Quote:
Originally Posted by Cold-co
...The first part, my answer is no. According to a trigonometric analysis, the force of vertical gravity diminishes as one progresses into the earth, but the force of horizontal gravity increases. My trigonometric analysis of a hot-core model agrees with the analysis done by Dziewonski (Harvard). I get the same results he gets. In an orb of constant density the relationship is: the absolute value of the force of vertical gravity plus the absolute value of force of horizontal gravity at any level within the orb is equal to twice the value of vertical gravity on the surface of the orb.
The second part, I have no problem with Newton's proof. The whole mass of an orb can be considered to be located at its center for mathematical purposes.
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I looked up the fella at Harvard you're referring to, and I see nothing of any weird horizontal or vertical gravity ideas.
I don't have much time today, so instead of trying to construct a proof of the spherical shell thing, I'll just quote a couple sources that have proofs. The idea, I will reiterate, is that an object in a hollow spherical shell of matter feels no gravitation force in any direction regardless of where the object is inside the shell.
This means you can pick a spot in the interior of the earth and imagine it splitting the earth into two pieces. The first piece being the solid sphere that exists below the point. If the point is 2,000 km from the center then that would be a sphere of 2,000 km. The second piece is the tick shell of matter above the point. Basically, everything that is further than the point under consideration from the center of the object.
In asking how much gravity this point feels we can dissect the earth like this and disregard the hollow shell part above it and treat the part below as a point mass, because (as the proofs below show) the part above has no net effect on the point gravitationally, and, as you agree, the part below is well-represented as a point mass. That's not to say there are no pressure effects from the matter above it—there certainly are, but no contribution as far as the force of gravity.
So... sources (I'm finding a bunch all over the internet)...
Quote:
Gravitational field due to rigid bodies:
... Gravitational field due to thin spherical shell...
Case 2 : The point “P” lies inside the shell. The total gravitational field is obtained by integrating the integral from x = a-r to x = a+r,...
This is yet another important result, which has been used to determine gravitational acceleration below the surface of Earth. The mass residing outside the sphere drawn to include the point below Earth’s surface, does not contribute to gravitational force at that point.
The mass outside the sphere is considered to be composed of infinite numbers of thin shells. The point within the Earth lies inside these larger shells. As gravitational intensity is zero within a shell, the outer shells do not contribute to the gravitational force on the particle at that point.
A plot, showing the gravitational field strength, is shown here for regions both inside and outside spherical shell :
Gravitational field due to rigid bodies
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I cut out all the pertinent math above, but the conclusions are what matters I suppose. Here's another:
Quote:
Field Inside a Spherical Shell:
This turns out to be surprisingly simple! We imagine the shell to be very thin, with a mass density p kg per square meter of surface. Begin by drawing a two-way cone radiating out from the point P, so that it includes two small areas of the shell on opposite sides: these two areas will exert gravitational attraction on a mass at P in opposite directions. It turns out that they exactly cancel.
This is because the ratio of the areas A1 and A2 at distances r1 and r2 are given by : A1/A2=r1^2/r2^2 since the cones have the same angle, if one cone has twice the height of the other, its base will have twice the diameter, and therefore four times the area. Since the masses of the bits of the shell are proportional to the areas, the ratio of the masses of the cone bases is also r1^2/r2^2. But the gravitational attraction at P from these masses goes as Gm/r^2, and that r^2 term cancels the one in the areas, so the two opposite areas have equal and opposite gravitational forces at P.
In fact, the gravitational pull from every small part of the shell is balanced by a part on the opposite side—you just have to construct a lot of cones going through P to see this. (There is one slightly tricky point—the line from P to the sphere’s surface will in general cut the surface at an angle. However, it will cut the opposite bit of sphere at the same angle, because any line passing through a sphere hits the two surfaces at the same angle, so the effects balance, and the base areas of the two opposite small cones are still in the ratio of the squares of the distances r1, r2.)
http://galileo.phys.virginia.edu/cla.../GravField.htm
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You can also use General Relativity to prove that the spacetime in a hollow sphere is flat, thus no field strength.
If you want to discuss this in more detail then I suggest we make a thread dedicated to your idea. Otherwise, unless you can find some source agreeing with your conclusions, I don't know what you mean to accomplish in this thread.
~modest
