Example 2 Clw suppase two objects M1 and m2 are a distance ol apart. Determine how the magnitude of the porce batween. them Charges if: 2.1 Both masses double and the distance between them is halved 2.2 The mass of mi doubles

To solve the problem, we need to apply Newton's law of universal gravitation, which states that the force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by the formula: \[ F = G \frac{m_1 m_2}{r^2} \] where \( G \) is the gravitational constant. ### Known Conditions: - Initial masses: \( m_1 \) and \( m_2 \) - Initial distance: \( r \) ### 2.1 Both masses double and the distance between them is halved 1. **New masses**: - \( m_1' = 2m_1 \) - \( m_2' = 2m_2 \) 2. **New distance**: - \( r' = \frac{r}{2} \) 3. **New force**: We can substitute the new values into the gravitational force formula: \[ F' = G \frac{m_1' m_2'}{(r')^2} \] Substituting the new values: \[ F' = G \frac{(2m_1)(2m_2)}{\left(\frac{r}{2}\right)^2} \] 4. **Simplifying the expression**: \[ F' = G \frac{4m_1 m_2}{\frac{r^2}{4}} = G \frac{4m_1 m_2 \cdot 4}{r^2} = 16 G \frac{m_1 m_2}{r^2} \] Thus, the new force \( F' \) is: \[ F' = 16F \] ### 2.2 The mass of \( m_1 \) doubles 1. **New mass**: - \( m_1' = 2m_1 \) - \( m_2' = m_2 \) (remains the same) 2. **Distance remains the same**: - \( r' = r \) 3. **New force**: Substitute the new mass into the gravitational force formula: \[ F' = G \frac{m_1' m_2'}{(r')^2} \] Substituting the new values: \[ F' = G \frac{(2m_1)(m_2)}{r^2} \] 4. **Simplifying the expression**: \[ F' = 2G \frac{m_1 m_2}{r^2} \] Thus, the new force \( F' \) is: \[ F' = 2F \] ### Summary of Results: - **2.1**: The new force when both masses double and the distance is halved is \( 16F \). - **2.2**: The new force when only \( m_1 \) doubles is \( 2F \).