A rod of mass m & length l = R is held very close to the earth so that it is aligned along the radial of earth. Its weight is equal to

mg/2

The force experted by earth on the differential segment of mass dm is equal to dF = $\frac{GM(\delta m)}{r^2}$

The force imparted on the total rod = F = $\int dF ⇒ F = GM \int\limits_{R}^{2 R} \frac{\delta m}{r^2}$

$=G M \int\limits_R^{2 R}\left(M . d r / r^2\right)$

$=\frac{G M m}{\ell} \int\limits_R^{2 R} \frac{d r}{r^2}$

$=\frac{G M m}{R}\left(\frac{1}{R}-\frac{1}{2 R}\right)=\frac{G M m}{2 R^2}=\frac{m g}{2}$    where g = $\frac{GM}{R^2}$