If the value of gravitational constant becomes double, what is the ratio of inertial mass to gravitational mass of Moon?

Answer

To solve the problem of determining the ratio of inertial mass to gravitational mass of the Moon when the gravitational constant \( G \) is doubled, we can follow these steps:

### Step-by-Step Solution

1. **Understanding Inertial and Gravitational Mass**:
- Inertial mass (\( m_i \)) is the mass that determines how much an object resists acceleration when a force is applied. It is defined by Newton's second law: \( F = m_i a \).
- Gravitational mass (\( m_g \)) is the mass that determines the strength of the gravitational force experienced by an object in a gravitational field. According to Newton's law of gravitation, the force \( F \) between two masses is given by \( F = \frac{G m_1 m_2}{r^2} \).

2. **Establishing the Relationship**:
- The gravitational force acting on the Moon can be expressed as:
\[ F = m_g g \]
- Where \( g \) is the acceleration due to gravity at the Moon's surface, which can be expressed as:
\[ g = \frac{G M}{R^2} \]
- Here, \( M \) is the mass of the Earth (or the attracting body) and \( R \) is the distance from the center of the Earth to the Moon.

3. **Finding the Ratio**:
- The ratio of inertial mass to gravitational mass is given by:
\[ \frac{m_i}{m_g} = \text{constant} \]
- From the equations of motion and gravitational force, we can derive that:
\[ m_i = \frac{F R^2}{G m_g} \]
- However, this leads us to the conclusion that:
\[ \frac{m_i}{m_g} = 1 \]
- This indicates that the ratio of inertial mass to gravitational mass is constant and equals 1, regardless of the value of \( G \).

4. **Considering the Change in \( G \)**:
- If the gravitational constant \( G \) is doubled, the equations governing the forces and masses remain consistent. The inertial mass and gravitational mass will still maintain their ratio:
\[ \frac{m_i}{m_g} = 1 \]

### Conclusion
Thus, when the gravitational constant \( G \) is doubled, the ratio of inertial mass to gravitational mass of the Moon remains:
\[ \frac{m_i}{m_g} = 1 \]