Suppose a tunnel is excavated along a chord of the earth at a...

Let the centre of the earth be the origin $O$. The tunnel is excavated along a chord at a perpendicular distance $\dfrac{R}{2}$ from $O$.

Let the centre of the tunnel be $C$. The distance $OC = \dfrac{R}{2}$.

When the particle is at a distance $x$ from $C$, its distance $r$ from the centre of the earth is given by:

$r = \sqrt{x^2 + \left(\dfrac{R}{2}\right)^2}$

The gravitational force exerted by the earth on the particle is directed towards the centre of the earth and its magnitude is:

$F_g = \dfrac{GMmr}{R^3}$

This force can be resolved into two components:
1. Along the tunnel (towards $C$): $F_{\parallel} = F_g \times \dfrac{x}{r} = \left(\dfrac{GMmr}{R^3}\right) \left(\dfrac{x}{r}\right) = \dfrac{GMm}{R^3}x$
2. Perpendicular to the tunnel: $F_{\perp} = F_g \times \dfrac{R/2}{r} = \dfrac{GMm}{R^3} \left(\dfrac{R}{2}\right)<br><br>Since the wall of the tunnel is frictionless, it provides a normal reaction$N$which exactly balances the perpendicular component of the gravitational force ($N = F_{\perp}$).<br><br>Therefore, the resultant force acting on the particle is solely the component of the gravitational force along the tunnel.<br><br>Magnitude of the resultant force =\dfrac{GMm}{R^3}$x