A hollow spherical shell is compressed to half its radius. The gravitational potential at the centre

Options

A. increases
B. decreases
C. remains same
D. it will depend on the mass of shell

Answer

To solve the problem of how the gravitational potential at the center of a hollow spherical shell changes when the shell is compressed to half its radius, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Gravitational Potential:**
The gravitational potential \( V \) at a distance \( r \) from a mass \( m \) is given by the formula:
\[ V = -\frac{Gm}{r} \]
where \( G \) is the gravitational constant.

2. **Identify the Initial Conditions:**
Let the initial radius of the hollow spherical shell be \( R \). The initial gravitational potential at the center of the shell (at radius \( R \)) can be expressed as:
\[ V_{\text{initial}} = -\frac{Gm}{R} \]

3. **Determine the New Radius:**
When the shell is compressed to half its radius, the new radius \( r \) becomes:
\[ r = \frac{R}{2} \]

4. **Calculate the New Gravitational Potential:**
Substitute the new radius into the gravitational potential formula:
\[ V_{\text{new}} = -\frac{Gm}{\frac{R}{2}} = -\frac{Gm \cdot 2}{R} = -\frac{2Gm}{R} \]

5. **Compare the Initial and New Potential:**
- The initial potential was \( V_{\text{initial}} = -\frac{Gm}{R} \).
- The new potential is \( V_{\text{new}} = -\frac{2Gm}{R} \).

Since \( -\frac{2Gm}{R} \) is more negative than \( -\frac{Gm}{R} \), we conclude that the gravitational potential has decreased.

6. **Conclusion:**
Therefore, when the hollow spherical shell is compressed to half its radius, the gravitational potential at the center decreases.

### Summary of the Solution:
The gravitational potential at the center of the hollow spherical shell decreases when the shell is compressed to half its radius.