To solve the problem of how the gravitational force of attraction between two masses changes when the distance between them is tripled, we can use Newton's Law of Universal Gravitation. This law states that the gravitational 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: - \( F \) is the gravitational force, - \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)), - \( m_1 \) and \( m_2 \) are the masses (in this case, both are 10 kg), - \( r \) is the distance between the centers of the two masses. ### Step-by-Step Explanation 1. **Initial Gravitational Force Calculation**: Let's denote the initial distance between the two masses as \( r \). The initial gravitational force \( F_1 \) can be calculated as follows: \[ F_1 = G \frac{m_1 m_2}{r^2} = G \frac{10 \, \text{kg} \times 10 \, \text{kg}}{r^2} = G \frac{100}{r^2} \] 2. **New Distance Calculation**: If the distance is tripled, the new distance \( r' \) becomes: \[ r' = 3r \] 3. **New Gravitational Force Calculation**: Now, we can calculate the new gravitational force \( F_2 \) using the new distance: \[ F_2 = G \frac{m_1 m_2}{(r')^2} = G \frac{10 \, \text{kg} \times 10 \, \text{kg}}{(3r)^2} = G \frac{100}{(3r)^2} = G \frac{100}{9r^2} \] 4. **Comparing the Forces**: To find out how the gravitational force changes, we can compare \( F_2 \) to \( F_1 \): \[ F_2 = \frac{1}{9} F_1 \] This means that the new gravitational force \( F_2 \) is one-ninth of the original gravitational force \( F_1 \). 5. **Conclusion**: Therefore, when the distance between the two masses is tripled, the gravitational force of attraction between them is reduced by a factor of \( \frac{1}{9} \). ### Why the Other Options are Incorrect - **Option A: One half** - This would imply that the force is reduced to \( \frac{1}{2} F_1 \). This is incorrect because the relationship between force and distance is quadratic, not linear. - **Option B: One third** - This suggests that the force is reduced to \( \frac{1}{3} F_1 \). Again, this is incorrect due to the quadratic relationship; tripling the distance does not reduce the force by a linear factor. - **Option C: One quarter** - This would imply that the force is reduced to \( \frac{1}{4} F_1 \). While this is a common misconception, the correct reduction factor when the distance is tripled is actually \( \frac{1}{9} \), not \( \frac{1}{4} \). ### Summary - The gravitational force between two masses is inversely proportional to the square of the distance between them. - Tripling the distance reduces the gravitational force by a factor of \( 1/9 \). - The correct answer is option D: one ninth. - Understanding the quadratic relationship in gravitational force is crucial for solving similar problems. This thorough understanding of the relationship between distance and gravitational force will help you tackle related questions in your exams effectively!