= Solution
In the <uniform-Earth chord oscillator>, neglect the <mass> removed by the narrow tunnel and any rotation of the <Earth>. Inside a uniform <sphere>, the <shell theorem> gives $\mathbf F=-(4\pi G\rho m/3)\mathbf r$. On the constrained chord $\mathbf r=(d,y)$, its along-tunnel component is
$$
\boxed{F_y=-m\omega_0^2y,\qquad\omega_0^2=\frac{4\pi G\rho}{3}.}
$$
The transverse component is balanced by the tunnel constraint. With release from $y_s=\sqrt{R^2-d^2}$ at rest,
$$
\ddot y+\omega_0^2y=0,\qquad \boxed{y(t)=y_s\cos\omega_0t.}
$$
The far end is $-y_s$, so \b[$t_{\rm transit}=\pi/\omega_0\approx2530\,\mathrm s=42.2\,\mathrm{min}$], independent of chord offset in this ideal model. Here $\rho=5515\,\mathrm{kg\,m^{-3}}$ and $G=6.67430\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ give $\omega_0=1.24171\times10^{-3}\,\mathrm{s^{-1}}$.
With linear drag, $\ddot y+(b/m)\dot y+\omega_0^2y=0$. Inserting a <complex exponential> $e^{-(\gamma+i\omega)t}$ gives
$$
\gamma^2-\omega^2-\frac bm\gamma+\omega_0^2+i\omega(2\gamma-b/m)=0,
$$
so in the <underdamped> regime
$$
\boxed{\gamma=\frac b{2m},\qquad\omega=\sqrt{\omega_0^2-\gamma^2}.}
$$
Real physical solutions combine conjugate modes. Release from rest specifically gives $y=y_se^{-\gamma t}[\cos\omega t+(\gamma/\omega)\sin\omega t]$; a lone exponential with prefactor $y_s$ does not impose that initial <velocity>.
The <critical gravity-tunnel settling time> follows from critical damping, the fastest non-oscillatory release:
$$
\boxed{b_c=2m\omega_0,\qquad b_c/m=2.48342\times10^{-3}\,\mathrm{s^{-1}},\qquad
y(t)=y_s(1+\omega_0t)e^{-\omega_0t}.}
$$
Interpret the 1% tolerance relative to the initial displacement from equilibrium. With $s=\omega_0t$, $(1+s)e^{-s}=0.01$ has $s=6.63835$. Hence
$$
\boxed{t_{1\%}=6.63835/\omega_0\approx5346\,\mathrm s=89.1\,\mathrm{min}.}
$$
The equilibrium is reached only asymptotically; using a pure $e^{-\omega_0t}$ would miss the critical-damping polynomial factor and underestimate this time.
Back to article page
