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` located at the center of a solid sphere of mass `M` and radius `R` to escape the gravitational field of the sphere, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Gravitational Potential Energy**:
The gravitational potential energy \( U \) of a mass \( m \) at a distance \( r \) from the center of a solid sphere of mass \( M \) is given by:
\[ U = -\frac{3GMm}{2R} \]
where \( G \) is the gravitational constant.

2. **Setting Up the Escape Condition**:
For the particle to escape the gravitational field, the total mechanical energy (kinetic energy + potential energy) must be greater than or equal to zero:
\[ K.E + U \geq 0 \]
The kinetic energy \( K.E \) of the particle when it has a velocity \( v \) is given by:
\[ K.E = \frac{1}{2}mv^2 \]

3. **Substituting the Energies**:
Substitute the expressions for kinetic energy and potential energy into the escape condition:
\[ \frac{1}{2}mv^2 - \frac{3GMm}{2R} \geq 0 \]

4. **Simplifying the Equation**:
Rearranging the inequality gives:
\[ \frac{1}{2}mv^2 \geq \frac{3GMm}{2R} \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2}v^2 \geq \frac{3GM}{2R} \]

5. **Solving for Minimum Velocity**:
Multiply both sides by 2:
\[ v^2 \geq \frac{3GM}{R} \]
Taking the square root of both sides gives:
\[ v \geq \sqrt{\frac{3GM}{R}} \]

6. **Conclusion**:
Therefore, the minimum velocity \( v_{min} \) required for the particle to escape the gravitational field of the sphere is:
\[ v_{min} = \sqrt{\frac{3GM}{R}} \]