91Ó°ÊÓ

Earth is uniform sphere and ignoring air drag and friction

The gravitational force at a radial distance r inside Earth (e.g., point A in the figure)

Formula:

$Fg=\frac{GMm}{R^3}r$

The component of the force along the tunnel is

$$\begin{gathered} Fx=Fgsin\mathrm{θ} \\ =(-\frac{GMm}{R^3}r)\frac{x}{r} \\ =\left(\frac{GMm}{R^3}\right)x \end{gathered}$$

This can be rewritten as

$$\begin{gathered} ax=\frac{d^{2}x}{d{t}^2}-\frac{GM}{R^3}x \\ =-{\mathrm{Ó\neg }}^{2}x \end{gathered}$$

Where,${\mathrm{Ó\neg }}^2=\frac{GM}{R^3}$. The equation is similar to Hooke’s law, in that the force on the train is proportional to the displacement of the train but oppositely directed. Without exiting the tunnel, the motion of the train would be periodic would a period given by$T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}$. The travel time required from Boston to Washington DC is only half that (one-way):

$$\begin{gathered} \mathrm{Δ}t=\frac{T}{2} \\ =\frac{\mathrm{Ï€}}{\mathrm{Ó\neg }} \\ =\mathrm{Ï€}\sqrt{\frac{R^3}{GM}} \\ =\mathrm{Ï€}\sqrt{\frac{{(6.37×{10}^6 m)}^3}{(6.67×{10}^{-11} m^3/kg.s^2)(5.98×{10}^{24}kg)}} \\ =2529 s \\ =42.1  min \end{gathered}$$

The travel time required from Boston to Washington DC is 42.1 min.