If the earth was a homogeneous sphere and a straight hole was bored through its centre, show that a body dropped into this hole will execute shm. Calculate the time period if the radius of the earth is 6400 km and g = `9.8 ms^(-2)`

Explanation

To show that a body dropped into a straight hole bored through the center of a homogeneous Earth will execute simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understanding the Force Acting on the Body
When a body is dropped into the hole, it experiences gravitational force. According to Newton's law of gravitation, the force acting on the body at a distance \( r \) from the center of the Earth is given by:

\[ F = -\frac{G M(r) m}{r^2} \]

where:
- \( G \) is the gravitational constant,
- \( M(r) \) is the mass of the Earth enclosed within radius \( r \),
- \( m \) is the mass of the body.

### Step 2: Finding the Mass Enclosed
For a homogeneous sphere, the mass \( M(r) \) of the Earth enclosed within radius \( r \) can be expressed as:

\[ M(r) = \rho \cdot \frac{4}{3} \pi r^3 \]

where \( \rho \) is the density of the Earth. The total mass of the Earth \( M \) can be expressed as:

\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

where \( R \) is the radius of the Earth.

### Step 3: Relating Density and Gravitational Acceleration
From the gravitational force at the surface of the Earth, we have:

\[ g = \frac{G M}{R^2} \]

Substituting for \( M \):

\[ g = \frac{G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} = \frac{4}{3} \pi G \rho R \]

From this, we can solve for \( \rho \):

\[ \rho = \frac{3g}{4 \pi G R} \]

### Step 4: Expressing the Force in Terms of Displacement
Now, substituting \( M(r) \) back into the force equation, we get:

\[ F = -\frac{G \left( \rho \cdot \frac{4}{3} \pi r^3 \right) m}{r^2} \]

Substituting for \( \rho \):

\[ F = -\frac{G \left( \frac{3g}{4 \pi G R} \cdot \frac{4}{3} \pi r^3 \right) m}{r^2} = -\frac{g m r}{R} \]

### Step 5: Identifying the SHM
The force can be rewritten as:

\[ F = -k r \]

where \( k = \frac{g m}{R} \). This shows that the force acting on the body is proportional to the displacement \( r \) from the center, indicating that the motion is simple harmonic.

### Step 6: Calculating the Time Period
The time period \( T \) of SHM is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{\frac{g m}{R}}} = 2\pi \sqrt{\frac{R}{g}} \]

Substituting \( R = 6400 \times 10^3 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \):

\[ T = 2\pi \sqrt{\frac{6400 \times 10^3}{9.8}} \]

Calculating the value:

\[ T = 2\pi \sqrt{65306.12} \approx 2\pi \times 255.5 \approx 1604.5 \, \text{s} \]

### Final Answer
Thus, the time period of the oscillation is approximately \( 1604.5 \, \text{s} \) or about \( 26.74 \, \text{minutes} \).

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