0. The gravitational attraction force between two ships with a mass of \( 6 \times 10^{4} \mathrm{~kg} \) and \( 2 \times 10^{5} \mathrm{~kg} \) respectively is 184 \( \times 10^{-2} \mathrm{~N} \). 10.1 Calculate how tar the ships must be from each other. 10.2 If the distance between the ships is halved, how will this influence the force between them2 10.1

To determine the distance between the two ships using the gravitational force formula, we can use Newton's law of universal gravitation, which states: \[ F = G \frac{m_1 m_2}{r^2} \] Where: - \( F \) is the gravitational force (184 × 10^{-2} N) - \( G \) is the gravitational constant (\( 6.674 × 10^{-11} \, \text{N m}^2/\text{kg}^2 \)) - \( m_1 = 6 × 10^4 \, \text{kg} \) - \( m_2 = 2 × 10^5 \, \text{kg} \) - \( r \) is the distance between the ships. Rearranging to find \( r \): \[ r = \sqrt{G \frac{m_1 m_2}{F}} \] Substituting the values: \[ r = \sqrt{ (6.674 × 10^{-11}) \frac{(6 × 10^4)(2 × 10^5)}{(184 × 10^{-2})} } \] Calculating this gives: \[ r \approx \sqrt{(6.674 × 10^{-11}) \frac{(1.2 × 10^{10})}{(1.84)}} \approx \sqrt{4.36 × 10^{-1}} \approx 0.66 \, \text{m} \] So, the ships must be approximately 0.66 meters apart. For the second part, if the distance between the ships is halved, we can see how the gravitational force changes. The gravitational force is inversely proportional to the square of the distance. Therefore, if the distance \( r \) is halved, the new force \( F' \) can be calculated as follows: \[ F' = G \frac{m_1 m_2}{(r/2)^2} = G \frac{m_1 m_2}{\frac{r^2}{4}} = 4 \cdot F \] Thus, the new force will be four times the original force. Therefore, if the distance is halved, the gravitational attraction between the ships will quadruple!