showpost.php?p=1321327&postcount=10".

Janus' statement is that time dilation is not a result of absolute gravitational force, it is a result of a difference in gravitational potential.

Alas, the discussion was never satisfactorily resolved.

 

[K^2]

I can tell you with certainty that it's not a function of gravitational strength. In a constant gravitational field, for example, time dilation will vary as you move along the field, despite field itself staying constant.

But I'm not sure it can be written as a function of potential either. Potential enters into rest mass, which is given by the norm of the 4-momentum. That norm will have contributions from both the time dilation/contraction and space dilation/contraction effects.

I'm going to take a look at some references, see if I can work out the equation at least for the perfectly uniform perfectly spherical body with no rotation and no pressure terms.

 

[Dickfore]

If you are in the center of the Earth, you would feel no gravitational force. However, as you move towards the Earth's surface, you would feel a certain gravitational force pulling you towards the center. Therefore, you need to do work against the gravitational forces to get to the Earth's surface. This means that the Earth's surface is surely at higher gravitational potential \varphi than the center of the Earth.

Finally, the potential at infinity is chosen to be zero. This means that the gravitational potential is negative everywhere. So, because the potential is smaller in the center than at the surface, it means it is actually greater by absolute value, but more negative.

According to the formula:

<br /> g_{00} = 1 + \frac{2 \, \varphi}{c^{2}}, \; |\varphi| \ll \frac{c^{2}}{2}<br />

and the relation between proper time and the zeroth component of space-time displacement:

<br /> d\tau = \frac{\sqrt{g_{00}}}{c^{2}} \, dx^{0}<br />

it means that the more negative \varphi is, the slower the proper time passes (smaller g_{00}. So, time passes slower in the center of the Earth.

 

[AJ Bentley]

I feel an uncontrollable urge to say 'When you find you're in a hole, stop digging'



(Sorry 'bout that)

 

[K^2]

According to the formula:

<br /> g_{00} = 1 + \frac{2 \, \varphi}{c^{2}}, \; |\varphi| \ll \frac{c^{2}}{2}<br />

That's the bit that's new to me. Do you have a link to any place that derives it?

 

[Dickfore]

Yes. L.D. Landau, E. M. Lifgarbagez, The Classical Theory of Fields (Course of Theoretical Physics vol.2), Fourth Revised English Edition, Eqn.(87.12)

 

[K^2]

So it is an approximation, rather than general result. I did not catch that from previous post. That makes more sense.

Certainly good enough for this problem, though. And thank you for the ref. to derivation. Good reading.

 

[Dickfore]

Of course it's an approximation, there is no meaning to define a gravitational potential in General Relativity.

 

[K^2]

Good point.

 

[DrGreg]

Of course it's an approximation, there is no meaning to define a gravitational potential in General Relativity.

Skimming through my copy of Rindler seems to suggest that, for some GR coordinates (static or stationary, I think), then there is such a thing as relativistic potential given not approximately but exactly by

g_{00} = e^{2 \varphi/c^2}​

I haven't read this in detail so I might be misunderstanding something.

 

[Dickfore]

Might be true, because, when solving the Einestein's equations for the Schwartzshild metric one makes a similar exponential approximation.

However, the case of a static field, strictly speaking can be achieved only in the presence of a single body, since two-body problems are time-dependent. Stationary fields have components g_{0 \alpha} which are of the same order of magnitude as g_{0 0} and, since they get multiplied by v_{\alpha}/c. contributed to the next order in the Lagrangian of a particle. Therefore, this gravitational potential does not really represent potential energy of anything.