Gravitation Inside Earth — Question 9

Question 9

A shorter chord tunnel has the same ideal transit time

In a uniform nonrotating sphere, $\vec g=-k\vec r$ with $k=g_s/R$. A narrow straight frictionless tunnel follows a chord at perpendicular distance $b<R$ from the center. Let $s$ measure distance along the chord from its midpoint. A particle is released from an endpoint; ignore removed mass.

1. Find its along-tunnel equation and endpoint-to-endpoint travel time, using $\ddot s=-\omega^2s\Rightarrow T=2\pi/\omega$.

2. Find the wall-force magnitude and explain why travel time does not depend on chord length.

Question 9 - Solution

A shorter chord tunnel has the same ideal transit time

(a) Derivation and calculation

Write $\vec r=s\hat s+b\hat b$. Projecting gravity along the tunnel, $$\ddot s=\vec g\cdot\hat s=-ks,\qquad s_0=\sqrt{R^2-b^2}.$$ Thus $$s(t)=s_0\cos(\sqrt k,t),\qquad \boxed{t_{\mathrm{transit}}=\pi/\sqrt k=\pi\sqrt{R/g_s}}.$$

(b) Result and interpretation

The particle has no transverse acceleration, requiring $$-mkb\hat b+\vec N=\vec0\quad\Longrightarrow\quad \boxed{\vec N=mkb\hat b,\quad N=mkb}.$$ The maximum speed is $$v_{\max}=\sqrt k\,s_0=\sqrt{k(R^2-b^2)}.$$ Distance and speed scale with the same amplitude $s_0$, leaving transit time independent of the chord length.

Physical check

At $b=0$ the wall force vanishes and the result reduces to the diameter tunnel.