Spherical shell theorem

A uniformly charged spherical shell produces zero field inside and the field of a point charge outside.

Place a fixed shell of radius $a$ and charge $-q$ at $(0,a/2)$ and a fixed shell of radius $2a$ and charge $+q$ at $(0,-a/2)$. On the positive $x$ axis their centre distances coincide. The electrostatic shell theorem makes the electric field vanish until $x=\sqrt{3}a/2$, gives magnitude $k_{e}q/(x^2+a^2/4)$ until $x=\sqrt{15}a/2$, and gives $k_{e}qa/(x^2+a^2/4{)}^{3/2}$ thereafter. At each ideal charged surface the field has distinct one-sided limits. The net charge is zero but the electric dipole moment is $-qa\hat{\mathbf{y}}$.

Inside a homogeneous spherical mass, gravity is proportional to position. Projection onto a fixed straight chord gives $\ddot{y}=-{\omega }_0^{2}y$, independently of its perpendicular offset from the sphere centre. Release from one surface endpoint at rest reaches the opposite endpoint after $\pi /{\omega }_0$. The transverse force is supplied by the tunnel constraint. This idealization neglects removed tunnel mass, rotation, density variation and dissipative forces.

Adding a velocity-proportional force $-b\dot{y}$ produces a damped harmonic oscillator. The fastest non-oscillatory release from rest is critical damping, with the repeated-root solution in the title. Reaching fraction $\epsilon$ of the initial displacement requires $(1+s)e^{-s}=\epsilon$, $s={\omega }_{0}t$. At one percent, $s=6.638352\dots$, rather than $\log 100$. Exact arrival at equilibrium takes infinite time; the tolerance reference must be stated.

For constant mass density $\rho$ on $a<r<b$, the shell theorem gives zero gravitational field for $r\le a$, inward magnitude $4\pi G\rho (r^3-a^3)/(3r^2)$ for $a<r<b$, and $4\pi G\rho (b^3-a^3)/(3r^2)$ outside. These formulas agree continuously at both boundaries. The empty cavity has zero gravitational field but generally nonzero constant gravitational potential.

For a spherical shell of uniform mass density $\rho$, inner radius $a$ and outer radius $2a$, suppose a negligible-width radial tunnel permits a small mass $m$ to fall from rest at the outer surface. The enclosed mass at radius $r$ within the material is $4\pi \rho (r^3-a^3)/3$. The spherical shell theorem gives inward gravitational acceleration $g(r)=4\pi G\rho (r-a^3/r^2)/3$. Integrating from $a$ to $2a$ gives released gravitational potential energy $4\pi G\rho m{a}^2/3$, hence speed $a\sqrt{8\pi G\rho /3}$ at the cavity. The cavity has zero gravitational field, so the diameter transit time is the title expression, independent of $a$ and $m$.