If a smooth tunnel is dug across a diameter of earth and a particle is released from the surface of earth, the particle oscillate simple harmonically along it
Maximum speed of the particle is

Options

A. a. `sqrt((2GM)/(R ))`
B. b. `sqrt((GM)/(R ))`
C. c. `sqrt((3GM)/(2R ))`
D. d. `sqrt((GM)/(2R ))`

Answer

To find the maximum speed of a particle oscillating in a smooth tunnel dug across the diameter of the Earth, we can use the principle of conservation of energy. Here's a step-by-step solution:

### Step 1: Understand the System
A smooth tunnel through the Earth allows a particle to oscillate back and forth. When the particle is released from the surface of the Earth, it will convert potential energy into kinetic energy as it moves towards the center.

### Step 2: Write the Conservation of Energy Equation
The total mechanical energy of the particle is conserved. Therefore, the change in kinetic energy (KE) is equal to the change in potential energy (PE):

\[ \Delta KE = -\Delta PE \]

### Step 3: Express the Kinetic Energy
The kinetic energy at the maximum speed (when the particle is at the center of the Earth) is given by:

\[ KE = \frac{1}{2} mv^2 \]

### Step 4: Express the Potential Energy
The gravitational potential energy at the surface of the Earth (initial position) is:

\[ PE_i = -\frac{GMm}{R} \]

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the particle, and \( R \) is the radius of the Earth.

At the center of the Earth, the potential energy (final position) is:

\[ PE_f = -\frac{3}{2} \frac{GMm}{R} \]

### Step 5: Calculate the Change in Potential Energy
The change in potential energy as the particle moves from the surface to the center is:

\[ \Delta PE = PE_f - PE_i = \left(-\frac{3}{2} \frac{GMm}{R}\right) - \left(-\frac{GMm}{R}\right) = -\frac{3}{2} \frac{GMm}{R} + \frac{GMm}{R} = -\frac{1}{2} \frac{GMm}{R} \]

### Step 6: Set Up the Energy Conservation Equation
Now we can set the change in kinetic energy equal to the change in potential energy:

\[ \frac{1}{2} mv^2 = -\Delta PE = \frac{1}{2} \frac{GMm}{R} \]

### Step 7: Solve for Maximum Speed
Cancelling \( m \) from both sides and solving for \( v \):

\[ \frac{1}{2} v^2 = \frac{GM}{R} \]

\[ v^2 = \frac{2GM}{R} \]

\[ v = \sqrt{\frac{2GM}{R}} \]

### Step 8: Conclusion
Thus, the maximum speed of the particle as it passes through the center of the Earth is:

\[ v = \sqrt{\frac{2GM}{R}} \]