Does something weigh less 1km under the ground?

[nonequilibrium]

Okay, the title sounds weird, let me explain:

The father of a friend of mine used to work in the mines about 800m under the ground and he always experienced that when he had to carry heavy gears, it felt as if the item was considerably lighter to carry underground than when he was aboveground carrying the same item.

Is there something that could explain this?

 

[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.

 

[Dickfore]

Inside a homogeneous sphere, using Gauss' Law, one can show that the gravitational field changes as:

$$g(r) = \frac{g_{0} \, r}{R}$$

where $g_{0}$ is the gravitational field on the surface of the sphere R is the radius of the sphere and r is the distance from the center. Taking:

$$r = R - h, \; h \ll R$$

then the relative change in the gravitational field is:

$$\delta = \frac{\Delta g}{g} = \frac{h}{R}$$

Because $R = 6.37 \times 10^{6} \, \mathrm{m}$, it means that $\delta = 1.3 \times 10^{-4}$.

I highly doubt that any human is sensitive enough to such small variations in the local gravitational field.

 

[K^2]

Actually, because the crust is less dense than the core, for the first kilometer, things keep getting heavier. I don't remember exact depth past which the things start getting lighter, but it is significantly deeper than that.

The effect might be due to higher atmospheric pressure allowing for more oxygen in the blood, making things feel lighter. At 800m bellow the surface, pressure is going to be 10% higher than at the surface. Unlike changes in gravity, that is significant.

 

[nonequilibrium]

Yeah thank you guys, so there's no real physical way to explain his sensations.

K^2 might be on the right track with his biological explanation -- too bad I hardly know enough biology to say how significant the change in apparent strength is due to oxygen increase. Should I ask the same question in the Biology forum?

 

[K^2]

You can try. I know that reduced pressure/oxygen makes you feel fatigued. But I don't know how significant the reverse is going to be.

 

[pallidin]

Actually, because the crust is less dense than the core, for the first kilometer, things keep getting heavier.

Not so.
As you move down into the earth, there is more mass above you, pulling you slightly upwards. Therefore, you would weigh less.

 

[K^2]

As you move down into the earth, there is more mass above you, pulling you slightly upwards. Therefore, you would weigh less.

Please tell me that words "Gauss Theorem" make the light bulb light up. I don't want to type it all out.

 

[pallidin]

Huh?

 

[nonequilibrium]

pallidin, it can be proven (rather elegantly that in a hollow sphere there's no gravitational attraction because the vectors cancel out exactly for every point inside the sphere

 

[pallidin]

pallidin, it can be proven (rather elegantly that in a hollow sphere there's no gravitational attraction because the vectors cancel out exactly for every point inside the sphere

Of course there is gravitational attraction!
Just because force vectors "cancel", this DOES NOT mean that the force no longer exists. In fact, for gravity, you can not shield it at all.
The "effect" is what is moderated, not gravity itself.

 

[nonequilibrium]

I get your point and it's a fair one; I was just using "no gravitational attraction" as shorthand for "no net gravitational attraction".

But anyway, come to think of it, in the theory of Einstein, it seems like there really is no gravitational attraction, but maybe I shouldn't say anything about that because I'm not even sure. But anyway, the theorem of Gauss says the effect of gravity is zero inside a hollow sphere, demolishing your earlier argument that you're pulled up underneath the Earth's crust.

 

[K^2]

pallidin, each spherical shell produces a gravitational field identical to point mass with the combined mass of the shell for any object outside the shell, and precisely zero anywhere in side the shell. Do you need that proven? Or can you look up Gauss Theorem on your own?

Now, given some density depending only on r, ρ(r), the equation for acceleration due to gravity at distance r from center of the planet can be given by the following.

$$g(r)=\frac{G}{r^2}\int_{0}^{r}\rho(z)dz$$

Is that equation clear? Do you need me to show plots to demonstrate that it keeps growing after you started going underground, if ρ is higher at the core then at the surface?

 

[pallidin]

I give up. There is no such thing as zero gravitational field regarding a hollow, massive object. Vectors could cancel? Yes, with regard to effect.
Does the field cease to exist? No.

 

[K^2]

It can't and screw 200 years of physics and mathematics that say otherwise?

 

[nonequilibrium]

pallidin, you're being stubborn to no cause: you're the one going into the debate of there actually being any force or not. That is not what we're saying or care to say: all we're saying is that the vectors cancel, and that is all that counts in this discussion. You were saying a person 1km below ground would experience a net upward force. This is false, the person experiences nothing what is above him.

 

[pallidin]

...the person experiences nothing what is above him.

With all due respect, that simply is not true at all.

 

[nonequilibrium]

I don't think you understand the concept of vector. Do you understand the vector sum of gravitational force is zero everywhere inside? If you accept that, there follows immediately that you won't experience any force, because what is experiencing a force? Accelerating. What is accelerating? Have a non-zero force vector act on you.

And if you don't accept the fact that the vector sum is zero, well, that's simply Gauss' theorem and that's been proven.

I don't know what else to say?

 

[K^2]

With all due respect, that simply is not true at all.

And you are wrong. Have you looked up Gauss Theorem yet? Do so. Really.

 

[pallidin]

Ok, you are defending a wrong perspective.
I'll do this gently...

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.
Get it?

 

[elfboy]

Gravity is diminished but to such a small extend it's beyond the sensitivity of human perception

 

[nonequilibrium]

pallidin, I may not be able to specificy what exactly the scientific method is, but one of them is bringing in arguments that are not self-made. I'm doing my best to be polite and build up a reasoning in my posts (like when I asked if you problem is in Gauss' theorem or rather how it all follows out of that), but you seem to ignore anything I say. Please specificy whether you agree the vector sum is zero or not, otherwise we're just talking past each other.

 

[K^2]

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.
Get it?

Wrong. The gravitational pull due to all of the outer shells is precisely zero.

Look up the Gauss' Theorem. You seriously aren't helping yourself by arguing something from pure ignorance and refusing to look up something that actually explains the physics behind it.

 

[pallidin]

pallidin, I may not be able to specificy what exactly the scientific method is, but one of them is bringing in arguments that are not self-made. I'm doing my best to be polite and build up a reasoning in my posts (like when I asked if you problem is in Gauss' theorem or rather how it all follows out of that), but you seem to ignore anything I say. Please specificy whether you agree the vector sum is zero or not, otherwise we're just talking past each other.

The vector sum "can" indeed be zero. BUT, the gravitational force does not cease to exist at all.
What happens, under this circumstance, is bloating of the object due to the spherical gravitational pulling, up-wards spherically.

 

[nonequilibrium]

The vector sum "can" indeed be zero. BUT, the gravitational force does not cease to exist at all.
What happens, under this circumstance, is bloating of the object due to the spherical gravitational pulling, up-wards spherically.

Okay great, good. Now I can copy paste my earlier post which you neglected to read

I don't think you understand the concept of vector. Do you understand the vector sum of gravitational force is zero everywhere inside? If you accept that, there follows immediately that you won't experience any force, because what is experiencing a force? Accelerating. What is accelerating? Have a non-zero force vector act on you.

That's all I have to say on this. You can't just make up thought experiments because you think they're right, that luckily has no scientific weight. You either use logical reasoning or empirism.

 

[pallidin]

Okay great, good. Now I can copy paste my earlier post which you neglected to read



That's all I have to say on this. You can't just make up thought experiments because you think they're right, that luckily has no scientific weight. You either use logical reasoning or empirism.

Wrong. The object will bloat, or experience a bloat gravitational influence.
Just because vectors cancel DOES NOT mean there is no influence. And that influence is substantial.

 

[nonequilibrium]

Wrong. The object will bloat, or experience a bloat gravitational influence.
Just because vectors cancel DOES NOT mean there is no influence. And that influence is substantial.

Please stop spamming my thread.

 

[elfboy]

Imagine falling into the earth. As you fall there is a semisphere pulling you lower and a semiphere pulling you up. So at a distance r inside the Earth you would need to find the gravitational pull of both semispheres.

this can be done by packing the semi spheres with tiny spheres and then finding gravitational pull of each of these. yea lot of work. You will have to find the average distance between you (standing on this semiphere) and the semi-sphere packed with tiny spheres. This requires some geometry and integration in 3d coordinates.

then take the difference the gravitational pull of the semi-sphere above and below you.Assuming we have an infinite quantity of these tiny spheres the problem boils down to finding the average distance between a person standing on the middle of the surface of a semipshere in space and any point intside the sphere.

 

[K^2]

Wrong. The object will bloat, or experience a bloat gravitational influence.
Just because vectors cancel DOES NOT mean there is no influence. And that influence is substantial.

No. Read Gauss' Theorem. How many times do I have to repeat this?

 

[pallidin]

...and a semiphere pulling you up...

Exactly. Thank you...

 

[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.