A particle would take a time t to move down a straight tube from the surface of earth (supposed to be a homogeneous sphere) to its centre. If gravity were to remain constant, then the time would be t. The ratio of t/t' will be

The force on particle for varying g is given by

$F=\frac{GMmx}{R^3}=m{\omega }^{2}x$

$\Rightarrow {\omega }^2=\frac{GM}{R^3}$

${\left(\frac{2\pi }{t}\right)}^2=\frac{GM}{R^3} \left(\because \omega =\frac{2\pi }{t}\right)$

$\Rightarrow t=2\pi \sqrt{\frac{R^3}{GM}}$

When g is constant,

$s = \frac{1}{2}gt{\text{'}}^2$               (For initial velocity, u = 0)

$\Rightarrow R=\frac{1}{2}\left(\frac{GM}{R^2}\right){\left(t^{\text{'}}\right)}^2 \left(\because g=\frac{GM}{R^2}\right)$

${\left(t^{\text{'}}\right)}^2=\frac{2R^3}{GM}\Rightarrow t^{\text{'}}=\sqrt{2}\sqrt{\frac{R^3}{GM}}$

$\therefore \frac{t}{t^{\text{'}}}=\frac{2\pi \sqrt{R^3/GM}}{\sqrt{2}\sqrt{R^3/GM}}=\frac{2\pi }{\sqrt{2}}=\sqrt{2}\pi$