A Tunnel so long!

In the solution,

$M\longrightarrow$ Mass of earth.

$R\longrightarrow Radius of earth$

$\rho \longrightarrow Density of earth$

Here, Since the gravity of earth decreases along with depth, we have to calculate the effective gravitational force acting on particle. ${ g }_{ effective }=\frac { GM(r) }{ { r }^{ 2 } }$

$M(r)$ is given by $M(r)=\rho \frac { 4 }{ 3 } \pi { r }^{ 3 }$ , where $\rho =\frac { M }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } }$

$\Rightarrow { g }_{ effective }=\frac { G\frac { M{ r }^{ 3 } }{ { R }^{ 3 } } }{ { r }^{ 2 } } =g\frac { r }{ R } \quad \quad \quad \quad$ {Note: $g=\frac { GM }{ { R }^{ 2 } }$ }

Taking positive $r$ as outward from center of earth, Force acting on the particle,

$F=m{ g }_{ effective }=-mg\frac { r }{ R } =-kr$

This is the same form as Hooke's Law for a mass on a spring.It would cause the particle to oscillate back and forth through the center of the Earth like a mass bobbing up and down on a spring. The period for this oscillation is $T=2\pi \sqrt { \frac { m }{ k } }$

For this case, the period of oscillation is,

$T=2\pi \sqrt { \frac { mR }{ mg } } =2\pi \sqrt { \frac { R }{ g } }$

Time taken for moving from one end to other is half the time taken for oscillation.

$\therefore t=\frac { T }{ 2 } =\pi \sqrt { \frac { R }{ g } } =\boxed{2513.27s}$