To tackle this problem, we need to analyze the motion of a particle in a tunnel that passes through the center of the Earth. The gravitational force acting on the particle will vary as it moves through the tunnel, which is a key aspect to consider. Let's break down each scenario step by step.
Understanding the Gravitational Force in the Tunnel
When a particle is inside the tunnel, the gravitational force it experiences is proportional to its distance from the center of the Earth. This is derived from the shell theorem, which states that a uniform spherical shell of mass exerts no net gravitational force on a particle located inside it. Therefore, the gravitational force acting on the particle at a distance \( r \) from the center is given by:
F = -\frac{GM(r)}{r^2} = -\frac{G \cdot M \cdot r}{R^3}
Here, \( M \) is the mass of the Earth, \( G \) is the gravitational constant, and \( R \) is the radius of the Earth. This results in a simple harmonic motion (SHM) scenario, where the acceleration \( a \) can be expressed as:
a = -\frac{g}{R} \cdot r
where \( g \) is the acceleration due to gravity at the surface of the Earth. The angular frequency \( \omega \) of this motion is given by:
\(\omega = \sqrt{\frac{g}{R}}\)
Case (a): Projected with Speed \(\sqrt{gR}\)
When the particle is projected into the tunnel with a speed of \(\sqrt{gR}\), it will undergo simple harmonic motion. The time period \( T \) of this motion can be calculated using the formula:
T = 2\pi \sqrt{\frac{R}{g}}
The time taken to cover the length of the tunnel (which is \( 2R \)) can be found by recognizing that the particle will take half the period to reach the other end:
Time = \frac{T}{2} = \pi \sqrt{\frac{R}{g}}
Case (b): Released from Height \( R \) Above the Tunnel
In this scenario, the particle is released from a height \( R \) above the tunnel. Initially, it will fall under the influence of gravity until it reaches the tunnel entrance. The time \( t_1 \) to fall this height can be calculated using the equation:
h = \frac{1}{2} g t_1^2
Solving for \( t_1 \), we get:
t_1 = \sqrt{\frac{2R}{g}}
Once it enters the tunnel, it will again undergo simple harmonic motion as described earlier. The time taken to traverse the tunnel is:
t_2 = \pi \sqrt{\frac{R}{g}}
The total time \( T_{total} \) is then:
T_{total} = t_1 + t_2 = \sqrt{\frac{2R}{g}} + \pi \sqrt{\frac{R}{g}}
Case (c): Thrown Vertically Upward with Speed \(\sqrt{gR}\)
When the particle is thrown upward along the length of the tunnel with a speed of \(\sqrt{gR}\), it will ascend until it comes to a stop due to gravity. The time \( t_{up} \) to reach the maximum height can be calculated using:
v = u - gt_{up}
Setting \( v = 0 \) (at the maximum height), we find:
t_{up} = \frac{\sqrt{gR}}{g} = \sqrt{\frac{R}{g}}
After reaching the maximum height, the particle will fall back down through the tunnel, taking the same amount of time to return to the starting point:
t_{down} = \pi \sqrt{\frac{R}{g}}
The total time \( T_{total} \) for this case is:
T_{total} = t_{up} + t_{down} = \sqrt{\frac{R}{g}} + \pi \sqrt{\frac{R}{g}}
Summary of Results
- For case (a): \( T = \pi \sqrt{\frac{R}{g}} \)
- For case (b): \( T_{total} = \sqrt{\frac{2R}{g}} + \pi \sqrt{\frac{R}{g}} \)
- For case (c): \( T_{total} = \sqrt{\frac{R}{g}} + \pi \sqrt{\frac{R}{g}} \)
These calculations illustrate the fascinating dynamics of motion in a gravitational field, especially in a unique setup like a tunnel through the Earth. Each scenario highlights different aspects of gravitational influence and motion, providing a deeper understanding of physics principles at play.
