Disproving
Newton
Part
2
Calculating
the Force for a Sphere
By Rogue Physicist, Mastervalver and Metafrizzics
(C) 1999, Updated Aug 24, 2005
Synopsis: Part 2 In this section we derive some tools and give a mathematical treatment of both hollow and solid spheres. Then follows an analysis, which tries to lay bare the assumptions supporting the bridge between the mathematics and the physical reality. In the following section, we show in detail how the Hollow Sphere Theorem breaks down at sizes smaller than about 10,000 times the diameter of an atom. The results also apply to spheres of any size having low numbers of charges, in the range of < 10,000 electrons or holes. This is obviously a significant finding, for which test by experiment is a reasonable expectation.
Table of
Contents
(A) Creating the Tools We Need 4
Introduction: Simplifying the Problem 4
1. Force between a Barbell and a Test-mass 5
2. Getting the Net Force for the Barbell Using the Cosine 6
3. The Equivalent Point Mass is NOT the Geometric Centre 7
4. The Force for a Ring of Negligible Thickness 8
5. Plotting the Force for the Ring 9
(B) The Force for a Hollow Sphere 10
1. Preliminaries for Calculation 10
2. Archimedes’ Theorem 11
3. Defining the Variables 12
4. Integrated Version for Hollow Sphere 14
5. Appendix: Integrating the Hollow Sphere Formula 15
(C) Force for a Solid Sphere 18
1. Force for a Uniform Disk 18
2. Graphing the Force for a Disk 19
3. Force for a Solid Sphere ( Disk Method ) 20
4. Formula and Graph for Solid Sphere ( Disk Method ) 21
5. Force for a Solid Sphere ( Shell Method ) 21
6. Integral and Graph for Solid Sphere ( Shell Method ) 23
7. Summary of Part 2 so far: 24
(D) Failure of Sphere Theorem at Molecular
Level 25
1. A Simple Quantization of Mass 25
2. General Method for Discrete Distributions of Mass 26
3. Discrete Form of Hollow Sphere Equation 28
4. Formula and Graph for Quantized Hollow Sphere 30
5. Sample Hollow Spheres of Discrete Particles 31
6. Summary and Conclusions with FAQs 33
(A) Creating the Tools We Need
Introduction: Simplifying the Problem
Part of solving any tough task is simplifying the problem any way we
can.
Newton's equation for the force looks like this, which seems
complicated:
For our purposes we can simplify this equation quite alot before we
even start:
- We can set the radius of the sphere we are going to measure to be '1' . We will measure all other distances in units of this radius. This is important, because the geometric features we'll find are all relative to sphere size.
- We can set the Gravitational Constant to be '1', by choosing the right units of mass. The Gravitational Constant is not an 'absolute' constant. It really coordinates mass with units of distance, and is set by our choice of units.
- We can set the mass of each body to be '1' as well. The mass of an object is usually constant. Later we can extend our findings to bodies with different masses and arbitrary sizes.
- We will centre the Sphere at the Origin of our coordinate axis, so the geometric centre (and centre of mass) will be: (x,y,z) = (0, 0, 0).
- Our Test-mass will just move along the X-axis. The position of our test-mass will just be the X-coordinate. This will greatly simplify our calculations.
Since G x ma
x
mb
= 1 x 1 x
1 = 1
Newton's equation is now simply the pure Inverse Square Law:
1. Force between a Barbell and a Test-mass
Take a barbell made of two equal uniform spheres, rigidly connected
by a rod of negligible mass, system A. A fixed
distance away from the barbell, is another uniform sphere, test-mass
B, the same mass as the barbell. 1
It should be obvious that the total force on B is just the
sum of the pull from each end of the barbell. We can work out the
pull from each end separately, and then just add them.
But since each end pulls in a slightly different direction the total
force is the vector sum of the forces for A1
and for A2. 2
r is the radius of system A, and d is the basic
centre-to-centre distance between the two systems. h is the
direct distance to each end, and the angle Theta q shows the actual direction of pull from each end relative to the
overall pull, which happens to be toward the geometric
centre of system A, due to symmetry.
____________________________________
1. The
geometric centre
(GC) of an object is
determined by its shape, i.e., distribution of volume and extension
in space. The centre of mass
(CM) is determined by
the distribution of mass. The CM
happens to coincide with the GC
in system A , since it is symmetrical in both shape and distribution
of mass . These two are not
the same as the centre of gravity,
which is only relevant in a uniform gravitational field.
Different
again is an equivalent point-mass position,
which we have to calculate.
By equivalent point-mass,
(or EPM), we mean a
point-mass having the same mass as the system we are replacing. By
EPM position,
we mean where we must put the EPM to generate exactly the same force
as the system. The Sphere Theorem (ST)
only claims that the
EPM is at the GC
for spheres, not
barbells or other shapes, so it is natural that we have to calculate
it.
2. If we were just solving
an example, we could combine the vectors using the parallelogram
rule on graph paper, but since we want a general
solution for all cases, we want to solve algebraically. This is
easily done by using trigonometry.
_____________________________________
2. Getting the Net Force for the Barbell Using the Cosine
Any vector can be split up into a vertical and horizontal component.
Here the vertical components for each sphere in system A are
opposing and equal so they cancel, leaving only the horizontal
components. We can ignore the vertical parts.
Multiplying by the Cosine function correctly scales down the force from each sphere to the net horizontal component only. We then just add these components to get the total force in the X direction.
Substituting the definition of cosine (adj / hyp = d / h), and using Newton,3
These are good given ( h and d
) or ( h and q ).
To get it for ( d and q
) we pick a multiplier and combine it in:
What is the
physical difference between equation (1.2) and (1.3) ?
In terms of h, if we spread the barbell ends apart in an
orbit around sphere B, keeping the distance h to
each ball constant, the force is proportional to the cosine.
But if we keep d constant (the distance between systems) and
just lengthen the barbell, the spheres spread apart vertically, and
the force drops as the cube of the cosine!
_________________
3. Note
that we use the hypotenuse for the direct
distance to each sphere from B, and then get the horizontal component
by multiplying by the cosine of the angle from the horizontal.
3. The Equivalent Point Mass
is NOT the Geometric Centre
The last result is practical, but if we want a version in terms of
d and r without having to deal with angles we can
take equation (1.1) and do this:
 |
What does it mean? As we
shorten the barbell the angle q shrinks,
r vanishes, h approaches d, d in the
numerator cancels out and we simply get Newton’s formula. If we
lengthen the barbell q
approaches 90o , 1
the force falls away to nearly zero, even though the GC and CM
hold position and the mass also stays constant.
Clearly, when the force is dropping
then the EPM must be moving further away, off of the
geometric centre and CM. So generally, the
centre of mass is not relevant for calculating the
force between the two systems.
|
The key here is that the EPM
is simply the viewpoint from sphere B. It is only
meaningful in relation to some fixed external position. If we
swung sphere B around to the other side, the EPM
would also move to the opposite side: A non-spherical object has
a different EPM for every possible position
around it, and each object will feel a force based upon its own
personal viewpoint.
_____________________________________
1
Keep in
mind that the
angle is not between the two spheres from B , but between the
horizontal line ( d
) and each sphere. The angle is always < 90o
and the cosine is always positive.
_______________________________________
4. The Force for a Ring of Negligible Thickness
Lets replace the shaft of the barbell with a perpendicular ring
around the axis. The spheres can be anywhere on the ring, as long as
they are opposite one another. The spheres can be replaced by
equivalent point-masses. We can even split each point-mass in half,
assign each ¼ the total mass, and spread them out equally around the
ring. We can do this without changing the total force, since all the
mass remains at the same distance and angle from sphere B.
Only the vertical components of the force are affected by
rearranging, but these stay balanced and cancel. Although the case
is now three-dimensional, the equations remain the same. This
process can be continued ad infinitum until the mass is equally
spread around the ring in increments as fine as we wish.
The real value of our
equations is now apparent. They also define exactly the force
of a uniform ring of negligible cross-section perpendicular
to sphere B. This is the kind of tool we can use to analyze hollow
cylinders, spherical shells, and other related shapes.
5. Plotting the Force for the Ring
We can now plot the force for
the ring with a radius of one unit, placed at (0,0,0) on our
coordinate axis. The test-particle is simply swept along the x-axis,
( d = x ) and we can plot the results by superimposing the graph and
the ring in space.
Summary:
It took a couple of pages, but it was worth it. We
have shown that the force weakens as a system’s mass spreads out
perpendicularly. We have also shown that the true EPM is
not at the GC but further away. And
we have an elegant and powerful equation for the force of a ring. No
fancy calculus was needed.
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