b
Let at some time particle is at a distance \(x\) from centre of Earth, then at that position field

\(E=\frac{G M}{R^3} x\)

\(\therefore \text { Acceleration of particle }\)

\(\vec{a}=-\frac{ GM }{ R ^3} \overrightarrow{ x }\)

\(\Rightarrow \omega=\sqrt{\frac{ GM }{ R ^3}}=\sqrt{\frac{ g }{ R }}\)

\(\text { Now } T =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{ R }{ g }}\)

\(\Rightarrow T =2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}}\)

\(=2 \times 3.14 \times 800\,sec \approx 1 \text { hour } 24 \text { minutes }\)