The weight of a body on the earth’s surface is, \[W=mg\] where, $m$ is the mass of the body and $g$ is the acceleration due to gravity.

The effective acceleration due to gravity at depth $x$ from the surface of earth is, \[g’=\left(1-\frac{x}{R}\right)mg\] where, $R$ is the radius of the earth.

Let $W_x$ be the weight of the body at depth $x$. Then,

\[W_x=mg’=m\left(1-\frac{x}{R}\right)g=\left(1-\frac{x}{R}\right)W\]

Half way down at the centre of earth, $x=\frac{R}{2},$

\[\therefore W_x=\left(1-\frac{R/2}{R}\right)W=\left(1-\frac{1}{2}\right)W=\frac{W}{2}\]

Hence, the weight of the body will be half of that on the surface of the earth.