A straight tunnel is dug into the earth as shown in figure at a distance $b$ from its center. A ball of mass $m$ is dropped from one of its ends. Find the time it takes to reach the other end is approximately.

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Solution

Potential energy at start of tunnel (1) $u_{1} = \frac{G M m}{R_{e}}$
potential energy (2)
$u_{2} = \frac{G ⎛ ⎜ ⎜ ⎜ ⎝ \frac{M}{\frac{4}{3} π {R^{3}}_{e}} ⎞ ⎟ ⎟ ⎟ ⎠ (\frac{4}{3} π b^{3}) \times m}{\frac{G M m b^{2}}{{R^{3}}_{e}}}$
Energy lost $\frac{G M m}{R_{e}} - \frac{G M m b^{2}}{{R^{3}}_{e}}$
$\frac{G M m}{R_{e}} [1 - \frac{b^{2}}{{R^{2}}_{e}}]$
[But this is an SHM motion as $\frac{d^{2} E}{d x^{2}} = 0$ ]
$\frac{G M m}{R_{e}} [1 - \frac{b^{2}}{{R^{2}}_{e}}] = \frac{1}{2} k x^{2}$
$x = \sqrt{{R^{2}}_{e} - b^{2}} \frac{G M m}{R_{e}} [1 - \frac{b^{2}}{{R^{2}}_{e}}] = \frac{1}{2} k ({R^{2}}_{e} - b^{2}) k = \frac{2 G M m}{{R^{3}}_{e}} \frac{2 π}{T} = \sqrt{\frac{k}{m}} \Rightarrow T = 2 π \sqrt{\frac{m}{k}} T = π \sqrt{\frac{m}{\frac{2 G M m}{{R^{3}}_{e}}}} = 2 π \sqrt{\frac{{R^{3}}_{e}}{2 G M}} = 2 π \sqrt{\frac{R_{e}}{2 g}}$
Time taken to reach other end is $2 π \sqrt{\frac{R_{e}}{2 g}}$

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