1. **State the problem:** Two asteroids attract each other with a gravitational force inversely proportional to the square of the distance between them. Given the force is 6 kN at 100 metres, find the force at 200 metres. 2. **Formula and explanation:** The force $F$ varies inversely as the square of the distance $d$, so: $$F \propto \frac{1}{d^2}$$ This means: $$F = \frac{k}{d^2}$$ where $k$ is a constant. 3. **Find the constant $k$ using the given data:** At $d=100$ m, $F=6000$ N (since 6 kN = 6000 N), so: $$6000 = \frac{k}{100^2} = \frac{k}{10000}$$ Multiply both sides by 10000: $$k = 6000 \times 10000 = 60,000,000$$ 4. **Find the force at $d=200$ m:** $$F = \frac{60,000,000}{200^2} = \frac{60,000,000}{40,000} = 1500$$ 5. **Interpretation:** The force at 200 metres is 1500 N or 1.5 kN. **Final answer:** The gravitational force at 200 metres is **1500 N**.