A Straight Frictionless Tunnel Is Bored Through The Earth From One Point Of Its Surface To Another Point. Show That The Object In The Tunnel Will Execute Simple Harmonic Motion. Calculate The Time Period.

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Motion of an object in the tunnel through the earth from one point to another point of the surface:

Let us consider the earth is a sphere of radius \( R \), mass \( M \), density \( \rho \) and \( O \) is the centre of the earth. \( PQ \) is a straight frictionless tunnel connecting two points \( P \) and \( Q \) on the surface of the earth as shown in the adjoining Fig. 1.

Let \( B \) is the centre of the tunnel \( PQ \) and a perpendicular line \( OB \) is drawn on the point \( B \) from the centre \( O \).

Let, an object of mass \( m \) is moving along the tunnel in the direction \( PQ \), at any instant of time the object is at a point \( A \) at a distance \( x \) from the point \( B \) and at a distance \( r \) from the centre \( O \).

Fig. 1

\( OA=r \) and \( AB=x \).

Now the gravitational force of attraction on the mass \( m \) at point \( A \) will be only due to the inner solid sphere of radius \( r \). The shell exterior to it will have no influence on \( m \).

Therefore the gravitational attraction force on \( m \) along \( AO \) is \( F=G\cdot\frac{(\frac{4}{3}\pi{r}^3\rho\cdot{m})}{r^2} \), where \( G \) is the gravitational constant.

or, \( F=G\frac{4}{3}\pi\rho{m}r \)

The component of the force \( F \) along the \( AB \) direction is,

\( F\cos\angle{OAB}= – G\frac{4}{3}\pi\rho{m}r\cdot\frac{x}{r} \\=- G\frac{4}{3}\pi\rho{m}x \)

The negative sign indicates that \( F \) and \( x \) are in opposite direction.

So we can write that \( F=-kx \), where \( k=G\frac{4}{3}\pi{m}\rho= constan \).

When an object moves along PBQ, it is acted by a force that is proportional to the distance from the mean position \( B \) and directed towards \( B \). So the motion of the object is a simple harmonic motion.

Time period of oscillation:

We know that the differential equation of such simple harmonic motion is,

\( m\frac{d^2{x}}{{dt}^2}=-kx \\or,\ \frac{d^2{x}}{{dt}^2}+\frac{k}{m}x=0\\ or,\ \frac{d^2{x}}{{dt}^2}+{\omega}^2{x}=0 \)

Where, \( \omega=\sqrt{\frac{k}{m}}=\sqrt{G\frac{4}{3}\pi\rho} \)

Now the time period of this simple harmonic motion is, \( T=\frac{2\pi}{\omega}=\frac{2\pi}{ \sqrt{G\frac{4}{3}\pi\rho} }\\ =\sqrt{\frac{3\pi}{G\rho}}\\=\sqrt{\frac{3\pi{4}\pi{R}G}{G3g}} \)

or, \( \displaystyle{T=2\pi\sqrt{\frac{R}{g}}} \)

Where, the acceleration due to gravity on the earth’s surface is,

\( g=\frac{GM}{R^2}=G\frac{\frac{4}{3}\pi{R}^3\rho{M}}{R^2} \\=G\frac{4}{3}\pi{R}\rho \)

or, \( \rho=\frac{3g}{4\pi{R}G} \).

Value of time period:

Time period \( T=2\pi\sqrt{\frac{R}{g}}\\=2\pi\sqrt{\frac{6.38\times10^8}{980}}\\=84.4\ minutes\ (approx) \)

where \( R=6.38\times10^6 \) m.

If the tunnel is bored through the centre of the earth, the time period of oscilation will be the same.

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