SOLUTION: Given the Earth's radius of 4000 miles, and a mass of 200 pounds: ------ The weight on the surface is 200 pounds. What is the weight in an elevator to the Earth's center at vari

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Question 1090222: Given the Earth's radius of 4000 miles, and a mass of 200 pounds:
------
The weight on the surface is 200 pounds.
What is the weight in an elevator to the Earth's center at various depths?
Depth of 500 miles, 1000 miles, 2000 miles, etc.
-\Ignore any effects of pressure and temperature.

Found 3 solutions by Fombitz, ikleyn, Alan3354:
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website!
The assumption here is that the Earth is a point mass.
All of the mass of the Earth is concentrated at the point where the radius is equal to zero (or a small
value compared to the radius of the Earth and the values where the measurements are taken).
Weight is inversely proportional to the distance from the center of the earth.

So,

Then for example, when

.
.
.
Uniform density solution:
So the gravitational force is a function of the two masses involved and the distance between them.

If you allow the mass of the Earth to change as a function of its radius (R) (assuming uniform density)

Substituting,

Where "a" is a constant (from all of the constants put together).
So then your mass doesn't change so it's also a constant.

You can then use the value at the earth to calculate the value of b.

So then,
valid for
Again that's a simple solution using uniform density model.

Answer by ikleyn(52810)   (Show Source):
You can put this solution on YOUR website!
.
When (and if) the body is placed inside the Earth at the radius "r" (the depth R-r), then the only the part of the Earth
which is inside of the sphere of the radius "r" creates (produces) the gravity.

The exterior spherical layers r < s < R produce ZERO gravity integral force.
In other words, they do not make gravity force.

This fact is known from the Newton times to all who know/knows Physics.

And this effect MUST be accounted for in the solution of problems like this one.

The situation (and the solution) is different comparing with the case when the body is lifted ABOVE the Earth surface.

Therefore, the solution by @Fombitz is incorrect and must be redone.

-----------

To get the feeling of this effect, imagine that the body is placed exactly at the center of the (spherical) Earth. What is the gravity force and what is its direction in this case ? The answer (quite unexpected ?) is: the gravitation force is ZERO in this case. Did I say "unexpected" ? - But if you think 5-10 seconds on it, you will get that is is correct and nothing else can happen (due to symmetry).

Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Given the Earth's radius of 4000 miles, and a mass of 200 pounds:
------
The weight on the surface is 200 pounds.
What is the weight in an elevator to the Earth's center at various depths?
Depth of 500 miles, 1000 miles, 2000 miles, etc.
-\Ignore any effects of pressure and temperature.
==================================
Inside a solid sphere of constant density, the gravitational force varies linearly with distance from the center, becoming zero by symmetry at the center of mass. This can be seen as follows: take a point within such a sphere, at a distance r from the center of the sphere. Then you can ignore all the shells of greater radius, according to the shell theorem. So, the remaining mass m is proportional to r^3, and the gravitational force exerted on it is proportional to m/r^2, so to r^3/r^2 = r, so it is linear in r.
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Weight = 200*(4000 - d)/4000 where d = the depth in miles.