Question

The gravitational force on a particle of mass$m$inside the earth at a distance$r$from the centre$(r<$the radius of the earth$R$) Is$F=-mgr/R$(Chapter 6, Section 8, Problem 21). Show that a particle placed in an evacuated tube through the centre of the earth would execute simple harmonic motion. Find the period of this motion.

Short answer

The solution of the differential equation is $\begin{array}{c}r=c_1{e}^{-i\omega t}+c_2{e}^{i\omega t}\\ =A\sin \omega t+B\cos \omega t\\ r=c\sin (\omega t+\gamma )\end{array}$

And the period is$T=5063.5\text{s}$

Step-by-step solution

Given information

The gravitational force on a particle of mass$m$ inside the earth at a distance$r$ from the centre$(r<$ the radius of the earth$R$ ) Is$F=-mgr/R$

Newton's second law

ewton's second law:

It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.$F=ma$

Solve for general solution and for time

According to Newton's second law, the force is given by$F=ma$here$m$ is the mass of the object, andis the second derivative of the position (i.e. the acceleration). Therefore, we can write the gravitational force inside the Earth surface as follow

$m\frac{d^{2}r}{d{t}^2}=-\frac{mg}{R}r$

butit is known that$g$ and$R$are constants, so assume,$g/R={\omega }^2$ so, the equation becomes

$\frac{d^{2}r}{d{t}^2}+{\omega }^{2}r=0$

To solve this second order differential equation, assume the differential operator$D=d/dt$ , to get the auxiliary equation

$(D^2+{\omega }^2)r=0$

$\begin{array}{c}\frac{dr}{dt}=-i\omega r\\ r=c_1{e}^{-i\omega t}\\ \frac{dr}{dt}=-i\omega r\\ r=c_2{e}^{i\omega t}\end{array}$

Solve further

Then take linear combination of the two independent solutions above, and get the general solution

$\begin{array}{c}r=c_1{e}^{-i\omega t}+c_2{e}^{i\omega t}\\ =A\sin \omega t+B\cos \omega t\\ =c\sin (\omega t+\gamma )\end{array}$

which is the same as the solution of the simple harmonic oscillator. So,It can be concludedthat, put a mass$m$ inside an evacuated tube which pass through the centre of the Earth it will execute simple harmonic motion. The period is the time required to leave a point and then return to it, it is also $T=2\pi /\omega$, so, for this mass, the period is

$\begin{array}{c}T=2\pi \sqrt{\frac{R}{g}}\\ =2\pi \sqrt{\frac{(6.371\times {10}^6)\text{m}}{9.81\text{m}/{\text{s}}^2}}\\ =5063.5\text{s}\end{array}$