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 a distance `R` from the hole along a line that passes through the hole and also through the centre of the shell. This mass subsequently moves under the gravitational force of the shell. How long does the mass take to travel from the hole to the point diametrically opposite.

Explanation

To solve the problem, we need to analyze the motion of the mass `m` as it travels from the hole in the spherical shell to the point diametrically opposite. The solution can be broken down into two parts: the motion until it reaches the hole and the motion inside the shell.

### Step 1: Analyze the motion until the hole

1. **Initial Setup**: The mass `m` is released from a distance `R` from the hole. The gravitational force acting on it is due to the mass `M` of the spherical shell.

2. **Conservation of Energy**: We can apply the conservation of mechanical energy. The initial potential energy (PE) when the mass is at distance `R` from the hole is given by:
\[ PE_i = -\frac{GMm}{R} \]
where `G` is the gravitational constant.

When the mass reaches the hole (at a distance `R/2` from the center), the potential energy is:
\[ PE_f = -\frac{GMm}{R/2} = -\frac{2GMm}{R} \]

3. **Change in Potential Energy**: The change in potential energy as the mass moves from the initial position to the hole is:
\[ \Delta PE = PE_f - PE_i = -\frac{2GMm}{R} + \frac{GMm}{R} = -\frac{GMm}{R} \]

4. **Kinetic Energy**: The change in kinetic energy (KE) is equal to the negative change in potential energy:
\[ KE = \Delta PE = -\frac{GMm}{R} \]
Since the mass starts from rest, its initial kinetic energy is zero:
\[ \frac{1}{2}mv^2 = \frac{GMm}{R} \]
Solving for `v`, we get:
\[ v = \sqrt{\frac{2GM}{R}} \]

### Step 2: Analyze the motion inside the shell

1. **Motion Inside the Shell**: Inside a thin spherical shell, the gravitational force is zero, and the potential energy is constant. Therefore, the mass will continue to move with the velocity it had when it reached the hole.

2. **Distance to Travel**: The distance from the hole to the diametrically opposite point is:
\[ d = 2R \]

3. **Time Calculation**: The time `t` taken to travel this distance at constant speed `v` is given by:
\[ t = \frac{d}{v} = \frac{2R}{\sqrt{\frac{2GM}{R}}} \]
Simplifying this, we find:
\[ t = 2R \cdot \sqrt{\frac{R}{2GM}} = 2\sqrt{\frac{R^3}{2GM}} \]

### Final Result

Thus, the total time taken for the mass `m` to travel from the hole to the point diametrically opposite is:
\[ t = 2\sqrt{\frac{R^3}{2GM}} \]