1. **State the problem:** We have the gravitational force formula $$F = \frac{GM_1M_2}{r^2}$$ where $G$, $M_1$, and $M_2$ are constants, and $r$ is the variable distance. 2. **Find the derivative $\frac{dF}{dr}$:** Using the power rule and constant multiple rule, $$\frac{dF}{dr} = \frac{d}{dr} \left( GM_1M_2 r^{-2} \right) = GM_1M_2 \cdot \frac{d}{dr} r^{-2} = GM_1M_2 (-2) r^{-3} = -\frac{2GM_1M_2}{r^3}.$$ 3. **Interpret the sign of $\frac{dF}{dr}$:** The derivative is negative, meaning as $r$ increases, the force $F$ decreases. This matches the physical intuition that gravitational force weakens with distance. 4. **Calculate the rate of change at $r=10000$ m given rate at $r=20000$ m:** Given $$\frac{dF}{dr} \bigg|_{r=20000} = -2 \text{ N/m}.$$ From the formula, $$\frac{dF}{dr} = -\frac{2GM_1M_2}{r^3}.$$ Let $$k = 2GM_1M_2,$$ then $$-2 = -\frac{k}{(20000)^3} \implies k = 2 \times (20000)^3.$$ Now find $$\frac{dF}{dr} \bigg|_{r=10000} = -\frac{k}{(10000)^3} = -\frac{2 \times (20000)^3}{(10000)^3} = -2 \times \left(\frac{20000}{10000}\right)^3 = -2 \times 2^3 = -2 \times 8 = -16.$$ So, the force changes at a rate of $-16$ N/m when $r=10000$ m. **Final answers:** (a) $$\frac{dF}{dr} = -\frac{2GM_1M_2}{r^3}$$ (b) The negative sign means the gravitational force decreases as the distance $r$ increases. (c) The force changes at a rate of $-16$ N/m when $r=10000$ m.