The Shell Theorem is not what I thought. This post needs rethinking.
In popular formulation of this theorem using elemental concentric rings, its derivation starts with a point mass outside of the shell. The elemental forces due to the rings do not change sign. The derivation then pushes the point inside the shell, where the total elemental force due to all ring above and below the point cancels to zero. Superposition is then used to construct a hollow planet with infinite number of infinitely thin shells. And so, conclude that a hollow planet has no gravity inside the hollow.
My understanding was wrong. Consider this part of the derivation inside the thin shell on the surface of the shell. (http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html)
\( F=-\cfrac { GM }{ 4Rr^{ 2 } } \int _{ s=R-r }^{ s=R+r }{ \left( 1+\cfrac { r^{ 2 }-R^{ 2 } }{ s^{ 2 } } \right) } ds.m\)
Obviously, when \(r=0\), at the center of the hollow sphere, the resultant \(F\) is zero. If we set now,
\( r\rightarrow R\)
\(\cfrac { r^{ 2 }-R^{ 2 } }{ s^{ 2 } } =\cfrac { (r-R)(r+R) }{ s^{ 2 } } \)
\(\because s=r-R\),
\(s\rightarrow0\) as \( r\rightarrow R\),
and also,
\( \cfrac { r^{ 2 }-R^{ 2 } }{ s^{ 2 } }=\cfrac { s(s+2R) }{ s^{ 2 } } =1+\cfrac { 2R }{ s } \rightarrow1+\lim\limits_{s\rightarrow0}{\cfrac { 2R }{ s }} \)
Note: \(s\) is the distance measured from the surface of the infinitely thin shell. \(r\) is measured from the center of the hollow shell. In Newton's formula for gravity, \(s\) is in the denominator (squared); used directly. As far as the formula is concerned, \(r\) is derived from \(s\). At \(s=0\), Newton's formula for gravity is not valid; this point on the surface of the infinitely thin shell must be considered separately and indirectly.
When we remove that singular explosive point (\(s=0\)) from the integral,
\( F=-\cfrac { GM }{ 4R^{ 3 } } \int _{ s=0 }^{ s=2R }{ 1 } ds.m\)
\( F=-\cfrac { GM }{ 4R^{ 3 } } 2R.m\)
Therefore,
\( g=-\cfrac{1}{2}\cfrac { GM }{ R^{ 2 } }=-\cfrac { G_cM }{ R^{ 2 } } \)
where \(G_c=\cfrac{G}{2}\).
Where we conclude that gravity in hollow earth due to all point on the shell except one immediately underneath, is halved of \(\cfrac { GM }{ R^{ 2 } }\) pointing towards the center of the shell.
The problem here is the continuity conditions on passing through the hollow sphere however thin it may be. We can similarly move the test point mass starting from inside the shell outwards. In this case, the integration will be from \(s^{'}=r-R\) to \(s^{'}=r+R\) over \(ds^{'}\) and still obtain a discontinuity factor of \(1/2\).
What happened? At \(r=R\), the expression for gravitational force due to the elemental ring at the point on the surface of the shell (\(\theta=0\) or \(\theta=\pi\)) is no longer valid (\(s=0)\). This point can be taken out of the integral and the effects of this point just below the test mass is considered (added back) by discussing boundary conditions at that point.
Obviously, gravity due to the point is,
\(g_p=\cfrac{1}{2}\cfrac { GM }{ R^{ 2 } }=\cfrac { G_cM }{ R^{ 2 } } \)
always pointing towards the surface of the hollow shell, because in passing from inside to outside and vice-versa, we have a loss by a factor of \(1/2\) always, whereas the shell is infinitely thin.
So, if normalize consistently,
There is equal and opposite gravity inside hollow earth, and gravity decreases upwards to zero at the center of the shell. And in the thickness of hollow earth there is a point where gravity reverses and is zero at that point.
And now I get a science prize from hollow earth.
