Quote:
Originally Posted by Boerseun
I suppose what Cold-co is saying with vertical and horizontal gravity, is the following:...For every kilometer you descend, there's a kilometer's worth of mass overhead, and one less kilometer's worth of mass between you and the Earth. ...you will experience gravity from mass distributed all around you - below you, overhead, etc. And the further you go down, the "less" you will weigh as you are now being attracted upwards as well as downwards. ...
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In an earlier post, I showed how to approach this problem.
Take a "shell" of the Earth. This shell will be a hollow sphere of some constant thickness. You may assume the outer surface of the shell is the surface of the Earth. We can relate this shell to the problem above by saying that the thickness of the shell is the same as the depth you have descended to in your vertical tunnel.
This shell, therefore, will contain ALL of the mass of the Earth that is "above" your depth. If you have descended 500 km down from the surface, then this shell will be 500 km thick.
Now, take any arbitrary point, P, anywhere
on the inside of the hollow shell. Take any arbitrary point, s, that is actually in the mass of the shell itself. The point s can be specified by its radius from the Earth's center, r; the longitudinal angle, phi, from some given zero longitude; and the latitudinal angle, theta, measured from the equator plane of the Earth, with positive theta being the Northern hemisphere and negative theta being the Southern.
Calculate the force of gravity on YOU (at point P), from a tiny element of mass at point s.
There's nothing really complicated so far, we're just using Newton's law of gravity. The tricky part is the distance between the tiny mass at point s, and YOU at point P, must be expressed in terms of r(s), phi(s), theta(s), and also r(P), phi(P) and theta(P). We will use standard Polar coordinate system.
Now we use integral calculus. We assume constant density in the shell. This is not a problem, as you can always divide your shell into hundreds of really thin, thin sub-shells and give them each their own unique (but still constant) density.
We integrate over theta(s) from -90 degrees to +90 degrees.
We integrate over phi(s) from 0 degrees to 360 degrees.
We integrate over r(s) from r = (inner radius of the shell) to r = (outer radius of the shell)
We have now accumulated (integrated) ALL the gravitational forces on YOU at point P, from ALL the tiny masses from ALL points s, within the ENTIRE shell.
The sum total gravitational force is zero. (0) Nothing. None Nada Zipity-doo-da.
It does not matter where you choose your original point P, as long as it is inside the hollow of the shell.
This makes the whole problem SOOOOO much easier.
You can always ignore ALL the mass of the Earth at ALL depths less than YOUR depth.
That is, you can always ignore ALL the mass of the Earth ABOVE you.
Because it ALL cancels out, ALL the time, at ANY depth you like.
And this includes ALL gravitational forces, including those that are horizontal, vertical or anywhere in between.
The
ONLY gravitational force you can possibly feel, will be from the sphere of the Earth at or below your depth. As you descend, that remaining sphere will get smaller of course, and will vanish as you reach the center of the Earth.
{{{ In my opinion, this is the most fascinating and awesome use of Integral Calculus in all of advanced freshman year physics. }}}
