Sol. Let at some time particle is at a distance x from centre of Earth, then at that position field
$$
E=\frac{G M}{R^3} x
$$
∴ Acceleration of particle
$$
\begin{aligned}
& \quad \vec{a}=-\frac{G M}{R^3} \vec{x} \\
& \Rightarrow \quad \omega=\sqrt{\frac{G M}{R^3}}=\sqrt{\frac{g}{R}} \\
& \text { Now } T=\frac{2 \pi}{\omega}=\pi \sqrt{\frac{R}{g}} \\
& \Rightarrow \quad T=2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} \\
& \quad \quad=2 \times 3.14 \times 800 \mathrm{sec} \approx 1 \text { hour } 24 \text { minutes }
\end{aligned}
$$
