A thin spherical shell of total mass M and radius R is held fixed. There is a small hole in the shell. A mass m is released from rest at a distance R from the hole along a straight line that passes through the hole and also through the centre of the shell. This mass subsequently moves under gravitational force of the shell. How long does the mass take to travel from the hole to the point diametrically opposite?

To solve the problem of how long it takes for a mass \( m \) to travel from a hole in a thin spherical shell to the point diametrically opposite, we need to consider the gravitational forces acting on the mass as it moves through the shell. The key principle here is that inside a uniform spherical shell, the gravitational force experienced by a mass is zero. However, as the mass moves towards the center and then out towards the opposite side, we can analyze the motion using concepts from classical mechanics.

Understanding the Motion Inside the Shell

When the mass \( m \) is released from a distance \( R \) from the hole, it starts moving towards the center of the shell. As it approaches the center, the gravitational pull from the shell cancels out due to symmetry. Therefore, the only force acting on the mass is the restoring force that pulls it back towards the center as it moves away from it.

Gravitational Force Inside the Shell

According to Newton's shell theorem, the gravitational force inside a uniform spherical shell is zero. This means that as the mass moves towards the center, it experiences no gravitational attraction from the shell itself. However, once it passes the center and moves towards the opposite side, the gravitational force starts to act on it again.

Modeling the Motion

We can model the motion of the mass \( m \) as simple harmonic motion (SHM). The mass will oscillate back and forth between the two points on the shell. The time period \( T \) of this oscillation can be derived from the properties of SHM. The formula for the time period of a mass-spring system, which is analogous to our situation, is given by:

• T = 2π√(L/g)

In our case, \( L \) is the distance from the center to the shell (which is \( R \)), and \( g \) is the effective gravitational acceleration experienced by the mass as it moves through the shell. However, since the mass experiences no gravitational force while inside the shell, we need to consider the motion from the hole to the center and then back to the opposite side.

Calculating the Time to Travel

The mass will take the same amount of time to travel from the hole to the center as it does to travel from the center to the opposite side. Therefore, we can find the total time \( T \) for the mass to travel from the hole to the diametrically opposite point by considering the time for half of the oscillation:

• T_half = π√(R/g)

Since the gravitational force is effectively zero inside the shell, we can simplify our calculations. The mass will accelerate towards the center due to the gravitational pull from the shell when it is outside, but it will decelerate as it approaches the center. The total time taken to travel from the hole to the opposite side is:

T_total = 2 * T_half = 2 * π√(R/g)

Final Result

In conclusion, the time taken for the mass \( m \) to travel from the hole to the point diametrically opposite is given by:

T = π√(R/g)

This result highlights the fascinating nature of gravitational forces and motion within a spherical shell, demonstrating how classical mechanics can be applied to understand complex systems.