Question

Imagine a frictionless tunnel along a chord of earth having length equal to the radius of earth $R$. Particle is thrown from surface of earth inside it along its length at a velocity $\sqrt{\frac{3gR}{4}}$ where $g$ is acceleration due to gravity on surface of earth. Find the time (in sec) taken by particle to cross the tunnel. Take $g=10{\text{m/s}}^2,\pi =3.14$ and $R=6400\text{km}$.

Answer 1

The particle will perform SHM
$(\omega =\sqrt{\frac{GM}{R^3}}=\sqrt{\frac{g}{R}})$

$\Rightarrow \omega \sqrt{A^2-\frac{R^2}{4}}=\sqrt{\frac{3gR}{4}}$

$\Rightarrow \sqrt{\frac{g}{R}}\sqrt{A^2-\frac{R^2}{4}}=\sqrt{\frac{3gR}{4}}$

$\Rightarrow A^2-\frac{R^2}{4}=\frac{3}{4}{R}^2{A}^2=R^2$
$A=R$
$t=\frac{\pi }{3\omega }=\frac{\pi }{3}\sqrt{\frac{R}{g}}$
$=\frac{3.14}{3}\times 800\Rightarrow 837.33\text{sec}$