Q 11-13-031NEETJEE MainMedium
A tunnel is dug along a diameter of the Earth. A ball dropped into it would perform SHM. Its period is about (radius of Earth $= 6.4 \times 10^6$ m, $g = 10\ \text{m/s}^2$)
Answer: (A) $84$ minutes
Inside the Earth $g' = g\dfrac{r}{R}$, so the force is proportional to $-r$: SHM with $\omega^2 = \dfrac{g}{R}$.
$$T = 2\pi\sqrt{\frac{R}{g}} = 2\pi\sqrt{6.4 \times 10^5} \approx 2\pi \times 800 \approx 5030\ \text{s} \approx 84\ \text{minutes}$$
(The same as the period of a satellite skimming the surface.)
Solution by Sreeraj P, M.Sc Physics