A particle of mass `m` is lying at the centre of a solid sphere of mass `M` and radius `R`. There is a turnel of negligible thickness, so that particle may escape. Find the minimum velocity required to escape the particle from the gravitational field of the sphere.

Answer

To find the minimum velocity required for a particle of mass `m` to escape from the gravitational field of a solid sphere of mass `M` and radius `R`, we can follow these steps:

### Step 1: Understand the gravitational potential inside the sphere
The gravitational potential \( V \) at a distance \( r \) from the center of a solid sphere of mass \( M \) and radius \( R \) is given by:
\[ V(r) = -\frac{GM}{R} + \frac{GM}{2R} \left(3r - \frac{r^2}{R}\right) \]
At the center of the sphere (\( r = 0 \)):
\[ V(0) = -\frac{3GM}{2R} \]

### Step 2: Set up the energy conservation equation
To escape the gravitational field, the particle must have enough kinetic energy to overcome the gravitational potential energy. The total mechanical energy at the center (initial) must equal the total mechanical energy at infinity (final). At infinity, the potential energy is zero, and we want the final kinetic energy to also be zero.

The conservation of energy can be expressed as:
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]
This simplifies to:
\[ \frac{1}{2}mv_0^2 + mV(0) = 0 \]

### Step 3: Substitute the potential energy at the center
Substituting \( V(0) \):
\[ \frac{1}{2}mv_0^2 - \frac{3GMm}{2R} = 0 \]

### Step 4: Solve for the minimum velocity \( v_0 \)
Rearranging the equation gives:
\[ \frac{1}{2}mv_0^2 = \frac{3GMm}{2R} \]
Dividing both sides by \( m \) (mass of the particle) and multiplying by 2:
\[ v_0^2 = \frac{3GM}{R} \]
Taking the square root:
\[ v_0 = \sqrt{\frac{3GM}{R}} \]

### Conclusion
The minimum velocity required for the particle to escape from the gravitational field of the sphere is:
\[ v_0 = \sqrt{\frac{3GM}{R}} \]