91Ó°ÊÓ

Given data can be listed below,

• Mass of Earth,$M_e$

• The radius of Earth, $R_e$

• Mass inside Earth, $m$

• Distance from the earth,$r$

• Acceleration due to the gravity,$g$

Period of oscillation, $T$

Acceleration due to the gravity of the earth is given by

$$\begin{gathered} g=\frac{G{M}_e}{R_e^2} \\ =9.8m/s^2 \end{gathered}$$

When we go toward the center. The person experiences weightless at the center of the earth, then the effective gravity given by

$g_{eff}=\frac{G{M}_r}{r^2}$

Where $M_r$is mass varying with the radius of the earth.

$\mathrm{ÒÏ}{M}_r=\mathrm{Ï€}R×\frac{4}{3}{R}_e^3$

Where $\mathrm{ÒÏ}$is the density of the earth.

$\mathrm{ÒÏ}=\frac{M_e}{{\displaystyle \frac{4}{3}}\mathrm{Ï€}{R}_e^3}$

Now effective gravity is

$$\begin{gathered} g_{eff}=\frac{G{M}_e{r}^3}{r^2{R}_e^3} \\ =\frac{gr}{R_e} \end{gathered}$$

The gravity force of the spherical shell is given as

$$\begin{gathered} F=-\frac{mgr}{R_e} \\ =-kr \end{gathered}$$

This is the same as the hook’s law for a spring-mass system

Expression for angular frequency and time period of the spring-mass system is

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt{\frac{k}{m}} \\ T=2\sqrt{\frac{m}{k}} \end{gathered}$$

The period of oscillation is

$$\begin{gathered} T=2\sqrt{\frac{m{R}_e}{mg}} \\ =2\mathrm{Ï€}\sqrt{\frac{R_e}{g}} \end{gathered}$$

The radius of the earth, $R_e=6.378×{10}^{6}m$and $g=9.8m/s$put in the above equation

$$\begin{gathered} T=2\sqrt{\frac{6.378×{10}^{8}m}{9.8m/s}} \\ =5068s \\ =84.5min \end{gathered}$$

Hence, the object travels toward the center of the earth and is weightless, and passes through the center.

$T=2\sqrt{\frac{R_e}{g}}$