In the approximation that the Earth is a sphere of uniform density, it can be shown that the gravitational force it exerts on a mass m inside the Earth at a distance r from the center is $\mathit{m}\mathit{g}\mathbf{(}\mathit{r}\mathbf{/}\mathit{R}\mathbf{)}$, where R is the radius of the Earth. (Note that at the surface, the force is indeed mg, and at the center it is zero). Suppose that there were a hole drilled along a diameter straight through the Earth, and the air were pumped out of the hole. If an object is released from one end of the hole, how long will it take to reach the other side of the Earth? Include a numerical result.

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.

$\rho M_r=\pi R\times \frac{4}{3}{R}_e^3$

Where $\rho$is the density of the earth.

$\rho =\frac{M_e}{{\displaystyle \frac{4}{3}}\pi R_e^3}$

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

$$\begin{gathered} \omega =\sqrt{\frac{k}{m}} \\ T=2\sqrt{\frac{m}{k}} \end{gathered}$$

The period of oscillation is

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

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

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

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

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