From: max@west.darkside.com (Erik Max Francis)
Newsgroups: rec.arts.startrek.tech
Subject: Uniform spherical shell law -- proof
Date: 4 Nov 92 18:36:22 GMT
Organization: The Dark Side of the Moon +1 408 245 SPAM

miy@cs.man.ac.uk (Martin York (JRG PhD)) writes:

> 	Now according to the majority of people a point on the  external surface
> will have an acceleration of 1g towards the sphere, while a point 10 mm away 
> 	Now this sounds exceedingly wrong to me.

Sounds counterintuitive, but it is true.

Here's an article I threw together for a person who wanted a mathematical 
proof of said property:

-----
Here's a little mathematical treatment of the spherical shell law:
 
The law states that the gravitational field inside a uniform, 
spherical shell is zero.  (Note that's the gravitational field caused 
by the shell itself; other massive objects can exert gravitational 
fields within the sphere, but the mass of the sphere itself takes no 
part.)
 
This is a rather simple treatment, without too much mathematics.
 
Consider a uniform spherical shell S.  We'll take a test point P 
somewhere within that sphere.  To find the gravitational acceleration 
(let's be Einsteinesque), we'll construct a double-cone C within that 
sphere with apex at P.  So both nappes (cones) of the double-cone 
project onto different parts of the sphere.  Pick one side 
arbitrarily, and we'll call the distance between the point P and where 
the axis of the cone intersects this side of the sphere r1.  We'll 
call the distance between P and the intersection of the axis on the 
other side r2.
 
We'll call the area of the sphere intersected by the first nappe A1, 
and the area intersected by the second A2.
 
Now we'll deal with the gravitational field at point P caused by these 
two sections of the sphere.  The mass m1 of the part of the sphere A1 is 
proportional to r1^2, and similarly, m2 is proportional to r2^2.  
(This is because the density of the shell is constant.)
 
The magnitude of the gravitational field at P is proportional to the 
difference between the gravitational accelerations (since they pull in 
opposite directions):
 
    a o= a1 - a2.
 
(I use the symbol o= to mean "is proportional to.")  We know that the 
acceleration caused by a body is proportional to the mass and 
inversely proportional to the square of the distance:
 
    a o= m1/r1^2 - m2/r2^2.
 
We also know, however, that m1 o= r1^2 and m2 o= r2^2, so:
 
    a o= r1^2/r1^2 - m2^2/m2^2;
 
      o= 1 - 1;
 
      o= 0.
 
So the gravitational field at P is zero.
 
Since we chose the position of P orientation of the double-cone C 
arbitrarily, we can choose _any_ possible position and orientation.  
Thus, we can choose any point within the sphere and any size or shape 
of the double-cone, and the gravitational field will still be zero at 
that point.  Therefore, the gravitational field anywhere within that 
sphere is zero.
 
This method basically points out that while the area of the sphere 
which is closer exerts a stronger gravitational field, the amount of 
the sphere on the other side is much bigger and thus cancels it out.  
So while the gravitational field follows the equation 1/r^2, the 
amount of the sphere intersected by that cone follows r^2, and they 
end up cancelling each other out.
 
You can also use a slightly more advanced approach, which is actually 
much simpler (but requires knowlege of vector calculus).  Knowing that 
the gravitational flux through a closed surface is proportional to the 
sum of masses inside and that the expression for gravitational flux is 
only dependent on the radial distance for a perfectly uniform 
spherical shell, you can choose a spherical Gaussian surface _just 
inside_ that spherical shell.  The algebraic sum of masses inside is 
zero (it's empty inside there), the flux through the surface is zero.  
Since PHI_g in this case = 4 pi r^2 g = 0, we know that g = 0.
 
Note that if you put the Gaussian surface _outside_ the shell, then 
you get a normal situation where the flux is proprortional to the 
masses inside, namely the spherical shell.  Then the gravitational 
field is directed toward the center of the sphere, as if all the mass 
were concentrated at a point.
 
Hope that's what people were looking for.
> 			+++++++++++
> 
> 	From the calculatuions I have done in my head, and a bit of intuition
> The gravitational force should be as follows.
> 
> 
> 
> 
>           Surface of the sphere
>                        |
>                        V
> F|
> O|                    /\
> R|                   /  \
> C|                  /    \
> E|                 /      \
>  |                /        \
>  |               /          \
>  |              /            \
>  |             /              \   The center of the sphere
>  |            /                \  |
>  |           /                  \ V
>  |          /                    \
> ___________________________________
> Distance
> 
> NB: The lines should be curves not straight ( to form a point).
> 
> Also the force acts to pull you to the surface of the sphere.
> 
> Well thats my 2 penys worth.
> 
> 
> Scotti.
> 
> +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>                 Work is the curse of the Drinking Classes 
> ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


----------
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