Now here's my midterm exam, also open book:
An infinitesimally thin uniform spherical shell of total mass $$M$$ and radius $$R$$ is placed in deep outer space, far away from any stars and planets.
1. Calculate the gravitational field strength produced by the shell as a function of distance ($$r$$) from the center of the shell:
Code:
a) Inside the shell ([tex]r<R[/tex])
b) Outside the shell ([tex]r>R[/tex])
2. Using your result from Part 1, or any other method you like, calculate the gravitational field strength as a function of $$r$$ in both regions, assuming that instead of a uniform spherical shell, we now have a uniform solid sphere, also with total mass $$M$$.
3. In 4 lines or less, calculate the electric field strength produced by a uniformly charged solid sphere with total charge $$Q$$ and radius $$R$$, again as a function of distance from the center ($$r$$), for both $$r>R$$ and $$r<R$$. You may use the results you found in Part 1 and Part 2.
You may drink whatever you want and smoke whatever you want, just make sure you share with your colleagues. Assume the classical laws of Newton gravitation and Coulomb electrostatics apply here (i.e. ignore any quantum/relativistic corrections).
For points r > R;
For an isolated thin shell of mass M = 1, then as r -> ∞ Fr -> 0. This tells me that in isolation there is no “gravitational potential field” at points r > R.
For R > r.
The force on a test particle inside the shell result in other than popularly believed conditions. With a test mass m, where m << M located a distance h from the surface along any radial effectively separates the mass of the shell affecting the test mass m unequally.
The surface area a spherical segment A = 2\pi h for R =1.
The area of the least SHELL segment (M1 ) defined by a line through m perpendicular to the radial through m, $$M_1 = 2\pi h$$ and $$M_2 = 2\pi (2 - h)$$for $$M = 1 = M_1 + M_2$$. The distances of the centers of mass of each segment are$$h/2$$. and $$(2 – h)/2$$ for $$M_1$$ and $$M_2$$ respectively.
Using the universal law of gravity,
$$F=\frac{GmM}{r^2}$$
and with ignoring [text2\pi[/tex], $$ G = 1; M_1 = h; M_2 = 2 – h$$ , then
$$F_1 = \frac{h(2 – h)}{[\frac {h}{2}]^2$$
and,
$$F2 =\frac{(h)(2 – h)}{[(2- h)/2]^2}}$$.
The ratio of the forces for $$2 > h > 0$$,
$$\frac{F_1}{F_2} = \frac{[h(2 - h)]^2}{h/2)^2}$$
For h = 1 the ratio of the forces
$$\frac{F_1}{F_2}=1$$
The ratio of the forces at h = 1 (the shell center) tells us that the forces on m attributed to each segment are equal,
at the shell center.
The heretical result here is that the integrated shell theorem before evaluation for $$K = \frac{GmM}{4Rr^2}$$ and $$k = r^2 - R^2$$ is,
$$F_r = K[\frac{-k}{x} + x]$$
If one evaluates this expression from R - r to R + r the reult will be zero.
However, evaluating from R - r to 1, then from 1 to R + r the result is not quite what dogmatic followers of Newton's Shell Theorem believe. Newton screwed up and those standing on Isaac's shoulders must bear their own consequences.
Notice that for $$K = \frac{GmM}{4Rr^2}$$ the universal law of gravity is embedded in the shell the theorem as a constant multiplied by 1/4R and that the evaluated expression for the shell integral returns a value of 4R, which is in units of distance. The shell theory re the evaluated shell integral is a manipulated and concocted sham.
Using the universal law of gravity question # 2 has a similar result.
Answer to question #4.There is no observable electric field strength in the absence of a test charge at r > R, and similar results are analogous to a point R > r. The expression for electric-forces use the same form of the mass-forces with a change in variable notation only.
The ULG expressed as,
$$)F=\frac{GmM}{r^2}$$
says only that the force on a test mass m from a this shell located a distance r from the shell center is F. The expression for F is that analyzed by Newton et al as reflecting that the shell behaves as if all the mass was concentrated at the shell center requires a bit more thought -
the development of the shell integral did not address the locations of the location of forces forces attributed to the infitesimal mass segment pairs.
Take any point on the circle of infitesimal mass in the nearest shell segment nearest to m and consider the mirror image infitesimal mass in the segment farthest from m and you will clearly see that all forces attributed to mass points in the nearest and furthest segments that $$F_n$$ > $$F_f$$. There is no way using ULG that m can presume the forces are centered at the shell center. The forces on m point to the shell along any extended radial (duh), but the mass force center is offset in the direction of m.