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

Let $O$ be the centre of the earth and $C$ be the centre of the tunnel.

The perpendicular distance from the centre of the earth to the tunnel is $OC = \dfrac{R}{2}$.

Let $P$ be the position of the particle in the tunnel. The distance of the particle from the centre of the tunnel is $CP = x$.

In the right-angled triangle $OCP$, the distance $r$ of the particle from the centre of the earth is given by:

$r = OP = \sqrt{OC^2 + CP^2} = \sqrt{\left(\dfrac{R}{2}\right)^2 + x^2} = \sqrt{x^2 + \dfrac{R^2}{4}}$

The magnitude of the gravitational force exerted by the earth on a particle of mass $m$ at a distance $r$ ($r < R$) from its centre is:

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

Substituting the value of $r$, we get:

$F = \dfrac{GMm}{R^3} \sqrt{x^2 + \dfrac{R^2}{4}}$