Does something weigh less 1km under the ground?

[D H]

I think what palladin was saying was that for a person below ground, there is some mass above them which pulls them upward, less mass below them to pull them downward, therefore they will weigh less, and that is a true and valid argument, which does not deny Gauss' theorem.

That apparently is what palladin was saying, and it is a false argument.

A correct statement is that at some depth below the surface of a spherical mass distribution, the spherical shell of mass above that depth contributes absolutely nothing to the gravitational force. The mass overhead does not pull them upward. All that matters is the sphere of below than the depth in question. In other words, with r=R-d, where R is the radius of the Earth and d is the depth below the surface,

$$g(r) = \frac{GM(r)}{r^2}$$With this, and with a model of density inside the Earth such as the Preliminary Reference Earth Model (http://geophysics.ou.edu/solid_earth/prem.html ), one can investigate the behavior of g inside the Earth. Note that for a spherical mass distribution the derivative of mass wrt radial distance is related to the local density via

$$\frac{dM(r)}{dr} = 4\pi r^2 \rho(r)$$

I'll define the mean density at some radial distance r from the center as

$$\bar{\rho}(r) \equiv \frac{M(r)}{V(r)} = \frac{M(r)}{4/3\pi r^3}$$

With this, differentiating gravitational force respect to radial distance r yields

$$\aligned \frac{dg(r)}{dr} &= \frac{G dM/dr}{r^2} - 2\frac{GM(r)}{r^3}\\[6pt] &= \frac{G 4\pi r^2 \rho(r)}{r^2} - 2\frac 4 3 \pi G \bar{\rho}(r)\\[6pt] &= 4\pi G\left(\rho(r) - \frac 2 3 \bar{\rho}(r)\right) \endaligned$$

In words, gravitational force increases with depth if 2/3 of the mean density of the stuff at greater depths exceeds the local density at the depth in question, decreases otherwise. There are two points inside the Earth where marked changes in density makes the difference $2/3 \bar{\rho}-\rho$ change from positive to negative with increasing depth. The core-mantle boundary (the D" zone) marks the transition from the rocky material of the mantle to the molten iron of the outer core. In the transition zone at the top of the inner mantle, the mantle rock changes crystalline structure and hence changes density. (The point in the lower mantle where this density difference changes from negative to positive is of no special interest.)

 

[Rap]

That apparently is what palladin was saying, and it is a false argument.

A correct statement is that at some depth below the surface of a spherical mass distribution, the spherical shell of mass above that depth contributes absolutely nothing to the gravitational force. The mass overhead does not pull them upward.

What I am saying is that both arguments are correct: Let's assume that the Earth is sphere with density that is a function of radius only. If you go underground, and then take a plane that passes through you, perpendicular to a line connecting you with the center of the earth, then that plane will divide the Earth into two parts - above you and below you. The net force that you experience will be the sum of the upward force due to the mass above you and downward force due to the mass below you, which will be less than the force you would experience if you were at the surface.

You could also take a sphere centered at the center of the earth, passing through you when you are underground. This will again divide the Earth into two pieces, the inner ball and the outer shell. The outer shell will have a thickness equal to the distance you are underground. The net force you experience will be the sum of the downward force from the mass of the inner ball, and the force from the mass of the outer shell which will be zero. This will be less than the force you experience at the surface, and equal to the force that you calculate by the first method.

 

[yuiop]

What I am saying is that both arguments are correct: Let's assume that the Earth is sphere with density that is a function of radius only. If you go underground, and then take a plane that passes through you, perpendicular to a line connecting you with the center of the earth, then that plane will divide the Earth into two parts - above you and below you. The net force that you experience will be the sum of the upward force due to the mass above you and downward force due to the mass below you, which will be less than the force you would experience if you were at the surface.

Both arguments are not correct because the conclusion of your partition plane argument contradicts the conclusion obtained by many here, that the force of gravity at the surface can be less than the force of gravity lower down if the density increases towards the centre. The partition plane argument fails to take into account that the inverse square (Gauss) law for the measurement at the surface.

 

[Subductionzon]

Rap, if you were familiar with Newton's shell method for calculating gravitational force you would realize it does not matter what is above you. This model assumes that the density is the same at anyone radius around the globe. If you do the calculations you will find that the mass above you cancels itself out. If you were within a hollow shell the force of gravity would be zero everywhere within that sphere.

 

[D H]

What I am saying is that both arguments are correct: Let's assume that the Earth is sphere with density that is a function of radius only. If you go underground, and then take a plane that passes through you, perpendicular to a line connecting you with the center of the earth, then that plane will divide the Earth into two parts - above you and below you.

You could do that, but who would want to do that? This involves solving for the gravitational acceleration due to two spherical caps with nonuniform density. Doable, but yech.

The net force that you experience will be the sum of the upward force due to the mass above you and downward force due to the mass below you, which will be less than the force you would experience if you were at the surface.

That is not necessarily true, and it is definitely not true for the Earth.

Take a look at the plot in post #69. As one proceeds from the surface toward the center of the Earth, gravitational acceleration initially increases. The acceleration reaches a local max of just over 10 m/s² at the bottom of the transition zone (the 670 km discontinuity). At this point gravitational acceleration starts a slight decline, reaching a local minimum about 800 km into the lower mantle. At this point it starts increasing again, reaching a global max of about 10.7 m/s² at the core-mantle boundary. Only then does gravitational acceleration start a monotonic decrease to 0 at the center of the Earth.

Also note that the gravitational acceleration inside the Earth remains greater than the surface value until one has gone more than half the distance to the center.

 

[Rap]

Both arguments are not correct because the conclusion of your partition plane argument contradicts the conclusion obtained by many here, that the force of gravity at the gravity at the surface can be less than the force of gravity lower down if the density increases towards the centre. The partition plane argument fails to take into account that the inverse square (Gauss) law for the measurement at the surface.

That is not necessarily true, and it is definitely not true for the Earth.

I stand corrected. My argument did not take into account the inverse square law for the different distances. I will amend it to say simply that if you are underground and take a plane passing through you, perpendicular to the radius, you will have divided the Earth into two parts, what is above you and what is below you. The net force you experience will be the sum due to the mass below you (downward) and the mass above you (upward). If you have a sphere centered at the center of the earth, passing through you, the net force you experience will be equal to the force of the inner ball only, since the force from the outer shell will be zero. This force will be equal to that calculated by the first method. I don't mean to say that the first method is mathematically convenient, only that it is true.