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

Let the centre of the earth be the origin $(0,0)$. Let the tunnel be parallel to the x-axis at a perpendicular distance $y = \dfrac{R}{2}$.

The position vector of the particle at a distance $x$ from the centre of the tunnel is given by $\vec{r} = x \hat{i} + \dfrac{R}{2} \hat{j}$.

The gravitational force on a particle of mass $m$ inside the earth at a position vector $\vec{r}$ from the centre is directed towards the centre and is given by:

$\vec{F} = -\dfrac{GMm}{R^3} \vec{r}$

Substituting the position vector $\vec{r}$ into the force equation, we get:

\vec{F} = -\dfrac{GMm}{R^3} \left( x \hat{i} +\dfrac{R}{2} \hat{j} \right)

$\vec{F} =$-\dfrac{GMm}{R^3}$x \hat{i} -$\dfrac{GMm}{2R^2} \hat{j}

The magnitude of the component of the force along the tunnel (x-direction) is:

$F_{\parallel} =$\dfrac{GMm}{R^3}$x$

The magnitude of the component of the force perpendicular to the tunnel (y-direction) is:

$F_{\perp} =$\dfrac{GMm}{2R^2}